Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap4 Units and estimation 6 Trusses and frames

Chapter 5 Statics of one object

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5.0Introductionstudy PDF
5.1Static equilibrium of a particlestudy PDF + solutions
5.2Equilibrium of one objectstudy PDF + solutions
5.3Equilibrium with frictional contactstudy PDF + solutions
5.4Internal forcesstudy PDF + solutions
5.53D staticsstudy PDF + solutions

One object is in equilibrium if the forces and moments balance. For a particle, force balance tells all. But for an extended object, moment balance is also essential. There are special shortcuts for an object that has exactly two or exactly three forces acting on it. If friction forces are relevant the possibility of motion needs to be taken into account. Many real-world problems are not statically determinate and thus yield either only partial solutions, or yield full solutions after you have made extra assumptions.

The goal here is to find unknown aspects of the forces acting on one object. The object is typically a part of a machine or structure. Such a part is also called a ‘body’. By ‘unknown’ we mean ‘unknown at the outset’ or ‘you-need-to-do-mechanics-calculations-to-find’. Most often ‘unknowns’ are tensions in ropes or rods, contact forces where one part presses and rubs against another, and the force on an object at a point of connection to another object. We will also find ‘unknown’ forces and moments that one part of an object applies to another part of the same object. Finally we might also find the ‘unknown’ direction or point of application of a force that has an a priori known magnitude.

Needed skills

Throughout this and all later chapters you need mastery of the vector and free-body diagram skills and concepts from chapters 2 and 3.

Statics is a subset of dynamics

Statics is the mechanics of things that don’t move. But everything does move, at least a little. So, strictly speaking, dynamics is always the applicable subject. For many practical problems, however, statics is a good approximation of dynamics, very good. With little loss of accuracy, sometimes very little loss, and a great saving of effort, usually a very great saving, statics can be used instead of dynamics. Statics is a useful model. Even for a fast-moving system, say an accelerating car, statics calculations are appropriate for many of the parts. Although statics is a subset of dynamics (See box 5.1 on page 5.1) typical engineers do more statics calculations than dynamics calculations. Statics is the core of structural and strength analysis. Statics is the central tool used to predict when a structure or part will or will not break. Finally, Statics is good preparation for Dynamics

^†^†margin: Ironically, for some people the main benefit from learning dynamics is the side effect of better mastery of the generally-more-useful statics.

For all of statics, we neglect the role of inertia. We assume forces and moments balance each other. We assume static equilibrium.

Two-dimensional and three-dimensional mechanics

The world we live in is three-dimensional and the theory of mechanics is a three-dimensional theory. But three dimensions are harder to understand than two. So most learning and much engineering analysis is done in two dimensions. You can’t critically judge the degree of simplification this involves until you understand 3D mechanics. But you aren’t ready to learn 3D mechanics until you understand 2D mechanics. We escape this catch-22 by being casual about the precise meaning of the 2D world view. For now we think of cylinders and spheres as circles, of boxes as rectangles, and of cars as things with two wheels (one in front, one in back).

Static equilibrium in a nutshell

The basic idea for all of statics is this: if the forces on a system (i.e., the forces showing on a free-body diagram of the system) satisfy eqs. (Ic) and (IIc) (inside cover) the system is said to be in static equilibrium or just in equilibrium^†^†margin: To be precise, static equilibrium requires that the system and all subsystems, all billion gazillion of them (all the different ways you could cut a piece out of your system), satisfy the equilibrium conditions. However, for simplicity at this point in the book, we don’t concern ourselves much with subsystems, just with a single whole object. .

A system is in static equilibrium if the applied forces and moments add to zero.

Another way to say this is that

A system in static equilibrium satisfies the linear and angular momentum balance neglecting the inertial ( $m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$ ) terms.

A final alternative description of statics is:

The full collection of forces on a system in static equilibrium are equivalent to (see Section 2.1 on page 2.1) a zero force and a zero couple.

The statics story is now, in-principle, complete. You have the tools (vectors and free-body diagrams) and you know the basic facts (the definition of statics, above). These are enough. But we’ll guide you through some of the subtleties, warn you away from common misconceptions, and teach you some of the tricks of the trade. You will see that the simply-stated laws of statics (above) allow you to accurately calculate useful things, things that most people who have not studied statics only vaguely understand.

5.1 Static equilibrium of a particle

What is a particle?

The word particle usually means something small. In mechanics a particle is an object for which we don’t worry about rotation, or the tendency of forces to cause rotation. A particle may or may not be small. Besides, smallness is in the eyes of the beholder. For some purposes a galaxy is well-modeled as a particle and for others a molecule is too big to be thought of as a particle. Big or small, the particle model of a system is defined by the lack of attention paid to the moment-balance equations

^†^†margin: Examples of particles. You might think of a galaxy as an immense thing, not just a dot. But its overall motion through a cluster of galaxies is probably well-described by thinking of it as a particle. The galaxy may rotate and distort in interesting ways, but one can ignore those rotations and distortions when calculating overall translation. That is, one might model an immense rotating and distorting galaxy as a particle. Similarly for an accelerating car, a block sliding on a ramp or a machine part, one may learn enough about the forces and motion using a particle model. Distortion and rotation might be there, and even large, but might not be important for understanding the overall motion or force balance.

Either moment balance is trivially satisfied or you can find what you need without worrying about how it is satisfied

^†^†margin: Equations are ‘satisfied’ when the right side is equal to the left. Take that as a definition or, if it helps you, think of it as a desire that you would like to accommodate.

\text{For statics of a particle, force-balance tells all:}\sum_{\rm All\ % forces\ \atop on\ the\ FBD}\kern-10.0pt{{\accentset{\rightharpoonup}{% \boldsymbol{F}}}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}\par\quad(Ic)

In two dimensions, this equilibrium equation makes up 2 independent scalar equations (2 components of the net force vector). In 3 dimensions we get 3 independent scalar equations. So we expect to be able to solve for 2 unknown quantities in 2D particle mechanics, and 3 in 3D.

The statics-of-a-particle recipe

For particle statics, we work with a simplified form of the general recipe from the inside back cover.

1)

Draw a free-body diagram (FBD) of the part of interest.
Use knowledge of the contact conditions (see Chapter 3) to draw known and unknown aspects of the forces appropriately (see fig. 2.64 on page 2.64);
2)

Set the sum of the forces on the FBD to zero: $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ .
(‘Equilibrium’, ‘force balance’, or ‘linear momentum balance in statics’);
3)

Solve the equations for unknowns.
Use vector manipulation skills (Chapter 2) to solve the force balance equation for unknowns of interest.

Scalar mechanics

In scalar mechanics — as opposed to vector mechanics — one takes the dot product of Eqn. (Ic) with unit vectors $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ and $\hat{\boldsymbol{k}}$ and write the three scalar component equations

^†^†margin: Actually, people who think in terms of scalar mechanics do the dot products, but implicitly (or unconsciously), and think of the component equations as fundamental.

\sum F_{x}=0,\qquad\sum F_{y}=0,\qquad\mbox{additionally, in 3D,}\qquad\sum F_% {z}=0.

Although you can do most problems plowing through with the ‘component’ or scalar approach, often there are shortcuts or insights that depend on the vectorial view.

1D statics of a particle

Let’s call the one dimension of interest the $x$ direction. The key governing equation is

\sum F_{x}=0.

You could call the special direction $y$ , $z$ , $x^{\prime}$ or $s$ if you like and then use, say $\sum F_{z}=0$ . The next two simple examples pretty much cover 1D particle statics.

Filename:tfigure8-rel-ang-vel

Figure 5.1: Equilibrium of a particle in 1D.

Example: Balance of two forces

For the particle in fig. 5.1, force balance gives

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}10\,% {\mbox{N}}\hat{\boldsymbol{\imath}}-F\hat{\boldsymbol{\imath}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}.

Either by equating $x$ components of both sides, or equivalently, by dotting both sides with $\hat{\boldsymbol{\imath}}$ , we get $F=10\,{\mbox{N}}$ . Or, we could have just jumped to scalar mechanics,

\sum F_{x}=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}10\,{\mbox{N}}-F=0\hbox to% 36.135pt{\hfil$\Rightarrow$\hfil}F=10\,{\mbox{N}}.

Most often we have to contend with forces which don’t show up until we draw a free-body diagram.

Filename:tfigure8-ang-vel-ex

Figure 5.2: Particle held by a string.

Example: Force pulling on a string.

For the particle in fig. 5.2, the quantity of interest, the tension in the cable, doesn’t show in the sketch. We need to draw a free body diagram of the particle which means cutting the string. This FBD is shown in fig. 5.1, where $F=T_{\rm AB}$ represents the tension in cable AB. So force balance gives $T_{\rm AB}=F=10\,{\mbox{N}}$ .

Filename:tfigure8-ang-vel

Figure 5.3: Weight hung from two lines with associated free-body diagram.

2D statics of a particle

The situation is less trivial when we go to 2D.

Example. A 100 pound (445 N) weight hangs from 2 lines in fig. 5.3.

We cut the strings, draw a free-body diagram and add the forces to get

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}\quad\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad 445\,{% \mbox{N}}(-\hat{\boldsymbol{\jmath}})+\underbrace{F_{A}\frac{(\hat{\boldsymbol% {\imath}}+\hat{\boldsymbol{\jmath}})}{\sqrt{2}}}_{{\accentset{\rightharpoonup}% {\boldsymbol{F}}}_{\!\!\mathrm{A}}}+\underbrace{F_{B}(-\frac{1}{2}\hat{% \boldsymbol{\imath}}+\frac{\sqrt{3}}{2}\hat{\boldsymbol{\jmath}})}_{{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{B}}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}.

(5.1)

This can be solved various ways (see below) to get $F_{A}=230.3\,{\mbox{N}}$ and $F_{B}=325.8\,{\mbox{N}}$ .

Although moment balance is technically superfluous in particle mechanics, when the forces are concurrent, moment balance can be used as a shortcut.

How to solve vector statics equations?

Method 1) Pull out both $x$ and $y$ components of the vectors to get 2 equations in 2 unknowns;

	$\displaystyle\sum F_{x}=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}$	$\displaystyle F_{A}/\sqrt{2}-F_{B}/2=0$
	$\displaystyle\sum F_{y}=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}$	$\displaystyle F_{A}/\sqrt{2}+F_{B}\sqrt{3}/2-445N=0.$

Method 2) Equivalently, dot both sides of the equation with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ to get 2 equations in 2 unknowns;

	$\displaystyle\{\mbox{eqn.~{}(\ref{eq:twolines})}\}\cdot\hat{\boldsymbol{\imath% }}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=F_{A}/\sqrt{2}-F_{B}/2$
	$\displaystyle\{\mbox{eqn.~{}(\ref{eq:twolines})}\}\cdot\hat{\boldsymbol{\imath% }}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=F_{A}/\sqrt{2}+F_{B}\sqrt{3}/2-445N.$

Method 3) dot both sides with a vector orthogonal to $r_{\rm AP}$ to get one equation in $F_{B}$ , similarly dot with a vector orthogonal to ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm BP}$ to get one equation in $F_{A}$ .

	$\displaystyle\{\mbox{eqn.~{}(\ref{eq:twolines})}\}\cdot(-\hat{\boldsymbol{% \imath}}+\hat{\boldsymbol{\jmath}})\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0$	$\displaystyle=-445N+F_{B}(\frac{1}{2}+\frac{\sqrt{3}}{2})$
	$\displaystyle\{\mbox{eqn.~{}(\ref{eq:twolines})}\}\cdot(\sqrt{3}\hat{% \boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}0$	$\displaystyle=-445N+F_{A}\frac{\sqrt{3}+1}{\sqrt{2}}$

^†^†margin: Note that if

{\accentset{\rightharpoonup}{\boldsymbol{v}}}=a\hat{\boldsymbol{\imath}}+b\hat% {\boldsymbol{\jmath}}

then

{\accentset{\rightharpoonup}{\boldsymbol{w}}}=-b\hat{\boldsymbol{\imath}}+a% \hat{\boldsymbol{\jmath}}

is orthogonal to it because

{\accentset{\rightharpoonup}{\boldsymbol{v}}}\cdot{\accentset{\rightharpoonup}% {\boldsymbol{w}}}=-ab+ba=0

. That is, in the plane to get a vector perpendicular to

{\accentset{\rightharpoonup}{\boldsymbol{v}}}

interchange the components and negate one of them (either one).

Method 4) cross both sides with ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{AP}$ to get one equation in $F_{B}$ , similarly cross with ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{BP}}$ to get $F_{A}$ .

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=\left(-\ell\hat{\boldsymbol{\imath}}/\sqrt{2}-\ell\hat{% \boldsymbol{\jmath}}/\sqrt{2}\right)\times\left(445\,{\mbox{N}}(-\hat{% \boldsymbol{\jmath}})+F_{A}\frac{(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{% \jmath}})}{\sqrt{2}}+F_{B}(-\frac{1}{2}\hat{\boldsymbol{\imath}}+\frac{\sqrt{3% }}{2}\hat{\boldsymbol{\jmath}})\right)$
		$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0=-445N+F_{B}(\frac{1}{2% }+\frac{\sqrt{3}}{2})$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle=\left(d\hat{\boldsymbol{\imath}}/2-d\sqrt{3}\hat{\boldsymbol{% \jmath}}/2\right)\times\left(445\,{\mbox{N}}(-\hat{\boldsymbol{\jmath}})+F_{A}% \frac{(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})}{\sqrt{2}}+F_{B}(-% \frac{1}{2}\hat{\boldsymbol{\imath}}+\frac{\sqrt{3}}{2}\hat{\boldsymbol{\jmath% }})\right)$
		$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}0=-445N+F_{A}\frac{\sqrt% {3}+1}{\sqrt{2}}.$

See section 1.5 starting on page 1.5 for more discussion about how to solve vector equations.

Another approach, mathematically equivalent to method 4 above, is to use moment balance.

Example: Moment balance

Consider again fig. 5.3. Moment balance about point A gives

\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}\quad\hbox to36.135pt{\hfil$\Rightarrow$\hfil% }\quad{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}\times% \left(445\,{\mbox{N}}(-\hat{\boldsymbol{\jmath}})\right)\ +\ {\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{P/A}}\times\left(F_{B}(-\frac{1}{% 2}\hat{\boldsymbol{\imath}}+\frac{\sqrt{3}}{2}\hat{\boldsymbol{\jmath}})\right% )\ +\ {\accentset{\rightharpoonup}{\boldsymbol{0}}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}.

Evaluating the cross products one way or another, we again get $F_{B}=325.8\,{\mbox{N}}$ . Similarly moment balance about B gives $F_{A}=230.3\,{\mbox{N}}$ .

Example: A kite.

A kite flying steadily in a breeze is roughly in static equilibrium. The three forces acting on it are from the air, pushing the kite downwind and up; from gravity, pulling the kite down; and from the string pulling the kite upwind and down. The three forces must add to zero.

A funny thing about kites is that they only stay up because you pull them down.

Whether force or moment balance is used, for concurrent force systems we only have two independent scalar equilibrium equations in 2D, and three in 3D .

Box 5.1 Existence and uniqueness

This is a relatively advanced aside.

The words existence and uniqueness may sound mathematically abstract and irrelevant to the real world. But if you translate ‘existence’ to ‘that’s possible’ and ‘non-existence’ to ‘no way’ you can see the relevance. Likewise ‘uniqueness’ and ‘non-uniqueness’ translate to ‘there’s just one way to do that’ and ‘there’s lots of ways to do that’.

In most homework problems there is an answer, and just one answer: the solution exists and is unique. But in projects and at work problems are often ill-posed. Non-existence or non-uniqueness may show up by a structure collapsing or a computer calculation giving erratic answers or error messages.

Existence. Sometimes equations have no solutions, they do not exist. For example the set

$x+2y$	$=7$
$2x+4y$	$=15$

has no solutions. They don’t exist. Why? Because subtracting twice the first equation from the second gives a contradiction: $0=1$ .

Example: Block on a slippery (frictionless) ramp. Use statics to find the normal force.

Filename:tfigure-blockcontradictiona

But the block slides and this is not a statics problem. So there is no statics solution. Even without intuition, the statics force-balance equations show that there is no value of $N$ that can make the force vectors add to zero. So no statics solution exists.

Such contradictions can be more subtle. Section 6.5 has some examples where even experts can’t intuitively see that there are no solutions.

Uniqueness. Sometimes statics problems have more than one solution, that is a non-unique solution: the equation $x+y=1$ has many solutions including $(x,y)=(1,0)$ , $(x,y)=(0,1)$ , $(x,y)=(10,-9)$ etc.

Structural mechanics problems often have non-unique solutions.

Example: Particle held by two strings. Find the tension in the strings to the sides of the point of application of a given load $F=10$ N.

Filename:tfigure-2strings

Force balance along the strings gives us one equation for the two unknown tensions.

\left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}\right\}\cdot\hat{\boldsymbol{\imath}}\hbox to% 36.135pt{\hfil$\Rightarrow$\hfil}-T_{1}+F+T_{2}=0

No other force balance or moment-balance equation gives more information. For any given $F$ this equation has many solutions. The pair ( $T_{1},T_{2}$ ) could be ( $10\,{\mbox{N}},0$ ) or ( $0,-10\,{\mbox{N}}$ ) ( $20\,{\mbox{N}},10N$ ) ( $17\,{\mbox{N}},7\,{\mbox{N}}$ ), etc.

Of course if you tie strings together like this and apply a force there is some actual tension in each string; reality, at any instant in time, is unique (as far as we know). For example, if you had tied the strings loosely together the right string gets slack and has $T_{2}=0$ and thus $T_{1}=10\,{\mbox{N}}$ . But it takes an extra assumption of this nature to get a unique solution.

And just because you can make an assumption that leads you to a unique solution doesn’t mean that this corresponds to reality. You might assume your friend had tied the strings together loosely and thus calculate $T_{2}=0$ and $T_{1}=10\,{\mbox{N}}$ . But really she tied them together tightly so $T_{2}=30\,{\mbox{N}}$ and $T_{1}=40\,{\mbox{N}}$ . Here is the same idea in 2D.

Example: Particle held by three strings.

Filename:tfigure-brach2wheels

Assume that $F_{x}$ and $F_{y}$ are given. What are the three tensions. Planar force balance gives two equations for the 3 unknown tensions. As in the previous example these equations have many solutions.

If you assume a) that one string goes slack and b) that no string can carry compression, then this problem has a unique answer. But you would have to know a priori that the strings were initially loose.

The same idea holds for 4 strings holding a particle in 3D. These string examples have a ‘one parameter family of solutions’; specifying one number (say that the tension in cable 2 is zero) determines the other tensions. But there can be more non-uniqueness than that by using more strings and then to get a unique solution you have to make more assumptions.

Counting equations and unknowns All of the uniqueness issues above could be detected by counting equations and unknowns. For the block on the ramp we had two equations for the one unknown $N$ . Whereas for the string problems we had more unknowns than equations. In summary,

•

If you have more equations than unknowns, existence is likely to be an issue; you probably can’t find any solutions.
•

If you have more unknowns than equations then uniqueness is likely an issue; any solution you find is probably non-unique

But, as for the block on ramp (2 equations with 2 unknowns), there are cases for which equation counting does not tell all about existence and uniqueness: see the lower right corner of the large table in section 6.5. Some simple counter examples: $x=3,2x=6$ has a unique solution; and you can solve the one equation $x^{2}+y^{2}=0$ for 2 unknowns.

Frictionless contact

As discussed in Chapter 3.1, engineered parts that slide often have bearings or lubrication to reduce the sliding resistance. To simplify analysis, that remaining resistance is often neglected, and we model the contact as ‘frictionless’ ( $\mu=0$ ). This makes the interaction force normal (perpendicular) to the contacting surfaces.

Example: Pull a wagon uphill

See fig. 5.4. From the free-body diagram we have

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}-1000\,{\mbox{N}}% \hat{\boldsymbol{\jmath}}+N\hat{\boldsymbol{e}}_{2}+T_{\rm AB}\hat{\boldsymbol% {e}}_{1}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.

(5.2)

where $\hat{\boldsymbol{e}}_{1}=\cos(30^{{}^{\circ}})\hat{\boldsymbol{\imath}}+\sin(3% 0^{{}^{\circ}})\hat{\boldsymbol{\jmath}}\hbox to36.135pt{\hfil and\hfil}\hat{% \boldsymbol{e}}_{2}=-\sin(30^{{}^{\circ}})\hat{\boldsymbol{\imath}}+\cos(30^{{% }^{\circ}})\hat{\boldsymbol{\jmath}}.$ $N$ and $T_{\rm AB}$ are unknown forces. Here are two ways to solve for the unknowns.

Method I. Substitute the expressions for $\hat{\boldsymbol{e}}_{1}$ and $\hat{\boldsymbol{e}}_{2}$ above into eqn. (5.2), extract $x$ and $y$ components to get 2 equations in two unknowns which you can solve to get $T_{\rm AB}=500\,{\mbox{N}}$ and $N=500\sqrt{3}\,{\mbox{N}}$ (note the font confusion that the force quantity $N$ and unit N have different meanings).

Method II. Using well chosen dot products can simplify the algebra. Take the dot products of both sides of eqn. (5.2) with $\hat{\boldsymbol{e}}_{1}$ and then separately with $\hat{\boldsymbol{e}}_{2}$ ,

to get two scalar equations. Dotting eqn. (5.2) with $\hat{\boldsymbol{e}}_{1}$ eliminates terms orthogonal to $\hat{\boldsymbol{e}}_{1}$ , namely $N\hat{\boldsymbol{e}}_{2}$ . And dotting eqn. (5.2) with $\hat{\boldsymbol{e}}_{2}$ ‘kills’ the $T_{\rm AB}\hat{\boldsymbol{e}}_{1}$ term. So the two equations each have only one unknown. See page 1.5 for more discussion of this method.

Filename:tfigure8-ang-accel

Figure 5.4: A wagon pulled uphill and associated free-body diagram. Because we are doing particle mechanics, we combine the two forces on the wheels into the single resultant

N

. Further, for particle mechanics we need not worry about where that

N

is applied.

Filename:tfigure8-ang-accel-ex

Figure 5.5: One unknown force

{\accentset{\rightharpoonup}{\boldsymbol{F}}}

Three-dimensional particle mechanics

The basic idea is the same in 3D as in 2D.

Example: One unknown force.

Assume 3 known forces and one unknown force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ are acting on a particle (fig. 5.5). Then from force balance

\begin{array}[]{rcl}{\accentset{\rightharpoonup}{\boldsymbol{0}}}&=&\sum{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}\\ \hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{\rightharpoonup}{% \boldsymbol{0}}}&=&(36\,{\rm lbf}\hat{\boldsymbol{\imath}}-16\,{\rm lbf}\hat{% \boldsymbol{\jmath}})+(-52\,{\rm lbf}\hat{\boldsymbol{k}}+5\,{\rm lbf}\hat{% \boldsymbol{\imath}})\\ &&+(-42\,{\rm lbf}\hat{\boldsymbol{k}}+20\,{\rm lbf}\hat{\boldsymbol{\imath}}-% 16\,{\rm lbf}\hat{\boldsymbol{\jmath}})+{\accentset{\rightharpoonup}{% \boldsymbol{F}}}\\[4.0pt] \hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{\rightharpoonup}{% \boldsymbol{F}}}&=&\left(-61\hat{\boldsymbol{\imath}}+32\hat{\boldsymbol{% \jmath}}+94\hat{\boldsymbol{k}}\right)\,{\rm lbf}.\end{array}

The new difficulties in 3D particle mechanics are

•

Visualization in 3D. (So practice making and reading 3D drawings.); and
•

The vector force-balance equation is 3D and thus equivalent to 3 scalar equations. Solving these is at the upper boundary of what most people can do reliably by hand or even with a non-programmable calculator. So methods that reduce the complexity of the solution are useful, as is the ability to set up the resulting equations on a computer or programmable calculator.

Hint: If the direction of a force is given (possibly implicitly) express the force as a scalar times a unit vector: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=F\hat{\boldsymbol{\lambda}}$ . (See top row, middle column of fig. 2.64 on page 2.64.)

Filename:tfigure8-ang-accel1

Figure 5.6: Mass A suspended by 3 ropes and a free-body diagram of the mass.

Example: Particle held by 3 ropes.

Say $m=100\,\mathrm{kg}$ and $g=10\,{\mbox{N}}/\,\mathrm{kg}$ in fig. 5.6. Force balance gives

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T_{% \rm AB}\hat{\boldsymbol{\lambda}}_{\rm AB}+\\ T_{\rm AC}\hat{\boldsymbol{\lambda}}_{\rm AC}+T_{\rm AD}\hat{\boldsymbol{% \lambda}}_{\rm AD}-mg\hat{\boldsymbol{k}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}

(5.3)

which is a 3D vector equation in 3 unknowns (3 scalar equations and 3 unknowns, good). The $\hat{\boldsymbol{\lambda}}$ ’s in eqn. (5.6) are known because the position vectors are given in the picture. For example,

\hat{\boldsymbol{\lambda}}_{\rm AB}=\frac{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\rm AB}}{|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AB% }|}=\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm B}-{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\rm A}}{|{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\rm B}-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm A}% |}.

To get to a numerical answer for the tensions you can use many methods such as (see Sample 5.17 on page 5.17)

1.

Brute force by hand.
2.

Systematically set up matrix equations for solution by some means.
3.

Set up and solve equations on a computer.
4.

Use an appropriate dot product to extract one equation in one unknown.
5.

Use moment about an axis to extract one equation in one unknown.

Refer to caption — Figure 5.7: The 900 ton platform is suspended 450 feet above the curved reflector surface. The platform hangs on eighteen cables from three towers, each about 300 feet high. *Courtesy of the NSF funded NAIC-Arecibo Observatory in Puerto Rico.*

Box 5.2 The simplification of dynamics to statics

Is statics good enough? The answer below is for your curiosity, not to help you with homework problems.

Classical mechanics, the equations on the inside cover of this book, is at least 99.99% accurate for at least 99.99% of mechanical engineering problems (see page 0.2 in the preface). Statics is a subset of classical mechanics. Statics is used for those dynamics situations where the statics approximation is reasonable (remember, no real situation is exactly static). Fortunately, for at least 90% of engineering mechanics calculations statics is at least 90% accurate.

In statics, we set the right hand sides of equations I and II (on inside cover) to zero. We are then thinking that these ‘inertial’ terms are small enough, compared to other forces, to be neglected. We replace the linear and angular momentum balance equations with their simplified statics forms

\sum_{\rm All\ external\atop forces}\kern-10.0pt{{\accentset{\rightharpoonup}{% \boldsymbol{F}}}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}\qquad\mbox{and% }\qquad\sum_{\rm All\ external\atop torques}\kern-10.0pt{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{C}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{C}$}}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}\quad(Ic,IIc).

which are sometimes called the force balance and moment balance and together are called the equilibrium equations. The forces to be summed (added) are the ones you see on a free-body diagram. The torques that are summed are those due to the same forces (by means of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i/C}}\times{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}}$ ) and any applied couples.

In dynamics forces ‘balance’ mass times acceleration.

In statics, the forces balance each other.

The approximation, the assumption, the ‘model’, see page 0.3) in statics is this:

Statics assumes that the forces on an object are much larger than the net force which accelerates it.

That is, in various ways of thinking:

•

The mass times acceleration is small. Small compared to what? Small compared to the individual forces.
•

The forces largely cancel. That is, the sum of the forces is much smaller than the individual terms.

Catch 22. Usually dynamics is too complicated for the trouble. So you use statics. But you are insecure and want to know if statics is accurate enough. To know this you have to do dynamics.

Estimating the errors from neglecting dynamics is a dynamics problem.

For each free-body diagram you have to know that

\left|m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}\right|% =\left|\sum_{\rm All\ forces\ \atop on\ the\ part}\kern-10.0pt{{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{i}}\right|\ \ll\ F_{\rm typical}.

To avoid Catch 22 we usually fake it by checking (usually in our heads) this rule of thumb.

Statics is probably accurate enough if

m_{\rm part}a_{\rm part}\ \ll\ \underbrace{m_{\rm load}(g+|a_{\rm load}|)+m_{% \rm part}g}_{F_{\rm typical}}.

(5.4)

Given this, the forces are more canceling each other out (balancing each other) than causing acceleration. So you can use statics to figure out just how these forces cancel each other.

Statics equations are often accurate-enough for

•

Things that a normal person would call “still” such as a building or bridge on a calm day, and a sleeping person;
•

Things that move with little acceleration, such as a tractor plowing a field or most of the parts in a smooth-flying airplane; and
•

Parts that mediate the forces needed to accelerate more massive parts, such as gears in a transmission, the rear wheel of an accelerating bicycle, the strut in the landing gear of an airplane, and the individual structural members of a building swaying in an earthquake.

Example: A bicycle wheel. The forces on a bicycle wheel are on the order of the weight of a person plus what is needed to accelerate the person up and down small hills, say about $2m_{\rm person}g$ in total, at most. As the bike rolls up and down small hills the wheel’s acceleration might also be about $g$ (at most) so that $|m_{\rm wheel}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm wheel}|% \approx m_{\rm wheel}g$ . We ask,

	$\displaystyle\left\|\,\mathrm{m}_{\rm wheel}{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{G}\right\|$	$\displaystyle\ll\ F_{\rm typical}\textrm{ \Large?}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m_{\rm wheel}g$	$\displaystyle\ll 2m_{\rm person}g$
	$\displaystyle\textrm{and indeed}\qquad m_{\rm wheel}$	$\displaystyle\ll 2m_{\rm person}.$

For a $50\,\mathrm{kg}$ rider on a bike with $1\,\mathrm{kg}$ wheels, the error from using statics instead of dynamics, for the forces on a bicycle wheel, is about 4%. The load on the wheel might be from acceleration of the rider above, but the analysis of the wheel itself can use statics with pretty good accuracy. Note, a similar argument also often shows that for structures holding other parts we can reasonably neglect, when studying the structure, not just the structure’s acceleration, but the weight of the structure too.

What if your statics calculation gives an inaccurate result? If your statics calculations make a bad prediction, one possible source of errors is your neglect of dynamic terms. But that is not the common case for things that seem relatively stationary. Rather,

Most bad statics predictions come from

•

Bad estimates of material properties (friction coefficient, failure strength, etc),
•

Wrong dimensions (or angles, etc), or
•

Math errors.

SAMPLE 5.1

Filename:sfig8-2-3a

Figure 5.8:

Equilibrium of a pin. Two rods, AB and BC, are pinned together at point B and to the ground as shown in the figure. A force $F=$ 100 N is applied at point B. For $\theta=45^{{}^{\circ}}$ , find the tension in the two rods.

Solution

Filename:sfig8-2-3b

Figure 5.9: The free-body diagram of the pin at B. Although we can see that rod AB is in compression, we use tension as positive and let the solution take care of the proper sign.

The free-body diagram of the pin at B is shown in fig. 5.9 where $T_{1}$ and $T_{2}$ are the tensions in rods AB and BC respectively. The static equilibrium of the pin at B requires that

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{F}}}+{\accentset{\rightharpoonup}{\boldsymbol{T}}% }_{1}+{\accentset{\rightharpoonup}{\boldsymbol{T}}}_{2}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}

\displaystyle\hbox to36.135pt{or\hfil}F(\sin\theta\hat{\boldsymbol{\imath}}-% \cos\theta\hat{\boldsymbol{\jmath}})-T_{1}\hat{\boldsymbol{\jmath}}+T_{2}(-% \sin\theta\hat{\boldsymbol{\imath}}-\cos\theta\hat{\boldsymbol{\jmath}})=0.

(5.5)

This is one vector equation in 2D in two unknowns $T_{1}$ and $T_{2}$ . We can solve for the unknowns in various ways.

Method-1: Separate out scalar equations in $x$ and $y$ directions.

The force equilibrium equation, eqn. (5.5), gives us two independent scalar equations in the $x$ and $y$ directions:

	$\displaystyle\sum F_{x}=0$	$\Rightarrow$	$\displaystyle F\sin\theta-T_{2}\sin\theta$	$\displaystyle=0$
	$\displaystyle\sum F_{y}=0$	$\Rightarrow$	$\displaystyle-F\cos\theta-T_{1}-T_{2}\cos\theta$	$\displaystyle=0.$

Solving these two equations simultaneously, we get

$\displaystyle T_{2}$	$\displaystyle=$	$\displaystyle F=100\,{\mbox{N}}$
$\displaystyle T_{1}$	$\displaystyle=$	$\displaystyle-(F+T_{2})\cos\theta$
	$\displaystyle=$	$\displaystyle-2F\cos\theta=-141.4\,{\mbox{N}}.$

Answer: $T_{1}=-141.4\,{\mbox{N}},\qquad T_{2}=100\,{\mbox{N}}$

^†^†margin: Note that the tension in rod AB,

T_{1}

turns out to be negative, that is, rod AB is in compression. This is expected since the other two forces at B,

F

and

T_{2}

, are pushing on rod AB. The equality of

F

and

T_{2}

is also expected from symmetry.

Method-2: Dot the equation with appropriate vectors.

The goal here is to dot the vector equation with appropriate vectors that give us one scalar equation in one unknown. Here, ${\accentset{\rightharpoonup}{\boldsymbol{T}}}_{1}$ acts in the $-\hat{\boldsymbol{\jmath}}$ direction; therefore, dotting the equation with $\hat{\boldsymbol{\imath}}$ gets rid of ${\accentset{\rightharpoonup}{\boldsymbol{T}}}_{1}$ and results in a scalar equation involving only $T_{2}$ :

	$\displaystyle\mbox{[eqn.~{}(\ref{eq:poletwocables})]}\cdot\hat{\boldsymbol{% \imath}}$	$\Rightarrow$	$\displaystyle F\sin\theta-T_{2}\sin\theta=0\hskip 108.405pt$
		$\Rightarrow$	$\displaystyle T_{2}=F=100\,{\mbox{N}}.$

Similarly, to get rid of ${\accentset{\rightharpoonup}{\boldsymbol{T}}}_{2}$ , dot the equation with a vector $\hat{\boldsymbol{n}}$ normal to ${\accentset{\rightharpoonup}{\boldsymbol{T}}}_{2}$ , i.e., with $\hat{\boldsymbol{n}}=\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{% \boldsymbol{\jmath}}$ :^†^†margin: How do we find the normal vector $\hat{\boldsymbol{n}}$ ? Well, we know the direction of ${\accentset{\rightharpoonup}{\boldsymbol{T}}}_{2}$ , which is $\hat{\boldsymbol{\lambda}}=-\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat% {\boldsymbol{\jmath}}$ . We can draw a unit vector $\hat{\boldsymbol{n}}$ normal to $\hat{\boldsymbol{\lambda}}$ and find its components from geometry (see figure below), or we can set $\hat{\boldsymbol{n}}=\hat{\boldsymbol{k}}\times\hat{\boldsymbol{\lambda}}$ which guarantees $\hat{\boldsymbol{n}}$ to be normal to $\hat{\boldsymbol{\lambda}}$ . In either case we get $\hat{\boldsymbol{n}}=\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{% \boldsymbol{\jmath}}$ .

Filename:sfig8-3-1

Figure 5.10: Finding a unit vector

\hat{\boldsymbol{n}}

normal to the direction

\hat{\boldsymbol{\lambda}}

T_{2}

$\displaystyle\mbox{[eqn.~{}(\ref{eq:poletwocables})]}\cdot\hat{\boldsymbol{n}}$	$\Rightarrow$	$\displaystyle[-T_{1}\hat{\boldsymbol{\jmath}}+F(\cos\theta\hat{\boldsymbol{% \imath}}-\sin\theta\hat{\boldsymbol{\jmath}})\cdot(\cos\theta\hat{\boldsymbol{% \imath}}-\sin\theta\hat{\boldsymbol{\jmath}})]=0$
	$\Rightarrow$	$\displaystyle T_{1}\sin\theta+F(\cos^{2}\theta+\sin^{2}\theta)=0$
	$\Rightarrow$	$\displaystyle T_{1}=-\frac{F}{\sin\theta}=-\frac{100\,{\mbox{N}}}{1/\sqrt{2}}=% -141.4\,{\mbox{N}}$

These are the same values of $T_{1}$ and $T_{2}$ , as they must be, obtained by Method-1.

Method-3: Use matrix equation and solve by hand or on a computer.

The two scalar equations obtained from eqn. (5.6) can be written in the matrix form as

\left[\begin{array}[]{rr}0&\sin\theta\\ 1&\cos\theta\end{array}\right]\left(\begin{array}[]{c}T_{1}\\ T_{2}\end{array}\right)=\left(\begin{array}[]{c}F\sin\theta\\ -F\cos\theta\end{array}\right).

Using $F=100\,{\mbox{N}}$ and $\theta=45^{\circ}=\pi/4$ , and solving the above matrix equation (see Sample 1.5 on page 1.5 and Sample 1.120 on page 1.120), we get

\left(\begin{array}[]{c}T_{1}\\ T_{2}\end{array}\right)=\left(\begin{array}[]{c}-141.4\\ 100\end{array}\right)\,{\mbox{N}}

which is, of course, the same result as we got above.

SAMPLE 5.2

Filename:sfig8-3-1a

Figure 5.11:

Filename:sfig8-5-wiper

Figure 5.12:

A mass held in equilibrium by unequal strings in 2D. A 10 kg block $m$ hangs from strings $A B$ and $A C$ in the vertical plane as shown in the figure. Find the tension in the strings.

Solution The free-body diagram of the block is shown in figure 5.12. The equation of force balance, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ , gives

\displaystyle T_{1}\hat{\boldsymbol{\lambda}}_{\mathrm{AB}}+T_{2}\hat{% \boldsymbol{\lambda}}_{\mathrm{AC}}-mg\hat{\boldsymbol{\jmath}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}},

(5.6)

where $\hat{\boldsymbol{\lambda}}_{\mathrm{AB}}$ and $\hat{\boldsymbol{\lambda}}_{BC}$ are unit vectors in the AB and AC directions:

	$\displaystyle\hat{\boldsymbol{\lambda}}_{AB}$	$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{AB}}{\|{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{AB}\|}=\frac{-2\,\mathrm{m}\hat{% \boldsymbol{\imath}}+2\,\mathrm{m}\hat{\boldsymbol{\jmath}}}{2\sqrt{2}\,% \mathrm{m}}=\frac{1}{\sqrt{2}}(-\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{% \jmath}})$
	$\displaystyle\hat{\boldsymbol{\lambda}}_{\mathrm{AC}}$	$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{AC}}{\|{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{AC}\|}=\frac{1\,\mathrm{m}\hat{% \boldsymbol{\imath}}+2\,\mathrm{m}\hat{\boldsymbol{\jmath}}}{\sqrt{5}\,\mathrm% {m}}=\frac{1}{\sqrt{5}}(\hat{\boldsymbol{\imath}}+2\hat{\boldsymbol{\jmath}}).$

Substituting in eqn. (5.6) and rearranging terms, we have

\left(-\frac{T_{1}}{\sqrt{2}}+\frac{T_{2}}{\sqrt{5}}\right)\hat{\boldsymbol{% \imath}}+\left(\frac{T_{1}}{\sqrt{2}}+\frac{2T_{2}}{\sqrt{5}}-mg\right)\hat{% \boldsymbol{\jmath}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.

Separating $x$ and $y$ components of this equation, we get the scalar equations

	$\displaystyle\sum F_{x}=0$	$\Rightarrow$	$\displaystyle-\frac{T_{1}}{\sqrt{2}}+\frac{T_{2}}{\sqrt{5}}$	$\displaystyle=0$
	$\displaystyle\sum F_{y}=0$	$\Rightarrow$	$\displaystyle\frac{T_{1}}{\sqrt{2}}+\frac{2T_{2}}{\sqrt{5}}-mg$	$\displaystyle=0.$

Solving these two equations simultaneously we get,

T_{1}=\frac{\sqrt{2}}{3}mg=46.24\,{\mbox{N}}\quad\mbox{and}\quad T_{2}=\frac{% \sqrt{5}}{3}mg=73.12\,{\mbox{N}}.

Answer: $T_{1}=46.24\,{\mbox{N}},\ T_{2}=73.12\,{\mbox{N}}$

Note:

Solving for $T_{1}$ and $T_{2}$

If you are comfortable with vector algebra, then solving for $T_{1}$ and $T_{2}$ from eqn. (5.6) is quite easy. Let us say, we find two unit vectors $\hat{\boldsymbol{n}}_{AB}$ and $\hat{\boldsymbol{n}}_{A}C$ normal to unit vectors $\hat{\boldsymbol{\lambda}}_{AB}$ and $\hat{\boldsymbol{\lambda}}_{AC}$ , respectively. Then dotting eqn. (5.6) with $\hat{\boldsymbol{n}}_{AB}$ and $\hat{\boldsymbol{n}}_{A}C$ , one at a time, we can solve for $T_{1}$ and $T_{2}$ in one step:

T_{2}=\frac{mg\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol{n}}_{AB}}{\hat{% \boldsymbol{\lambda}}_{AC}\cdot\hat{\boldsymbol{n}}_{AB}},\hbox to36.135pt{% \hfil and\hfil}T_{1}=\frac{mg\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol{n}% }_{AC}}{\hat{\boldsymbol{\lambda}}_{AB}\cdot\hat{\boldsymbol{n}}_{AC}}.

For computing the values, we need to carry out the dot products. Noting that $\hat{\boldsymbol{n}}_{AB}=\frac{1}{\sqrt{2}}(i+j)$ and $\hat{\boldsymbol{n}}_{AC}=\frac{1}{\sqrt{5}}(-2\hat{\boldsymbol{\imath}}+\hat{% \boldsymbol{\jmath}})$ (you can write these vectors by looking at $\hat{\boldsymbol{\lambda}}_{AB}$ and $\hat{\boldsymbol{\lambda}}_{AC}$ ), we can carry out the dot product and get the values of $T_{1}$ and $T_{2}$ .

Matrix equation

The two scalar equations obtained from eqn. (5.6) can be written in the matrix form as

\left[\begin{array}[]{rr}-\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{5}}\\ \frac{1}{\sqrt{2}}&\frac{2}{\sqrt{5}}\end{array}\right]\left(\begin{array}[]{c% }T_{1}\\ T_{2}\end{array}\right)=\left(\begin{array}[]{c}0\\ mg\end{array}\right).

Using $mg=(10\,\mathrm{kg})\cdot(9.81\,\mathrm{m}/\mathrm{s}^{2})=98.1\,{\mbox{N}}$ , and solving the above matrix equation (see Sample 1.5 on page 1.5 and Sample 1.120 on page 1.120), we get

\left(\begin{array}[]{c}T_{1}\\ T_{2}\end{array}\right)=\left(\begin{array}[]{c}46.24\\ 73.12\end{array}\right)\,{\mbox{N}}

which is, of course, the same result as we got above.

SAMPLE 5.3

Filename:sfig8-5-wiper-a

Figure 5.13:

Tensions in a paraglider’s ropes. A paraglider is held by two ropes that in turn connect to the parachute with many ropes. The angles that the two ropes make with the horizontal (or vertical) are not necessarily the same for each rope. Assume that the right rope makes an angle $\theta_{1}$ with the horizontal and the left one makes an angle $\theta_{2}$ (in flight, these angles will vary with time). Also assume that the paraglider is descending at some uniform velocity. Find the tensions in the two ropes in terms of the weight of the person and the harness (say, $m g$ ), and the angles $\theta_{1}$ and $\theta_{2}$ .

Solution

Filename:sfig8-5-2disks

Figure 5.14: The free-body diagram of the paraglider considering the ropes to be at different angles with the horizontal.

The free-body diagram of the glider is shown in fig. 5.14. We assume that the tensions $T_{1}$ , $T_{2}$ , and the weight $m g$ are all in one vertical plane. Since the paraglider has a uniform velocity (no acceleration), we can use the force-balance equation for static equilibrium, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ :

T_{1}\hat{\boldsymbol{\lambda}}_{1}+T_{2}\hat{\boldsymbol{\lambda}}_{2}-mg\hat% {\boldsymbol{\jmath}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}

(5.7)

where $\hat{\boldsymbol{\lambda}}_{1}=\cos\theta_{1}\hat{\boldsymbol{\imath}}+\sin% \theta_{1}\hat{\boldsymbol{\jmath}}$ and $\hat{\boldsymbol{\lambda}}_{2}=-\cos\theta_{2}\hat{\boldsymbol{\imath}}+\sin% \theta_{2}\hat{\boldsymbol{\jmath}}$ are the unit vectors along the two ropes, respectively. Thus the vector equation of equilibrium is:

T_{1}(\cos\theta_{1}\hat{\boldsymbol{\imath}}+\sin\theta_{1}\hat{\boldsymbol{% \jmath}})+T_{2}(-\cos\theta_{2}\hat{\boldsymbol{\imath}}+\sin\theta_{2}\hat{% \boldsymbol{\jmath}})-mg\hat{\boldsymbol{\jmath}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}.

(5.8)

From this equation, we get two independent scalar equations by separating the $x$ and $y$ components:

	$\displaystyle\sum F_{x}=0$	$\Rightarrow$	$\displaystyle T_{1}\cos\theta_{1}-T_{2}\cos\theta_{2}$	$\displaystyle=0$
	$\displaystyle\sum F_{y}=0$	$\Rightarrow$	$\displaystyle T_{1}\sin\theta_{1}+T_{2}\sin\theta_{2}-mg$	$\displaystyle=0.$

Solving these two equations simultaneously we get,

T_{1}=\frac{\cos\theta_{2}}{\sin(\theta_{1}+\theta_{2})}mg\quad\mbox{and}\quad T% _{2}=\frac{\cos\theta_{1}}{\sin(\theta_{1}+\theta_{2})}mg.

Answer: $T_{1}=\frac{mg\cos\theta_{2}}{\sin(\theta_{1}+\theta_{2})}$ and $T_{2}=\frac{mg\cos\theta_{1}}{\sin(\theta_{1}+\theta_{2})}$

Another vector method: We can find $T_{1}$ and $T_{2}$ directly from eqn. (5.7) by taking cross products with $\hat{\boldsymbol{\lambda}}_{2}$ and $\hat{\boldsymbol{\lambda}}_{1}$ , respectively:

$\displaystyle\hat{\boldsymbol{\lambda}}_{2}\times(T_{1}\hat{\boldsymbol{% \lambda}}_{1}+T_{2}\hat{\boldsymbol{\lambda}}_{2}-mg\hat{\boldsymbol{\jmath}})$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T_{1}\underbrace{(\hat{% \boldsymbol{\lambda}}_{2}\times\hat{\boldsymbol{\lambda}}_{1})}_{-\sin(\theta_% {1}+\theta_{2})\hat{\boldsymbol{k}}}-mg\underbrace{(\hat{\boldsymbol{\lambda}}% _{2}\times\hat{\boldsymbol{\jmath}})}_{-\cos\theta_{2}\hat{\boldsymbol{k}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T_{1}$	$\displaystyle=$	$\displaystyle\frac{\cos\theta_{2}}{\sin(\theta_{1}+\theta_{2})}mg$

Similarly, $\lambda_{2}\times$ [eqn. (5.7)] gives us

T_{2}=\frac{\cos\theta_{1}}{\sin(\theta_{1}+\theta_{2})}mg.

Thus we get the same results as we got above by solving the two scalar equations simultaneously.

SAMPLE 5.4

Filename:sfig8-4-1

Figure 5.15: A mass-particle on an inclined plane.

A single string holding a mass on a frictionless incline. A block of mass $m$ rests on a frictionless inclined plane with the help of a string that connects the mass to a fixed support at A. Find the force in the string.

Solution The free-body diagram of the mass is shown in Fig. 5.16. The string force $F_{s}$ and the normal reaction of the plane $N$ are unknown forces.

Filename:sfig8-4-1a

Figure 5.16: Free-body diagram of the mass and the geometry of force vectors.

The force-balance equation, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ , is

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{s}}+{\accentset{% \rightharpoonup}{\boldsymbol{N}}}+m{\accentset{\rightharpoonup}{\boldsymbol{g}% }}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.

We can express the forces in terms of their components in various ways and then dot the vector equation with appropriate unit vectors to get two independent scalar equations. For example, we write the force-balance equation using mixed basis vectors $\hat{\boldsymbol{e}}_{t}$ and $\hat{\boldsymbol{e}}_{n}$ , and $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ :

F_{s}\hat{\boldsymbol{e}}_{t}+N\hat{\boldsymbol{e}}_{n}-mg\hat{\boldsymbol{% \jmath}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.

(5.9)

We can now find $F_{s}$ directly by taking the dot product of the above equation with $\hat{\boldsymbol{e}}_{t}$ since the other unknown $N$ is in the $\hat{\boldsymbol{e}}_{n}$ direction and $\hat{\boldsymbol{e}}_{n}\cdot\hat{\boldsymbol{e}}_{t}=0$ :

	$\displaystyle\{\mbox{eqn.~{}(\ref{eq2.1.13.1})}\}\cdot\hat{\boldsymbol{e}}_{t}$	$\Rightarrow$	$\displaystyle F_{s}-mg\overbrace{(\hat{\boldsymbol{\jmath}}\cdot\hat{% \boldsymbol{e}}_{t})}^{\sin\theta}=0$
		$\Rightarrow$	$\displaystyle F_{s}=mg\sin\theta.$

Answer: $F_{s}=mg\sin\theta$

Note: We can also find $N$ from a single equation by taking the dot product of eqn. (5.9) with $\hat{\boldsymbol{n}}$ :

	$\displaystyle\{\mbox{eqn.~{}(\ref{eq2.1.13.1})}\}\cdot\hat{\boldsymbol{e}}_{n}$	$\Rightarrow$	$\displaystyle N-mg\overbrace{(\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol{e% }}_{n})}^{\cos\theta}=0$
		$\Rightarrow$	$\displaystyle N=mg\cos\theta.$

Filename:sfig8-6-2

Figure 5.17: (a) Components of

m g

along

t

and

n

directions. (b) The mixed basis dot products:

\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol{e}}_{t}=\sin\theta

and

\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol{e}}_{n}=\cos\theta

Scalar approach: We resolve all forces into their $\hat{\boldsymbol{e}}_{t}$ and $\hat{\boldsymbol{e}}_{n}$ components and then sum the forces. Here, $F_{s}$ is along the plane and therefore, has no component perpendicular to the plane. Force $N$ is perpendicular to the plane and therefore, has no component along the plane. We resolve the weight $m g$ into two components: (1) $mg\cos\theta$ perpendicular to the plane (along $\hat{\boldsymbol{e}}_{n}$ ) and (2) $mg\sin\theta$ along the plane (along $\hat{\boldsymbol{e}}_{t}$ ). Now we can sum the forces:

	$\displaystyle\sum F_{t}=0$	$\Rightarrow$	$\displaystyle F_{s}-mg\sin\theta=0;$
	$\displaystyle\mbox{and}\quad\sum F_{n}=0$	$\Rightarrow$	$\displaystyle N-mg\cos\theta=0$

which, of course, is essentially the same as the equations obtained above.

SAMPLE 5.5

Filename:sfig8-6-2a

Figure 5.18:

A particle in 3D. A particle of mass 1 $\,\mathrm{kg}$ is attached to two strings tied at points C and D shown in the figure. Another string, AB, attached to the particle, passes over a pulley and is used to hold the particle in equilibrium under gravity such that it loses contact with the ground at point A. Find the tension in string AB.

Solution

Filename:efig1-2-28

Figure 5.19:

The free-body diagram of the particle is shown in fig. 5.19. Assuming the tensions in strings AB, AC, and AD to be $T_{\rm AB}$ , $T_{\rm AC}$ , and $T_{\rm AD}$ respectively, we can represent the string forces acting on the particle as $T_{\rm AB}\hat{\boldsymbol{\lambda}}_{\mathrm{AB}}$ , $T_{\rm AC}\hat{\boldsymbol{\lambda}}_{\mathrm{AC}}$ , and $T_{\rm AD}\hat{\boldsymbol{\lambda}}_{\mathrm{AD}}$ , where the $\hat{\boldsymbol{\lambda}}$ ’s are the unit vectors along the strings.

The force balance on the particle gives us

T_{\rm AB}\hat{\boldsymbol{\lambda}}_{\mathrm{AB}}+T_{\rm AC}\hat{\boldsymbol{% \lambda}}_{\mathrm{AC}}+T_{\rm AD}\hat{\boldsymbol{\lambda}}_{\mathrm{AD}}-mg% \hat{\boldsymbol{k}}=0.

(5.10)

This is the equation we need to solve to find $T_{\rm AB}$ . We show various methods below that you can use to get $T_{\rm AB}$ .

1.

Brute force (by hand).

From the given figure, the unit vectors are:

$\displaystyle\hat{\boldsymbol{\lambda}}_{\mathrm{AB}}$ $\displaystyle=$ $\displaystyle\frac{-4\hat{\boldsymbol{\imath}}+3\hat{\boldsymbol{\jmath}}+12% \hat{\boldsymbol{k}}}{\sqrt{4^{2}+3^{2}+12^{2}}}=-\frac{4}{13}\hat{\boldsymbol% {\imath}}+\frac{3}{13}\hat{\boldsymbol{\jmath}}+\frac{12}{13}\hat{\boldsymbol{% k}}$

$\displaystyle\hat{\boldsymbol{\lambda}}_{\mathrm{AC}}$ $\displaystyle=$ $\displaystyle-\hat{\boldsymbol{\jmath}}$

$\displaystyle\hat{\boldsymbol{\lambda}}_{\mathrm{AD}}$ $\displaystyle=$ $\displaystyle\frac{12\hat{\boldsymbol{\imath}}+5\hat{\boldsymbol{k}}}{\sqrt{12% ^{2}+5^{2}}}=\frac{12}{13}\hat{\boldsymbol{\imath}}+\frac{5}{13}\hat{% \boldsymbol{k}}.$

Substituting these vectors in eqn. (5.10) and equating the $x$ , $y$ and $z$ components of the equation to zero separately, we get

$\displaystyle-\frac{4}{13}T_{\rm AB}+\frac{12}{13}T_{\rm AD}$ $\displaystyle=$ $\displaystyle 0$

$\displaystyle\frac{3}{13}T_{\rm AB}-T_{\rm AC}$ $\displaystyle=$ $\displaystyle 0$ (5.11)

$\displaystyle\frac{12}{13}T_{\rm AB}+\frac{5}{13}T_{\rm AD}$ $\displaystyle=$ $\displaystyle mg.$

We can solve the three equations simultaneously to get

$T_{\rm AB}=\frac{39}{41}mg,\quad T_{\rm AC}=\frac{9}{41}mg,\quad\qquad\mbox{% and}\qquad\quad T_{\rm AD}=\frac{13}{41}mg.$

Substituting $m=1\,\mathrm{kg}$ and $g=9.81\,\mathrm{m}/\mathrm{s}^{2}$ , we get the required values.

Answer: $T_{\rm AB}=9.33\,{\mbox{N}},\quad T_{\rm AC}=2.15\,{\mbox{N}},\quad T_{\rm AD}% =3.11\,{\mbox{N}}$
2.

Systematically set up matrix equations. Eqn. 5.10 can be written in matrix form as

$\underbrace{\left[\begin{array}[]{ccc}\left[\hat{\boldsymbol{\lambda}}_{\rm AB% }\right]_{xyz}^{\prime}&\left[\hat{\boldsymbol{\lambda}}_{\rm AC}\right]_{xyz}% ^{\prime}&\left[\hat{\boldsymbol{\lambda}}_{\rm AD}\right]_{xyz}^{\prime}\end{% array}\right]}_{\hbox{$3\times 3$ matrix}}\left[\begin{array}[]{c}T_{\rm AB}\\ T_{\rm AB}\\ T_{\rm AB}\end{array}\right]=\left[\begin{array}[]{c}0\\ 0\\ mg\end{array}\right]$

where $\left[\hat{\boldsymbol{\lambda}}_{\rm AB}\right]_{xyz}^{\prime}$ is a column of 3 numbers, namely the $x, y,$ and $z$ components of $\hat{\boldsymbol{\lambda}}_{\rm AB}$ ; similarly for the other two columns of the $3\times 3$ matrix. This matrix equation is then ready to hand to a calculator or computer for a matrix solution. Thus, eqn. (5.11) can be written as,

$\left[\begin{array}[]{rrr}-4/13&0&12/13\\ 3/13&-1&0\\ 12/13&0&5/13\end{array}\right]\left(\begin{array}[]{r}T_{\rm AB}\\ T_{\rm AC}\\ T_{\rm AD}\end{array}\right)=\left(\begin{array}[]{r}0\\ 0\\ mg\end{array}\right).$

Using the pseudo code shown on the side ^†^†margin: Pseudo-code:
Let m=1, 9=9.81
A = [ -4/13 0 12/13
          3/13 -1 0
          12/13 0 5/13 ]
b = [ 0 0 m*g]’
solve A*T = b for T we solve the equations on a computer and get,
                  T = [9.33 2.15 3.11]
which is the solution that we obtained above by hand calculation.

Computer solution. All the math can be handed to a computer by a sequence of commands like this, working from the knowns to the unknowns (see page The order of calculation is often backwards from the order of thinking), all in consistent units:

 % Get all the knowns into the computer
   rA = [0 0 0]’ ;   rB = [-4 3 12]’
   rC = [0 -15 0]’ ;   rD = [12 0 5]’
   m = 1 ;  g = 9.81 ;
 % Make relative position vectors
   rAB = rB- rA;   rAC = rC - rA;  rAD = rD - rA
 % Make unit vectors
   lamdaAB = rAB/magnitude(rAB);
   lamdaAC = rAC/magnitude(rAC);
   lamdaAD = rAD/magnitude(rAD);
 % Set up and solve the matrix equation
   M = [lamdaAB  lamdaAC  lamdaAD]  %3x3 matrix
   F = [0 0 mg]’   %column vector of known force
   solve {M T = F} for T

The column of numbers T will be the tensions in the 3 cables. Using this pseudo-code on a computer, we get T = [9.33 2.15 3.11] again.

Be tricky to get one equation in one unknown. Since we are interested only in $T_{\rm AB}$ , we can get rid of the terms we don’t know or care about. ^†^†margin: Another way of doing this is by taking Moment about an axis. This approach is similar in spirit to the previous approach. Instead of the equilibrium eqn. (5.10) we could have used moments about axis CD to ‘kill off’ the tensions in ropes AC and AD (they have no moment about that axis), like this, $\sum M_{\rm axis\ CD}=0$ ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm CD}\cdot\left\{{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\rm DA}\times\left\{T_{\rm AB}\hat{% \boldsymbol{\lambda}}_{\rm AB}-mg\hat{\boldsymbol{k}}\right\}\right\}=0$ $T_{\rm AB}=\frac{mg{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm CD}\cdot% \left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm DA}\times\hat{% \boldsymbol{k}}\right)}{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm CD}% \cdot\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm DA}\times\hat{% \boldsymbol{\lambda}}_{\rm AB}\right)}.$ Again we have found one equation for one unknown, $T_{\rm AB}$ . All the quantities on the right can be evaluated to give $T_{\rm AB}$ . The vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AC}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\rm AD}$ is orthogonal to both ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AC}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AD}$ , so it is orthogonal to $\hat{\boldsymbol{\lambda}}_{\rm AC}$ and $\hat{\boldsymbol{\lambda}}_{\rm AD}$ . So taking the dot product of both sides of eqn. (5.10) with ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AC}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\rm AD}$ , we get

\begin{array}[]{c}\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AC}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AD}\right)\cdot\left% \{T_{\rm AB}\hat{\boldsymbol{\lambda}}_{\rm AB}+T_{\rm AC}\hat{\boldsymbol{% \lambda}}_{\rm AC}+T_{\rm AD}\hat{\boldsymbol{\lambda}}_{\rm AD}-mg\hat{% \boldsymbol{k}}\right\}=\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% \rm AC}\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AD}\right)% \cdot{\accentset{\rightharpoonup}{\boldsymbol{0}}}\\[4.0pt] \left(\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AC}\times{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AD}\right)\cdot\hat{% \boldsymbol{\lambda}}_{\rm AB}\right)T_{\rm AB}=mg\left(\left({\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\rm AC}\times{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\rm AD}\right)\cdot\hat{\boldsymbol{k}}\right)\\[4.0pt] T_{\rm AB}=\frac{mg\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AC% }\times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AD}\right)\cdot\hat{% \boldsymbol{k}}}{\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AC}% \times{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AD}\right)\cdot\hat{% \boldsymbol{\lambda}}_{\rm AB}}.\end{array}

Since, ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AC}\times{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\rm AD}=(-15\hat{\boldsymbol{\jmath}})\,% \mathrm{m}\times(12\hat{\boldsymbol{\imath}}+5\hat{\boldsymbol{k}})\,\mathrm{m% }=(180\hat{\boldsymbol{k}}-75\hat{\boldsymbol{\imath}})\,\mathrm{m}^{2}$ , substituting this cross product and other known quantities, we get

$\displaystyle T_{\rm AB}$	$\displaystyle=$	$\displaystyle\frac{mg\cdot 180}{180\cdot 12/13+75\cdot 4/13}$
	$\displaystyle=$	$\displaystyle 0.95\,mg$
	$\displaystyle=$	$\displaystyle 9.33\,{\mbox{N}}.$

Answer: $T_{\rm AB}=9.33\,{\mbox{N}}$

Problems for 5.1 Static equilibrium of a particle

Preparatory Problems

5.1.1 What is a particle?

5.1.2 What are the equations of equilibrium for a particle (also called “equilibrium conditions”, “force balance”, or “linear momentum balance for statics”?

5.1.3 The particle shown in the figure is in static equilibrium. Find the unknown force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ .

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Figure 5.20:

5.1.4 Four forces act on a block as shown in the figure and hold it in static equilibrium. Assume that the magnitude of force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ is known and it is $F$ .

(a)

Write the scalar equations of equilibrium in the $x$ and $y$ directions.
(b)

Solve for ${\accentset{\rightharpoonup}{\boldsymbol{N}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{r}}$ in terms of $F$ , $m g$ , and $\theta$ .
(c)

What is ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{r}}$ when $\theta=90^{\circ}$ ?

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Figure 5.21:

5.1.5 A particle of mass $m=2\,\mathrm{kg}$ hangs from strings AB and AC as shown. AB is horizontal and $\theta=45^{{}^{\circ}}$ . Find the tension in the two strings.

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Figure 5.22:

5.1.6 What force should be applied to the end of the string over the pulley at $C$ so that the mass at $A$ is at rest in the configuration shown?

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Figure 5.23:

5.1.7 $N$ small blocks each of mass $m$ hang vertically as shown, connected by $N$ inextensible strings. Find the tension $T_{n}$ in string $n$ . Answer: $T_{N-1}=(N-1)Nmg$ , $T_{2}=2mg$ , $T_{1}=mg$ , and in general $T_{n}=nmg$

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Figure 5.24:

5.1.8 For each situation below, assume static equilibrium under the applied force and find the tensions in the two rods.

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Figure 5.25:

5.1.9 A particle of mass $\,\mathrm{m}=5\,\mathrm{kg}$ at the end of a horizontal massless rod CB of length $1.2\,\mathrm{m}$ is held in place with the help of a string AB that makes an angle $\theta=45^{{}^{\circ}}$ with the vertical in the equilibrium position. Find the tension in the bar CB (it is ok to have negative tension).

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Figure 5.26:

5.1.10 For each structure shown below, find the tension in each rod. (Note the tension can be less than zero.)

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Figure 5.27:

5.1.11 In the following structures, a pin connects two thin bars that are very nearly either horizontal or vertical. Find the tensions in each rod under the applied loads. (Note the tension is less than zero for some of the rods.)

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Figure 5.28:

5.1.12 For each situation shown below, equilibrium is not possible. Write the vector equation for force balance and show that it has no solutions (i.e., leads to an equation like $7=0$ ).

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Figure 5.29:

5.1.13 Assume no sliding friction ( $\mu=0$ ). Assume equilibrium. Find all reactions, tensions, and forces. Answer: (a) is nonsense; others are fine

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Figure 5.30:

5.1.14 In each of the two cases given below, find the tension in the string AB assuming the block to be at rest and the ramp to be frictionless.

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Figure 5.31:

5.1.15 Find the unknown forces and tensions in each structure shown below.

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Figure 5.32:

5.1.16 A block of mass $m=5\,\mathrm{kg}$ rests on a frictionless inclined plane as shown in the figure. Let $\theta=\alpha=30^{{}^{\circ}}$ . Find the tension in the string.

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Figure 5.33:

More-Involved Problems

5.1.17 For small $\delta$ what is the relation between $F$ and $\delta$ (and $g$ and $\ell$ ) for a static pendulum?

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Figure 5.34:

5.1.18 In the situations shown in the figures, find the value of $\theta$ that minimizes $F$ . What is the corresponding value of $F$ in each case?

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Figure 5.35:

5.1.19 An object of weight $W=10\,{\mbox{N}}$ is held in equilibrium in the vertical plane by two strings AC and BC. Let $\theta=30^{{}^{\circ}}$ and $0\leq\phi\leq 90^{{}^{\circ}}$ . Find and plot the tension in the two strings against $\phi$ and comment on the variation in the tensions.

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Figure 5.36:

5.1.20 Find the tensions in the three strings shown in the figure.

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Figure 5.37:

5.1.21 Find the tensions in the three strings shown in the figure. String CD is horizontal and the force at D is $100\,{\mbox{N}}$ straight down. [Hint: this problem has a trick to it.] Answer: The cables happen to be co-planar and the force is not in that plane. So there is no solution.

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Figure 5.38:

5.1.22 Show that the particle acted upon by the given force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=(3\hat{\boldsymbol{\imath}}+4% \hat{\boldsymbol{\jmath}}+5\hat{\boldsymbol{k}})\,{\mbox{N}}$ , and held by the two bars as shown in the figure cannot be in equilibrium.

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Figure 5.39:

5.1.23 In the figure shown, the force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ , in the $x$ - $z$ plane, acts on the particle (weighing $100\,{\mbox{N}}$ ). Find $F$ as a function of $\theta$ for equilibrium of the particle. Find the value of $\theta$ for which the required force is smallest.

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Figure 5.40:

5.1.24 For the three cases (a), (b), and (c), below, find the tension in the string AB. In all cases the strings hold up the mass $m=3\,\mathrm{kg}$ . You may assume the local gravitational constant is $g=10\,\mathrm{m}/\mathrm{s}^{2}$ . In all cases the winches are pulling in the string so that the velocity of the mass is a constant $4\,\mathrm{m}/\mathrm{s}$ upwards (in the $\hat{\boldsymbol{k}}$ direction). [ Note that in problems (b) and (c), in order to pull the mass up at constant rate the winches must pull in the strings at an unsteady speed.] Answer: (a) $T_{AB}=30\,{\mbox{N}}$ , (b) $T_{AB}=\frac{300}{17}\,{\mbox{N}}$ , (c) $T_{AB}=\frac{5\sqrt{26}}{2}\,{\mbox{N}}$

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Figure 5.41:

5.1.25 A block of weight W , held by two strings AC and DC, rests on a slippery plane AEH. String CD is parallel to EH. Find the tensions in the two strings and the reaction of the plane. You may approximate AC to lie in the plane AEH.

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Figure 5.42:

5.2 Equilibrium of one object

For particle statics, we used that the forces acting on an object in equilibrium have no net push or pull; the forces add to zero. Now we will use that the forces have no tendency to cause rotation; the moments add to zero. These are not two in a long list of facts about equilibrium, but the whole story. As stated on the inside front cover and this chapter’s introduction (page 5)

An object is in static equilibrium if and only if the force balance and moment-balance equations hold.

\displaystyle\underbrace{\sum_{\rm All\ external\atop forces}\kern-10.0pt{{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}}_{\mbox{force balance}}

and

\displaystyle\underbrace{\sum_{\rm All\ external\atop torques}\kern-10.0pt{% \accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{i/C}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}}_{\mbox{moment balance}}

(Ic,IIc)

The total force system acting on the object is then equivalent to a zero force and zero moment acting at C.

By supplementing the force-balance equation with moment balance we can determine more about the forces that act on an object.

Rigid-body statics

To start with, one often thinks of the object of interest as one piece, for example a whole car, a wheel, a person, a limb, a chair or a derrick. We often think of such an object as rigid, meaning that the object’s shape and size only change negligibly due to the forces of interest. Thus the phrase rigid-body mechanics. Actually, however, the equilibrium equations apply just as well to all things with little acceleration, whether or not they are stiff and solid. For a first pass at the subject, one thinks of applying the principles of statics to single, rather-solid, simply-defined objects. And such will be our main initial concern in this section. But really the delineation of an ‘object’ is up to you. And in later chapters we apply the same statics equations to clearly-non-rigid systems like water and rope. For statics the only concern is the delineation of the system at the instant of interest.

Once you know its shape, whether an object is rigid or not is irrelevant for statics.

The reference point C in moment balance.

The moment-balance equation is calculated by calculating the moments of forces relative to a point C using

\ {\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{i}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i/C}}\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{i}.

C is any convenient point, possibly the origin O of your coordinate system. C is not a special point. As discussed in Section 2.1, if a force system is equivalent to zero force and zero couple at C, it is equivalent to a zero force and zero couple at any and every point D, E, Q, etc.

Example. As you sit still reading, gravity is pulling you down and forces from the floor on your feet, the chair on your seat, and the table on your elbows hold you up. All of these forces add to zero. The net moment of these forces about the front-left corner of your desk adds to zero. And the net moment of these forces about the mole near your left elbow is also zero.

The freedom to use any point you like for moment balance provides an oft-used shortcut.

Number of equations and number of unknowns

In two dimensions the equilibrium equations make up 3 independent scalar equations. These could be:

•

2 components of force balance and the one non-trivial component of moment balance; or
•

moment balance about any two points and force balance in any direction (except in the direction orthogonal to the line connecting the two moment-balance points).
•

moment balance about 3 points (any three points not on a straight line suffice).

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Figure 5.43: A set of forces acting concurrently on an object. All force lines of action intersect at one point.

Note that moment balance necessarily is part of the equilibrium equations, but that force balance can be finessed. With one 2D free-body diagram, the equilibrium equations can be solved to find three unknown scalars, for example,

•

The magnitudes of three forces whose directions are known a priori; or
•

One unknown force vector (two components, or angle and magnitude) and one unknown magnitude; or
•

Some other three scalars associated with the forces on the free-body diagram. Besides force components and magnitudes these could include a force angle $\theta$ , a friction coefficient $\mu$ , or the location of force application.

Once you have three independent equations any additional equations you write, say moment about still another point, contains no new information^†^†margin: A fourth equilibrium equation may superficially look different from an equation already written, but it can always be derived from the other equations. .

In some problems the forces shown on a free-body diagram automatically satisfy one or more of the equilibrium equations; in making the drawing you may have implicitly solved some equilibrium equations. The equilibrium equations then offer less new information, and sometimes none at all (see 2-force bodies below).

In 3 dimensions, the equilibrium equations make up 6 independent scalar equations. Most directly these are 3 components of force and 3 components of moment. But there are many combinations of equilibrium equations that yield 6 independent scalar equations.

Special cases: concurrent forces, two-force bodies, three-force bodies

We now discuss some special loading situations for which there are special insights or problem-solution tricks. In principle, you don’t need to know any of them because force balance and moment balance spell out the whole statics story. In practice it is best to know these special cases.

Concurrent forces

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Figure 5.44: (a) Two forces acting on a body

\cal B

. (b) force balance implies that the forces are equal in magnitude and opposite in direction (

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{P}}=-{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{Q}}

. (c) moment balance implies that the forces are collinear. Body

\cal B

is a two-force member; the equilibrium equations imply that the two forces must be equal in magnitude, opposite in direction, and collinear. If the free-body diagram shows equal-in-magnitude, opposite-in-direction and collinear forces, then the equilibrium equations add no new information.

In the special case when the lines of action of all applied forces intersect at one point, moment balance is trivially satisfied (because none of the forces has a moment about the intersection point). Such a system of forces is called concurrent (fig. 5.43) and the particle model is appropriate^†^†margin: If forces are not concurrent the particle model may still be useful, as demonstrated in the previous section. . In such a case, the 2D equilibrium equations only provide two independent scalar equations and one can only use them to solve for two unknown scalars. In 3D one gets three independent scalar equations for a concurrent force system.

One-force body

Let’s first treat “one-force” bodies. Consider a finite body with only one force acting on it. Assume it is in equilibrium. Force balance says that the sum of forces must be zero. So that one force must be zero.

If only one force is acting on a body in equilibrium, that force is zero.

That was too easy. But a count to 3 wouldn’t feel complete if it didn’t start at 1.

Two-force body

When only two forces act on an object, the situation is also simplified, though not so drastically as the case with one force. An object with only two forces acting on it is called a two-force body or two-force member.

If a body in static equilibrium is acted on by two forces, then those forces are equal in magnitude, opposite in direction, and have a common line of action (the line connecting the two points of application).

This result is shown in fig. 5.44 and explained in box 5.2. If you recognize a two-force body you can draw it in a free-body diagram as in fig. 5.44c and the equations of force and moment balance provide no new information. The two-force-body shortcut is especially useful for systems with several parts, some of which are two-force members. Springs, dashpots, struts, and strings are generally idealized as two-force bodies.

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Figure 5.45: An object balancing on one point of support is a two-force body so has to have its center-of-mass over the support point.

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Figure 5.46: Dragging a heavy stone on rolling round logs. If there is only point contact between the logs and the ground and stone, each log is a two-force body. If the logs are also round there is no resistance to forward motion. The situation with ball bearings is identical.

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Figure 5.47: A cart is propelled by forces not shown. We are concerned about light strut BC and heavy tower AB. The strut is probably well-modeled as a two-force body.

Example: Tower and strut

Consider an accelerating cart (fig. 5.47) holding up massive tower $A B$ which is pinned at $A$ and braced by the light strut $B C$ . The rod $B C$ qualifies as a two-force member. The rod $A B$ does not because it has three forces and is also not in static equilibrium (non-negligible accelerating mass). Thus, the free body diagram of rod $B C$ shows the two equal and opposite collinear forces at each end parallel to the rod and the tower AB does not.

Example: Logs as bearings

Consider an ancient Egyptian dragging a big stone (fig. 5.46). If the stone and ground are flat and rigid, and the log is round, rigid and much lighter than the stone we are led to the free-body diagram of the log shown. With these assumptions there can’t be any resistance to rolling. Note that this effectively frictionless rolling occurs no matter how big the friction coefficient between the contacting surfaces. That the Egyptian got tired comes from logs not being perfectly round, the ground or stone not being perfectly flat, and, most importantly, the ground, log or stone not being perfectly rigid. In any case, it takes effort to pick up the logs in the back and move them to the front.

Example: Pliers

The pliers of fig. 2.61a on page 2.61, when considered as a whole (with the pencil they are squeezing), are a two-force body. Thus $F_{HE}=F_{EH}$ and these two forces must act on a common line. Assuming the forces are large enough that gravity can be neglected, and the motions are slow enough that statics is accurate, the person has no choice but to apply the forces in this way.

Example: One point of support

If an object with weight is supported at just one point (fig. 5.45), that point must be directly above or below the center-of-mass. Why? The gravity forces are equivalent to a single force at the center-of-mass. The body is then a two-force body. Since the direction of the gravity force is down, the support point and center-of-mass must be above one another.

Similarly,

If a body is suspended from one point, the center of gravity must be directly above or below that point.

Three-force body

If a body in equilibrium has only three forces on it, the equilibrium equations again restrict the forces in a geometrically describable manner.

The simplification is not as great as for two-force bodies but is remarkably useful for both calculation and intuition. In box 5.2 on page 5.2 moment balance about various axes is used to prove that

If exactly three forces act on a body (2D and 3D), the body is in equilibrium only if

1.

the three force vectors are coplanar,
2.
and either
1. (a)
  
  have lines of action which intersect at a single point (i.e., they are concurrent), or
2. (b)
  
  they are parallel.

One could imagine three random forces acting on a body. But, for equilibrium they must be coplanar and either concurrent or parallel. Unlike the case for 2-force bodies where the 2-force-body conditions imply the satisfaction of all equilibrium equations, for 3-force bodies planar concurrency still leaves two independent equilibrium equations possibly unsatisfied (for both 2D and 3D). That is, one still needs the equations of force balance in the plane (or, in the special case of three parallel forces, one scalar force balance and one moment-balance equation).

Example: Hanging book box

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Figure 5.48: A book is in a box which hangs from two ropes: a three-force body.

A box with a book inside is hung by two strings so that it is in equilibrium when level. The system is a three-force body so the lines of action of the two strings must intersect on the vertical line that goes through the center-of-mass of the box/book system.

Example: Which way do the forces go?

The spread of directions (the smallest angle that includes all force directions) in a 3-force body can be (a) greater than, (b) equal to, or (c) less than $180^{{}^{\circ}}$ (see the figure below). In each case we can know something about the directions of the forces. Call the point of force concurrency D.

(a)

Forces spread over less than $180^{{}^{\circ}}$ . Force balance perpendicular to the middle force implies that the outer two forces are both directed toward D or both directed away from D. Force balance in the direction of the middle force shows that it has to have the opposite sense than the outer forces. If the others are pushing in then it is pulling away. If the outer forces are pulling away then it is pushing in.
(b)

Forces spread exactly $180^{\circ}$ . Force balance in the direction perpendicular to line ADC shows that the odd force must be zero. The other forces must obviously oppose each other.
(c)

Forces spread over more than $180^{\circ}$ . Force balance perpendicular to the force at C shows that the other two forces must both pull away towards D or both push in. Then force balance along C shows that all three forces must have the same sense. All three forces are pulling away from D or all three are pushing in.

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Figure 5.49: If exactly 3 forces act on a body the lines of action of the forces intersect at a single point and are coplanar. The point of intersection does not have to lie within the body. A special case is when that point is at infinity and the three forces are parallel.

Box 5.3 Two-force bodies

Here we derive the ubiquitously-used result that, if only two forces act on a body, the two forces must be equal in magnitude, opposite in direction, and on a common line of action. You can (and will) use this result even if you do not master the reasoning in this box. But learning this reasoning may help your intuition.

Consider the free-body diagram of a body $\cal B$ in fig. 5.44a. Forces ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{P}$ and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{Q}$ are acting on $\cal B$ at points $P$ and $Q$ . Let’s apply the equilibrium equations.

First, we have that the sum of all forces on the body is zero,

	$\displaystyle\sum_{\rm All\ external\atop forces}\kern-10.0pt{{\accentset{% \rightharpoonup}{\boldsymbol{F}}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{P}+{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{Q}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}\quad\hbox to36.135% pt{\hfil$\Rightarrow$\hfil}\quad{\accentset{\rightharpoonup}{\boldsymbol{F}}}_% {P}=-{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{Q}.$

Thus, the two forces must be equal in magnitude and opposite in direction. So, thus far, we can conclude that the forces must be parallel as shown in fig. 5.44b.

But the forces still seem to have a net turning effect, thus still violating the concept of static equilibrium. The sum of all external torques on the body about any point are zero. So, summing moments about point $P$ , we get,

	$\displaystyle\sum_{\rm All\ external\atop torques}\kern-10.0pt{{\accentset{% \rightharpoonup}{\boldsymbol{M}}}}_{/P}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{Q/P}\times{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{Q}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$		( ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{P}$ produces no torque about $P$ )
	$\displaystyle\left\|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{Q/P}\right\|% \left(\hat{\boldsymbol{\lambda}}_{Q/P}\times{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{Q}\right)={\accentset{\rightharpoonup}{\boldsymbol{0}}}$		( $\hat{\boldsymbol{\lambda}}_{Q/P}=\frac{{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{Q/P}}{\|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{Q/P}\|}% =-\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{P/Q}}{\|{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{P/Q}\|}$ )

So ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{Q}}$ has to be parallel to the line connecting P and Q. Similarly, taking the sum of moments about point $Q$ , we get

-\hat{\boldsymbol{\lambda}}_{Q/P}\times{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{Q}={\accentset{\rightharpoonup}{\boldsymbol{0}}}

and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{P}$ also must be parallel to the line connecting P and Q. So, not only are ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{P}$ and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{Q}$ equal and opposite, they are collinear as well since they are parallel to the axis passing through their points of action (see fig. 5.44c).

Box 5.4 Three-force bodies

Here is a brief derivation of the result for three-force bodies. The derivation is not needed for problem-solving. However understanding the derivation may help build intuition.

Consider a body in static equilibrium with just three forces on it; ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{1}$ , ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{2}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{3}$ acting at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{1}$ , ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{2}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{3}$ . Taking moment balance about the axis through points at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{2}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{3}$ implies that the line of action of ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{1}$ must pass through that axis. Similarly, for equilibrium to hold, the line of action of ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{2}$ must intersect the axis through points at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{1}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{3}$ and the line of action of ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{3}$ must intersect the axis through ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{1}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{2}$ . So, the lines of action of all three forces are in the plane defined by the three points of action and the lines of action of ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{2}$ and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{3}$ must intersect. Taking moment balance about this point of intersection implies that ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{1}$ has a line of action passing through the same point. A special case is when ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{1}$ , ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{2}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{3}$ are parallel and have a common plane of action (equivalent to the concurrency point being at infinity).

The idealized massless pulley

Both real machines and mechanical models are built of various building blocks. One of the standards is a pulley. We often draw pulleys schematically something like in fig. 5.50a which shows that we believe that the tension in a string, line, cable, or rope that goes around an ideal pulley is the same on both sides, $T_{1}=T_{2}=T$ . An ideal pulley is

(i.)

Round,
(ii.)

Has frictionless bearings,
(iii.)

Has negligible inertia, and
(iv.)

Is wrapped with a line which only carries forces along its length.

We now show that these assumptions lead to the result that $T_{1}=T_{2}=T$ . First, look at a free-body diagram of the pulley with a little bit of string at both ends (fig. 5.50b).

Since we assume the bearing has no friction, the interaction between the pulley-bearing shaft and the pulley has no component tangent to the bearing.

To find the relation between tensions, we apply angular momentum balance (equation II) about point O

\left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\rm O}=\dot{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm O}\right\}\cdot\hat{% \boldsymbol{k}}.

(5.12)

Evaluating the left hand side of eqn. 5.12

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{O}\cdot\hat{% \boldsymbol{k}}$	$\displaystyle=$	$\displaystyle R_{2}T_{2}-R_{1}T_{1}+\underbrace{\mbox{bearing friction}}_{0}$
		$\displaystyle=$	$\displaystyle R(T_{2}-T_{1}),\mbox{ since $R_{1}=R_{2}=R$}.$

Because there is no friction, the bearing forces acting perpendicular to the round bearing shaft have no moment about point O (see also the short example on page 2.1). Because the pulley is round, $R_{1}=R_{2}=R$ .

When mass is negligible, dynamics reduces to statics.

Putting these assumptions and results together gives

	$\displaystyle\left\{\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\rm O}=% {\accentset{\rightharpoonup}{\boldsymbol{0}}}\right\}\cdot\hat{\boldsymbol{k}}$
	$\displaystyle\Rightarrow R(T_{2}-T_{1})=0$
	$\displaystyle\Rightarrow T_{1}=T_{2}$

Thus, the tensions on the two lines of an ideal massless pulley are equal.

Lopsided pulleys are not often encountered, so it is usually satisfactory to assume round pulleys. But, in engineering practice, the assumption of frictionless bearings is often suspect. In dynamics, you may not want to neglect pulley mass.

Filename:pfigure-blue-68-1

Figure 5.50: (a) An ideal massless pulley, (b) FBD of idealized massless pulley, detailing the frictionless bearing forces and showing forces at the cut strings, (c) final FBD after analysis.

Lack of equilibrium as a sign of dynamics

Surprisingly, statics calculations often give useful information about dynamics. If, in a given problem, you find that forces or moments cannot be balanced this is a sign that the related physical system will accelerate in the direction of imbalance (See the example ‘block on ramp’ on page 5.3). For more about nonexistence of a statics solution, see box 5.1 on page 5.1.

Linearity and superposition

For a given geometry, the equilibrium equations are linear: If you know a set of forces that is in equilibrium and you also know a second set of forces that is in equilibrium, then the sum of the two sets is also in equilibrium.

Example: A bicycle wheel

Filename:pfigure-blue-58-1

Figure 5.51: A bicycle wheel.

The free-body diagram of an ideal massless bicycle wheel with a vertical load is shown in (a) above. The same wheel driven by a chain tension but with no weight is shown in equilibrium in (b) above. The sum of these two load sets (c) is therefore in equilibrium.

That you can add solutions of linear equations is called the principle of superposition, also called the principle of superimposition

^†^†margin: Superimposition. Here’s a bad pun to help your memory. When talkative Sam comes over, you get bored. When hungry Sally comes over, you reluctantly go get a snack for her. When Sam and Sally come over together, you get bored and reluctantly go get a snack. Each one of them is imposing. When they come over together, their effects add. They are super imposing.

The principle of superposition provides a useful shortcut for some mechanics problems.

Box 5.5 Moment balance about 3 points is sufficient in 2D

This is a theoretical aside showing that moment balance can totally replace force balance.

In 2D one can solve any statically determinate problem using moment balance about any 3 non-collinear points. Force balance adds no information.

Here we show the math behind this useful trick. The derivation here is only for logical completeness, it does not help with problem solving.

Consider two points A and B. Moment balance about these two points gives

\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i/A}}\times{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}\qquad\mbox{and}\qquad\sum{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{i/B}}\times{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{i}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.

Subtracting one of these equations from the other gives:

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i/A}% }\times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}-\sum{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i/B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{\rightharpoonup}{\boldsymbol% {0}}}$
	$\displaystyle\sum\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{i/A}}-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i/B}}% \right)\times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$
	$\displaystyle\sum\left(({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{i}}-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}})-({% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i}}-{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}})\right)\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{\rightharpoonup}{\boldsymbol% {0}}}$
	$\displaystyle\sum\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{B}}-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}\right% )\times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$
	$\displaystyle\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}% }-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}}\right)\times% \sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$

Dotting both sides with a vector $\hat{\boldsymbol{k}}$ normal to the plane we get (recalling the mixed triple product identity from page 1.3 in Section 1.3 that $({\accentset{\rightharpoonup}{\boldsymbol{A}}}\times{\accentset{% \rightharpoonup}{\boldsymbol{B}}})\cdot{\accentset{\rightharpoonup}{% \boldsymbol{C}}}=({\accentset{\rightharpoonup}{\boldsymbol{C}}}\times{% \accentset{\rightharpoonup}{\boldsymbol{A}}})\cdot{\accentset{\rightharpoonup}% {\boldsymbol{B}}}$ ) we can re-arrange terms to get

	$\displaystyle\left(({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B% }}-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}})\times\sum{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}\right)\cdot\hat{\boldsymbol{k% }}={\accentset{\rightharpoonup}{\boldsymbol{0}}}\cdot\hat{\boldsymbol{k}}$
	$\displaystyle\left(\hat{\boldsymbol{k}}\times({\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{B}}-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_% {\!\mathrm{A}})\right)\cdot\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{% i}=0.$

Thus moment balance about the points A and B implies force balance in the direction $\hat{\boldsymbol{k}}\times({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{B}}-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A}})$ . This is force balance in the direction normal to the line AB (and in the plane).

Now consider a third point C. By the same reasoning moment balance about B and C implies force balance in the direction orthogonal to BC. So long as BC is not parallel to AB then we have force balance in two independent directions. So

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}.

The result only goes sour if the two directions are parallel, which occurs when two of the points A, B, and C are on a line. If A, B, and C are not on a line, moment balance about them implies force balance. So use of moment balance replaces the force-balance equilibrium equations.

Moment balance about convenient points A, B, and C can simplify the equilibrium equations if the points are picked so that, by inspection, some forces have no moment.

Box 5.6 How to hold something in place statically determinately?

Often something is held in place, loads are applied and you are to calculate the constraint forces (the forces that hold the object in place). You do this with the equilibrium equations. In 2D, force and moment balance for one free-body diagram give you 3 independent scalar equations. So you can find 3 scalar forces and moments.

Some examples of how to hold an extended object statically determinately in 2D:

•

A hinge joint and a rod or string (the line of the string cannot intersect the hinge);
•

A hinge joint and a frictionless contact (the normal line at the contact point cannot intersect the hinge);
•

Three rods (their three lines of action cannot intersect at one point);
•

Three frictionless contacts (the three normals to the contact points cannot intersect at one point, and the applied load must be such that all of the contacts are compressive); and
•

A welded joint, this carries two reaction forces and a moment.

SAMPLE 5.6

Filename:pfigure-blue-157-1

Figure 5.52:

Find force $F$ for equilibrium of the L-shaped rigid object. The dimensions of the angle are $d=0.3\,\mathrm{m}$ and $a=0.2\,\mathrm{m}$ .

Filename:summer95f-5

Figure 5.53:

Solution The free-body diagram of the angle is shown in fig. 5.53. Since we are interested in force $F$ , we can write the scalar moment-balance equation (in $\hat{\boldsymbol{k}}$ direction) about point C (and thus get rid of the other unknown force ${\accentset{\rightharpoonup}{\boldsymbol{R}}}$ ):

$\displaystyle-Fa-(100\,{\mbox{N}})d$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F$	$\displaystyle=$	$\displaystyle-(100\,{\mbox{N}})\frac{d}{a}$
	$\displaystyle=$	$\displaystyle-100\,{\mbox{N}}\cdot\frac{0.3\,\mathrm{m}}{0.2\,\mathrm{m}}$
	$\displaystyle=$	$\displaystyle-150\,{\mbox{N}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=-(150\,{\mbox{N}})\hat{% \boldsymbol{\imath}}$

SAMPLE 5.7

Filename:pfigure-blue-127-2

Figure 5.54:

Consider the angle shown in the figure with the applied forces. Can the angle be in equilibrium for some value of $F$ ? Explain.

Solution Let us assume that the angle is in equilibrium. Then the forces acting on the angle must satisfy the force and moment-balance equations. Now the force balance in the $\hat{\boldsymbol{\jmath}}$ direction gives

	$\displaystyle F+(100\,{\mbox{N}})$	$\displaystyle=$	$\displaystyle 0$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F$	$\displaystyle=$	$\displaystyle-100\,{\mbox{N}}.$

The moment balance about point A gives

$\displaystyle Fd$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F$	$\displaystyle=$	$\displaystyle 0$
Thus,
$\displaystyle-100\,{\mbox{N}}$	$\displaystyle=$	$\displaystyle 0$

which is a contradiction. Thus the angle cannot be in equilibrium with the applied forces.

It is easy to see that no matter which way $F$ acts (up or down), it cannot simultaneously balance the applied force at A and its moment. If $F=100\,{\mbox{N}}$ acts upwards at B, the angle will accelerate up because it has a net force in the $\hat{\boldsymbol{\jmath}}$ direction. If $F=100\,{\mbox{N}}$ acts downwards at B, the two equal and opposite forces at A and B produce a net moment on the angle and therefore the angle will start spinning about the $\hat{\boldsymbol{k}}$ direction. In fact, no matter what the value or direction of $F$ is, as long as it acts at point B, the angle cannot be in equilibrium. This is because the angle, as given, is a two-force body, and for equilibrium, the two applied forces must be equal, opposite and collinear.

Answer: Equilibrium not possible.

SAMPLE 5.8

Filename:pfigure-s94h13p4

Figure 5.55:

A bar as a 2-force body: A 4 $\,\mathrm{ft}$ long horizontal bar AC supports a load of 60 $\,{\rm lbf}$ at one end and is pinned to a wall at the other end. The bar is also supported by a string BC as shown in the figure. Find the forces applied by the pin and the string on the bar.

Solution Let us do this problem two ways — using equilibrium equations without much thought, and using those equations with some insight.

Filename:pfigure-blue-90-2

Figure 5.56:

The free-body diagram of the bar is shown in Fig. 5.56. The moment balance about point A, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}=0$ , gives

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}% \times T\hat{\boldsymbol{\lambda}}+{\accentset{\rightharpoonup}{\boldsymbol{r}% }}_{\!\mathrm{C/A}}\times(-P\hat{\boldsymbol{\jmath}})$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\underbrace{\ell\hat{\boldsymbol{\imath}}\times T(-\cos\theta\hat% {\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{\jmath}})}_{\ell T\sin\theta% \hat{\boldsymbol{k}}}+\underbrace{\ell\hat{\boldsymbol{\imath}}\times(-P\hat{% \boldsymbol{\jmath}})}_{-\ell P\hat{\boldsymbol{k}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle(T\ell\sin\theta-P\ell)\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$	(5.13)
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad T=\frac{P}{\sin% \theta}=\frac{60\,{\rm lbf}}{3/5}$	$\displaystyle=$	$\displaystyle 100\,{\rm lbf}.$

The force equilibrium, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=0$ , gives

\displaystyle(A_{x}-T\cos\theta)\hat{\boldsymbol{\imath}}+(A_{y}+T\sin\theta-P% )\hat{\boldsymbol{\jmath}}

\displaystyle=

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}

(5.14)

Separating out $x$ and $y$ components of this equation, we get

	$\displaystyle\quad A_{x}=T\cos\theta=(100\,{\rm lbf})\cdot\frac{4}{5}$	$\displaystyle=$	$\displaystyle 80\,{\rm lbf}$
	$\displaystyle\qquad\qquad\quad A_{y}=\,P-T\sin\theta$	$\displaystyle=$	$\displaystyle 0$

where the last equation, $A_{y}=\,P-T\sin\theta=0$ follows from eqn. (5.13) or directly from moment balance about point C. Thus, the force in the rod is ${\accentset{\rightharpoonup}{\boldsymbol{A}}}=(80\,{\rm lbf})\hat{\boldsymbol{% \imath}}$ , i.e., a purely compressive force, and the tension in the string is 100 $\,{\rm lbf}$ .

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{A}}}=(80\,{\rm lbf})\hat{\boldsymbol{% \imath}},\quad T=100\,{\rm lbf}$

Alternate Solution:

Filename:s92f1p7

Figure 5.57:

From the free-body diagram of the rod (see fig. 5.57), we realize that the rod is a two-force body, since the forces act at only two points of the body, A and C. The reaction force at A is a single force ⇀ A , and the forces at end C, the tension ⇀ T and the load ⇀ P , sum up to a single net force, say ⇀ F . So, now using the fact that the rod is a two-force body, the equilibrium equation requires that ⇀ F and ⇀ A be equal, opposite, and collinear (along the longitudinal axis of the bar). Thus,

{\accentset{\rightharpoonup}{\boldsymbol{A}}}=-{\accentset{\rightharpoonup}{% \boldsymbol{F}}}=-F\hat{\boldsymbol{\imath}}.

Now,

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{P}}}+{\accentset{% \rightharpoonup}{\boldsymbol{T}}}$
	$\displaystyle-F\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle-P\hat{\boldsymbol{\jmath}}+T\sin\theta\hat{\boldsymbol{\jmath}}-% T\cos\theta\hat{\boldsymbol{\imath}}$		(5.15)

Separating out $x$ and $y$ components of this equation, we get

	$\displaystyle-F+T\cos\theta$	$\displaystyle=$	$\displaystyle 0$		(5.16)
	$\displaystyle P-T\sin\theta$	$\displaystyle=$	$\displaystyle 0.$		(5.17)

Solving these two equations simultaneously, we get $T=P/\sin\theta=100\,{\rm lbf}$ and $F=T\cos\theta=80\,{\rm lbf}$ . The answers, of course, are the same.

SAMPLE 5.9

Filename:twodisks-ang-mom

Figure 5.58:

A bottle holder: A clever design of a bottle holder (a plank with a hole) is shown in the figure. Note that the holder is not fixed to the support; it stands freely, but only when the bottle is in. Assume that the mass of the bottle is 1 $\,\mathrm{kg}$ and that the center-of-mass of the bottle is at 3/5th of its length ( $h=35\,\mathrm{cm}$ ) from the neck support point. The bottle in its rest position is slightly tipped down ( $\alpha=15^{{}^{\circ}}$ ). Assuming the mass of the stand to be negligible and $\ell=30\,\mathrm{cm}$ , find the angle $\theta$ of the stand so that the bottle and the stand can be in equilibrium together as shown.

Solution

Filename:pfigure-blue-90-1

Figure 5.59:

Let us draw the free-body diagram of the bottle and the stand together as one system. The forces acting are shown in fig. 5.59.

Since the only forces acting on the system are $R$ and $m g$ , they must be equal, opposite and collinear. Thus the line of action of the weight, $m g$ , must pass through the center of the stand’s footprint. From the given geometry, then, we must have,

$\displaystyle\ell\cos\theta$	$\displaystyle=$	$\displaystyle\frac{3h}{5}\cos\alpha$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\theta$	$\displaystyle=$	$\displaystyle\cos^{-1}\left(\frac{3h}{5\ell}\cos\alpha\right)$
	$\displaystyle=$	$\displaystyle\cos^{-1}\left(\frac{3\cdot 35\,\mathrm{cm}}{5\cdot 30\,\mathrm{% cm}}\cos 15^{{}^{\circ}}\right)$
	$\displaystyle=$	$\displaystyle 47.5^{{}^{\circ}}.$

Answer: $\theta=47.5^{{}^{\circ}}$

Note: The latitude in design of the angle $\theta$ depends on the width of the base of the stand. The two forces acting on the system must be collinear and must pass through the base. Therefore, a wider base (perhaps at the expense of elegance) provides more freedom for the forces to move sideways, giving a range of $\theta$ and $\alpha$ for design. (see fig. 5.60.)

Filename:pfigure-blue-110-1

Figure 5.60: A bottle holder with a wider base.

SAMPLE 5.10

Filename:pfigure-blue-107-1

Figure 5.61:

Reactions at fixed ends. For the bent bar shown in the figure, find the reaction forces at the fixed end for $F=10\,{\rm kN}$ .

Solution

Filename:pfigure-s94h14p4

Figure 5.62:

The free-body diagram of the rod is shown in fig. 5.62. Note that in addition to the reaction force ${\accentset{\rightharpoonup}{\boldsymbol{R}}}$ , there is a reaction moment ${\accentset{\rightharpoonup}{\boldsymbol{M}}}=M\hat{\boldsymbol{k}}$ acting on the rod because of the fixed support.

The force-balance equation, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ , gives us

	$\displaystyle F\hat{\boldsymbol{\imath}}+{\accentset{\rightharpoonup}{% \boldsymbol{R}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{R}}}$	$\displaystyle=$	$\displaystyle-F\hat{\boldsymbol{\imath}}=-(10\,{\rm kN})\hat{\boldsymbol{% \imath}}.$

Now, we can write the moment-balance equation about point C, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{C}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , to give

	$\displaystyle M\hat{\boldsymbol{k}}-Fd\hat{\boldsymbol{k}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}M$	$\displaystyle=$	$\displaystyle Fd=20\,{\rm kN}\mathord{\cdot}\,\mathrm{m}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{R}}}=-10\,{\rm kN}\,\hat{\boldsymbol{% \imath}},\quad{\accentset{\rightharpoonup}{\boldsymbol{M}}}=20\,{\rm kN}% \mathord{\cdot}\,\mathrm{m}\,\hat{\boldsymbol{k}}$

SAMPLE 5.11

Filename:pfigure-blue-112-1

Figure 5.63:

Consider the structure (a rocker arm) shown in the figure. Assume that bar CD can only take axial load (tension or compression). If a horizontal force, $F=2\,{\rm kN}$ is applied at point A, what is the tension in rod CD?

Solution

Filename:tfigure8-alt-app2c

Figure 5.64: Free-body diagram of the structure with rod CD cut just below C.

Let $T$ be the tension in the rod (although intuitively you can see that the rod must be under compression). Then, the free-body diagram of the rocker arm ABC is as shown in fig. 5.64. We need to find $T$ . We can do so using either moment balance or force balance as shown below.

Method-1: Using moment balance

The easiest way to solve this problem is to apply moment balance, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{B}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , about point B. Taking moments about this point gets rid of the unknown reaction force $R_{B}$ and relates $T$ to $F$ directly:

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/B}}\times{% \accentset{\rightharpoonup}{\boldsymbol{F}}}+{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{C/B}}\times{\accentset{\rightharpoonup}{% \boldsymbol{T}}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}

We can evaluate the cross products vectorially or use the scalar form of the moment calculation (force times the lever arm) to give

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/B}}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle-F\ell\sin\theta\hat{\boldsymbol{k}}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/B}}% \times{\accentset{\rightharpoonup}{\boldsymbol{T}}}$	$\displaystyle=$	$\displaystyle-T\ell\cos\theta\hat{\boldsymbol{k}}.$

So, the scalar moment-balance equation in the $\hat{\boldsymbol{k}}$ direction is

	$\displaystyle-F\ell\sin\theta-T\ell\cos\theta$	$\displaystyle=$	$\displaystyle 0$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T$	$\displaystyle=$	$\displaystyle-F\tan\theta.$

Now substituting the given values, $F=2\,{\rm kN}$ , and $\theta=30^{{}^{\circ}}$ , we get

T=-(2\,{\rm kN})\cdot(\tan 30^{{}^{\circ}})=-1.15\,{\rm kN}.

Thus the rod is under compression, not tension. It is also clear from the picture that if we push at A, ABC will try to rotate clockwise about B, thus pushing down on the rod at C.

Answer: $T=-1.15\,{\rm kN}$

Method-2: Using force balance

Filename:sfig4-6-4a

Figure 5.65: The rocker arm is a three-force body and, therefore, the three forces have to be concurrent (their lines of action intersect at one point, C). Thus the support reaction

{\accentset{\rightharpoonup}{\boldsymbol{R}}}_{B}

acts in the

\hat{\boldsymbol{\lambda}}

direction. From geometry,

\hat{\boldsymbol{\lambda}}=\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}}

. A unit vector normal to

\hat{\boldsymbol{\lambda}}

is, therefore,

\hat{\boldsymbol{n}}=-\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{% \boldsymbol{\jmath}}

(you can guess it from geometry or find it from

\hat{\boldsymbol{n}}=\hat{\boldsymbol{k}}\times\hat{\boldsymbol{\lambda}}

We can also use the force-balance equation, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ to find $T$ . However, force balance will involve two unknown forces $T$ and $R$ . The force balance gives

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}+{\accentset{% \rightharpoonup}{\boldsymbol{T}}}+{\accentset{\rightharpoonup}{\boldsymbol{R}}% }_{B}=0$
	$\displaystyle\hbox to36.135pt{or\hfil}F\hat{\boldsymbol{\imath}}-T\hat{% \boldsymbol{\jmath}}+R_{B}\hat{\boldsymbol{\lambda}}=0$		(5.18)

where $\hat{\boldsymbol{\lambda}}$ is a unit vector in the direction of ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{B}$ and is not known yet. However, we know that the rocker arm is a three-force body, and therefore, all the three forces must be concurrent (they cannot be parallel here). From geometry it is clear that the lines of action of all the three forces must pass through point C. This realization immediately gives us the direction of ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{B}$ , that is, $\hat{\boldsymbol{\lambda}}=\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{% \boldsymbol{\jmath}}$ . So, now we can write out eqn. (5.18), separate out $x$ and $y$ components and solve the two scalar equations simultaneously to find both $T$ and $R_{B}$ . But we are not interested in finding $R_{B}$ . So why not get use an appropriate dot product with eqn. (5.18) to get rid of $R_{B}$ and get one scalar equation relating $T$ to $F$ . Let $\hat{\boldsymbol{n}}$ be normal to $\hat{\boldsymbol{\lambda}}$ . Thus, $\hat{\boldsymbol{n}}=-\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{% \boldsymbol{\jmath}}$ . Now, dotting with $\hat{\boldsymbol{n}}$ gives

$\displaystyle\mbox{[eqn.~{}(\ref{eq:s4.10.FB})]}\cdot\hat{\boldsymbol{n}}$	$\Rightarrow$	$\displaystyle F\underbrace{\hat{\boldsymbol{\imath}}\cdot\hat{\boldsymbol{n}}}% _{-\sin\theta}-T\underbrace{\hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol{n}}% }_{\cos\theta}+R_{B}\underbrace{\hat{\boldsymbol{\lambda}}\cdot\hat{% \boldsymbol{n}}}_{0}=0$
	$\Rightarrow$	$\displaystyle-F\sin\theta-T\cos\theta=0$
	$\Rightarrow$	$\displaystyle T=-F\tan\theta=-1.15\,{\rm kN}$

as obtained by moment balance.

Problems for 5.2 Static equilibrium of one body

Preparatory Problems

5.2.1 For problems below, assume a 2D free-body diagram has been drawn where forces ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}},{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}},\ldots,{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{5}}$ are applied at positions ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}},{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}},\ldots,{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{5}}$ relative to the origin. Use this information in the answers below.

(a)

What is the force-balance equation?
(b)

What is the moment-balance equation about the origin?
(c)

What are equilibrium conditions?
(d)

Write equilibrium conditions in as many different ways as you can.
(e)

How many independent scalar equations can one write using various force and moment-balance equations?
(f)

If force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{4}}$ is moved to a new position along its direction, which equilibrium equations are changed and which are not?
(g)

If force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{4}}$ is displaced sideways relative to its direction, which equilibrium equations are changed and which are not?

5.2.2 What is the meaning of the line of action of a force?

5.2.3 If only two forces, ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}$ , act on a body at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}$ , what do the equilibrium conditions tell you about the two forces?

5.2.4 If only three forces, ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}},{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{3}}$ , act on a body at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}},{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{3}}$ , what do the equilibrium conditions tell you about the three forces?

5.2.5 Which of the bars below cannot possibly be in equilibrium and which ones can? (Where the center of mass is indicated, assume non-zero weight acting vertically downwards. Assume dimensions as needed.)

i)

Explain in words.
ii)

Explain using equations.

Note that scalars (e.g., $F$ , $F_{1}$ , etc.) can be positive or negative.

Filename:sfig4-6-3

Figure 5.66:

5.2.6 Which of the objects below cannot possibly be in equilibrium and which ones can? (Where the center of mass is indicated, assume non-zero weight acting vertically downwards. Assume dimensions as needed.)

(a)

Explain in words.
(b)

Explain using equations.

Note that scalars (e.g., $F$ , $F_{1}$ , etc.) can be positive or negative unless mentioned otherwise.

Filename:sfig4-6-3a

Figure 5.67:

5.2.7 In the problems shown below, find $F$ for equilibrium.

Filename:pfigure4-4-rp12

Figure 5.68:

5.2.8 A straight uniform $1000\,{\mbox{N}}$ beam is $6\,\mathrm{m}$ long. It rests on a flat stack of boards with a $2\,\mathrm{m}$ overhang. How far out the overhang can an $800\,{\mbox{N}}$ person walk without the beam tipping over?

Filename:pfigure-blue-38-2

Figure 5.69:

5.2.9 The uniform bar AB is $5\,\mathrm{m}$ long and weighs $100\,{\mbox{N}}$ . It is pinned at A and supported by the horizontal cord BC attached at end B. A $50\,{\mbox{N}}$ weight hangs from end B.

(a)

Find the tension in cord C.
(b)

Find the magnitude and direction of the force exerted on the pin at A by the bar.

Filename:pfigure-blue-49-2

Figure 5.70:

5.2.10 For static equilibrium of the system and the configuration shown in the figure, find the support reaction at end A of the bar.

Filename:pfigure4-4-rp13

Figure 5.71:

5.2.11 A $400\,{\mbox{N}}$ child stands on the end of a uniform $800\,{\mbox{N}}$ diving plank which is pinned on one end and which also rests on a log (idealized as frictionless). Find the force of the log on the plank and of the pin on the plank.

Filename:f92h7p1

Figure 5.72:

5.2.12 A negligible weight $6\,\mathrm{m}$ rod is pinned at one end and leans over a frictionless wall a third of the way up from the bottom. Find the forces of the wall and the pin on the rod.

Filename:pfigure-blue-52-2

Figure 5.73:

More-Involved Problems

5.2.13 The uniform boom AB is $20\,\mathrm{ft}$ long and weighs $150\,{\rm lbf}$ . A $1500\,{\rm lbf}$ weight is suspended from a point $5\,\mathrm{ft}$ from end B. The boom is pinned at A and supported by the cable BC attached at end B.

(a)

Find the tension in the cable.
(b)

Find the force exerted on the boom by the pin at A.

Filename:pfigure4-4-rp16

Figure 5.74:

5.2.14 The $30\,{\mbox{N}}$ uniform rectangular plate is supported by a pin at A and cable BC attached at corner B. A $65\,{\mbox{N}}$ weight hangs from corner D.

(a)

Find the tension in the cable.
(b)

Find the force exerted on the plate by the pin at A.

Filename:pfigure4-4-rp14

Figure 5.75:

5.2.15 A uniform door of width $1\,\mathrm{m}$ and weight $200\,{\mbox{N}}$ is supported by two hinges a distance $2\,\mathrm{m}$ apart.

(a)

Find the horizontal component of the force by the door on the upper hinge.
(b)

Find the horizontal component of the force by the door on the lower hinge.
(c)

Can you find the vertical force of the door on the upper or lower hinge? If not, what do you know about these forces?

Filename:pfigure4-4-rp15

Figure 5.76:

5.2.16 In the mechanism shown, find the maximum force $F$ that can be applied at A normal to the link AB such that the magnitude of the force in rod CD does not exceed $10\,{\rm kN}$ .

Filename:tfigure4-spherical-rotaxis

Figure 5.77:

5.2.17 For biomechanics purposes muscles are commonly modeled as massless cables and joints (elbow, shoulder, hip, ankle, etc) as frictionless hinges connecting rigid bones. You will find that the muscle tension and joint reaction forces are large compared to the loads being carried. This is a general feature in biomechanics because muscles usually have short lever-arms relative to the bone lengths.

A human forearm weighs $14\,{\mbox{N}}$ and supports a $100\,{\mbox{N}}$ weight. Find the muscle tension and the force of the upper arm on the forearm at the elbow.

Filename:tfigure5-7

Figure 5.78:

5.2.18 See Problem 5.77. An arm weighs $7$ pounds and supports a 12 pound weight. Find the tension in the deltoid muscle and the force of the body on the arm at the shoulder joint.

Filename:tfigure5-gen-rigid-body

Figure 5.79:

5.2.19 A $240\,{\mbox{N}}$ roller is $1\,\mathrm{m}$ in diameter. It is being pulled over a $0.1\,\mathrm{m}$ curb with a horizontal rope. The roller does not slide on the curb.

(a)

What is the force required to lift the roller over the curb with the rope attached at the middle?
(b)

What is the force required if the rope is instead wrapped around the roller as shown?

Filename:tfigure5-term1-a

Figure 5.80:

5.2.20 What are the forces on the disk due to the groove? Define any variables you need.

Filename:tfigure5-term1-b

Figure 5.81:

5.2.21 A solid sphere of mass $m=5\,\mathrm{kg}$ and radius $R=250\,{\rm mm}$ rests between two frictionless inclined planes. Let $\alpha=60^{{}^{\circ}}$ . Find the magnitudes of normal reactions of the plane as functions of $\beta$ and plot normalized reactions ( $N_{1}/mg\mbox{ and }N_{2}/mg$ for $0<\beta\leq 90^{{}^{\circ}}$ ). Comment on the plot.

Filename:sfig1-2-12

Figure 5.82:

5.2.22 Assuming the spool is massless and that there is no friction at point A, find the force on the spool at point B in order to maintain equilibrium. Note, there is friction at B. Answer in terms of some or all of $r,R,g,\theta$ , and $m$ .

Filename:sfig4-6-5

Figure 5.83:

5.2.23 Find the tension in cord AB.

Filename:sfig4-6-5a

Figure 5.84:

5.3 Equilibrium with frictional contact

Filename:sfig4-6-5b

Figure 5.85: Pulling a block on a frictional ground, with related FBD. The complicated distribution of normal forces and friction forces on the block from the ramp have been replaced with the equivalent pair

N

and

F

. For particle mechanics, we don’t worry about where

N

and

F

are applied.

Contacting objects are prevented from passing through each other by pressing against each other. Generally there is also some frictional resistance to relative slip. We have neglected friction so far for simplicity and because the neglect of friction is a reasonable approximation for some lubricated contact problems. On the other extreme, in some situations we have assumed that friction so well resists slip that we assumed ‘no slip’ and that frictional contact acts like a hinge or weld. Either way, with friction negligibly small, or reliably large, we have not worried about it.

However, for some purposes friction forces are not reasonably neglected during slip. Or, when there is no slip, sometimes we have to worry about whether the frictional bond is strong enough to prevent slip.

Although slip means motion and motion sounds like dynamics (contradicting the premise of statics), there are many situations where there is enough motion for friction to be important but not so much acceleration that inertial terms ( $m{\accentset{\rightharpoonup}{\boldsymbol{a}}}$ ) are important.

How friction forces are represented on free-body diagrams was discussed in Section 2.3 which you should review before proceeding further here. We will now consider friction forces in equilibrium conditions.

For simplicity, and because of the relatively high accuracy to complexity ratio, we consider only Coulomb friction with a single coefficient of friction $\mu$ ^†^†margin: If you have studied friction before, our approximation is that $\mu=\mu_{static}=\mu_{dynamic}$ adequately captures the complex and hard to quantify reality of frictional forces .

Example: Drag a block with friction.

Consider the block with friction (fig. 5.85). You want to pull it slowly to the right with rod AB. Say $m=100\,\mathrm{kg}$ , $g=10\,\mathrm{m}/\mathrm{s}^{2}$ , and $\mu=0.3$ .
Force balance, using the forces on the free-body diagram gives:

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}-mg% \hat{\boldsymbol{\jmath}}+T_{\rm AB}\hat{\boldsymbol{\imath}}+N\hat{% \boldsymbol{\jmath}}-F\hat{\boldsymbol{\imath}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}

(5.19)

These, with the friction relation $F=\mu N$ , are 3 scalar equations in $T_{\rm AB}$ , $F$ and $N$ with solution $N=mg=1000\,{\mbox{N}},F=\mu mg=300\,{\mbox{N}},$ and $T_{\rm AB}=\mu mg=300\,{\mbox{N}}$ .

Example: Drag a block on a ramp with friction.

Filename:sfig4-6-5c

Figure 5.86: Pulling a block up a frictional slope with related FBD. As in the previous example, the pair

N

and

F

represent the net normal and frictional forces.

Consider the block with friction on a slope (fig. 5.86). You want to hold it with rod AB. Maybe you want to (i) slide it up slowly, or (ii) down slowly or (iii) hold it still. Say $m=100\,\mathrm{kg}$ , $g=10\,\mathrm{m}/\mathrm{s}^{2}$ , $\theta=45^{{}^{\circ}}$ , and $\mu=0.3$ .
Force balance, using the forces on the free-body diagram, gives:

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}-mg% \hat{\boldsymbol{\jmath}}+T_{\rm AB}\hat{\boldsymbol{e}}_{1}+N\hat{\boldsymbol% {e}}_{2}-F\hat{\boldsymbol{e}}_{1}={\accentset{\rightharpoonup}{\boldsymbol{0}}}

(5.20)

These, with the friction relation, are 3 scalar equations in $T_{\rm AB}$ , $F$ and $N$ .

Summing forces in the rope direction and normal to the plane we get:

	$\displaystyle\{({\rm Eqn.~{}}\ref{eq:blockslope})\}\cdot\hat{\boldsymbol{e}}_{1}$	$\Rightarrow$	$\displaystyle-mg\sin\theta+T_{\rm AB}-F=0$
	$\displaystyle\{({\rm Eqn.~{}}\ref{eq:blockslope})\}\cdot\hat{\boldsymbol{e}}_{2}$	$\Rightarrow$	$\displaystyle mg\cos\theta+N=0$		(5.21)

or, for the quantities given $N=(100\,\mathrm{kg})(10\,\mathrm{m}/\mathrm{s}^{2})(\cos 45^{{}^{\circ}})=707% \,{\mbox{N}}$ .

We assume that $F$ and $N$ are related by friction described with the standard Coulomb’s friction model

^†^†margin: Caution: A common beginner’s mistake is to assume the equation

F=\mu N

applies when there is friction. Rather, if the friction is preventing slip,

F

could be anything so long as

|F|\leq\mu N

. And if the slip is opposite in direction from that implicitly assumed in the free-body diagram, then

F=-\mu N

(see case (ii) in the example above).

i)

$F=\phantom{-}\mu N$ if the block is sliding up;
ii)

$F=-\mu N$ if the block is sliding down; or
iii)

$-\mu N\leq F\leq\mu N$ if the block is not sliding.

Solving eqn. (5.20) with the friction relations gives

^†^†margin: We could be tricky and get a single equation for the scalar

T_{\rm AB}

by dotting both sides of eqn. (5.20) with a vector orthogonal to the resultant of

N\hat{\boldsymbol{e}}_{2}-F\hat{\boldsymbol{e}}_{1}

. For the case of uphill sliding such a vector would be

\hat{\boldsymbol{e}}_{1}+\mu\hat{\boldsymbol{e}}_{2}

i)

$T_{\rm AB}=mg\left(\mu\cos\theta+\sin\theta\right)$ if the block is sliding up;
ii)

$T_{\rm AB}=mg\left(-\mu\cos\theta+\sin\theta\right)$ if the block is sliding down;
Note that if $\tan\theta<\mu$ then $T_{\rm AB}<0$ and it then takes a push to slide down; or
iii)

$mg\left(-\mu\cos\theta+\sin\theta\right)\leq T_{\rm AB}\leq mg\left(\mu\cos% \theta+\sin\theta\right)$
if the block is not sliding.
If $\tan\theta<\mu$ then $T_{\rm AB}=0$ is amongst the solutions for, so no sliding and the block can sit still on the slope with no pull on the rope.

Note that the tension $T_{\rm AB}$ scales with $m g$ . So doubling $m$ or $g$ doubles all the forces in all of these answers, as you might guess from dimensional considerations.

The mathematically-abstract-sounding issues of existence and uniqueness often show up in friction problems. For example, sometimes there is no statics solution (non-existence).

Example: Block on ramp.

Filename:sfig4-6-5d

Figure 5.87: Block on steep ramp and related FBD.

A statics problem without a solution. A block with coefficient of friction $\mu=0.5$ is in static equilibrium sliding steadily down a $45^{{}^{\circ}}$ ramp (fig. 5.87). Not! If there is constant velocity motion, then statics would apply. But the forces in the free-body diagram cannot add to zero (because the resultant of the friction and normal force is tipped up and to the left and thus cannot be parallel to the vertical gravity force). The assumptions are not consistent with statics (actually this is a dynamics problem, the block accelerates down the ramp). If you saw a block just sitting there on a ramp, then you can be sure that the slope and friction coefficient are not those given above.

Friction problems might be studied with a particle model, as above, or also with moment balance.

Filename:sfig4-5-6

Figure 5.88: Dragging a block, taking account (in a simple way) the distribution of contact forces from the ground. Assume slip to the right is occurring.

Example: Dragged block as an extended body.

This is a repeat of the first example on page 5.85. One might wonder if the dragging causes an uneven distribution of force up on the block. Does the block dragging back, for example, cause a bigger pressure on the back? As a simple model, assume all the ground force is at the front and back edge of the block. Force balance gives basically the same information as for the particle model, namely that:

N_{C}+N_{D}=W\qquad\mbox{and}\qquad\underbrace{F_{C}}_{\mu N_{C}}+\underbrace{% F_{D}}_{\mu N_{D}}=T_{AB}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T_{AB}=\mu W.

One can find more with moment balance about any point you like, say C, with force balance giving

\sum M_{/C}=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}N_{D}=\frac{W}{2}+\frac{% \mu hW}{2\ell}\qquad\mbox{and}\qquad N_{C}=\frac{W}{2}-\frac{\mu hW}{2\ell}

So there is more pressure on the front than the back. This difference goes away if either the friction or the height of the string attachment vanishes.

Conditional contact, consistency, and contradictions

There is a natural hope that a subject will reduce to the solution of some well defined equations. For better and worse, things are not always this simple. For better, because it means that the recipes are still not so well defined that computers can easily steal the subject of mechanics from people. For worse, because it means you have to think hard to do some mechanics problems.

One source of these difficulties is the conditional nature of the equations that govern contact. For example:

•

The ground pushes up on something to prevent interpenetration if the pushing is positive, otherwise the ground does not push up.
•

The force of friction opposes motion and has magnitude $\mu N$ if there is slip, otherwise the force of friction is something less than $\mu N$ in magnitude.
•

The distance between two points is kept from increasing by the tension in the string between them if the tension is positive, otherwise the tension is zero.

These conditions are, implicitly or explicitly, in the equations that govern these interactions. One does not always know which of the alternative contact conditions, if either, apply when one starts a problem. Sometimes multiple possibilities need to be checked

^†^†margin: All of the combinatorics in this recipe follows from the inability of our mathematics to deal easily with relations between variables with the step shape of Coulomb friction (fig. 2.106 on page 2.106).

On a FBD at every point of frictional contact

•
If the direction of slip or impending slip is known, either
- –
  
  Draw a normal force $N$ and a friction force $F=\mu N$ opposing the relative slip, or
- –
  
  Draw a single force $R$ at an angle $\phi$ from the normal of the contact in the direction which resists slip (with $\tan\phi=\mu$ )
•
If there is no slip, either
- –
  
  Draw a normal force $N$ and tangential force $F$ or
- –
  
  Draw a single force vector ${\accentset{\rightharpoonup}{\boldsymbol{R}}}$ with unknown components
•
If you don’t know whether or not there is slip, first
- –
  
  Guess that there is no slip, then
- –
  Solve the equilibrium equations, then
  - *
    
    If $F\leq\mu N$ : you guessed right and have found a solution to both the equilibrium and friction equations.
  - *
    
    If $F>\mu N$ : you guessed wrong and have to guess that there is slip in one direction (guess which), then
    
    ·
    
    see if you can solve the equilibrium equations, if not then
    
    ·
    
    assume slip in the opposite direction and try to solve the equilibrium equations, if you can’t, then
    
    ·
    
    the problem has no solution

Example: Robot hand

Filename:tfigure-erdmannpalms

Roboticist Michael Erdmann has designed a palm-like robot ‘hand’ that manipulates objects without squeezing them. The flat robot palms just move around and the object consequently slides. Determining whether the object slides on one or the other, or possibly on both hands in a given movement is a matter of case study. The computer checks to see if the equilibrium equations can be solved with the assumption of sticking or slipping at one or the other contact.

Once you find a solution to a problem with friction there remains the possibility of multiple solutions, in this case for different reasons than the usual static indeterminacy. The following problem shows a case where a statics problem has multiple solutions due to frictional effects.

Box 5.7 Undriven wheels and two-force bodies

One often hears whimsical reverence for the “invention of the wheel.” Now, using elementary mechanics, we can gain some appreciation for this revolutionary way of sliding things.

Without a wheel, the force it takes to drag something is about $\mu W$ . Since $\mu$ ranges between about .1 for teflon, to about .6 for stone on ground, to about 1 for rubber on pavement, you need to pull with a force that is on the order of a half of the full weight of the thing you are dragging.

You have seen how rolling on round logs cleverly take advantage of the properties of two-force bodies (page 5.2). But that good idea has the major deficiency of requiring that logs be repeatedly picked up from behind and placed in front again. So we invent the wheel.

The simplest wheel design uses a dry “journal” bearing consisting of a non-rotating shaft protruding through a near close fitting hole in the wheel. Here is shown part of a cart rolling to the right with a wheel rotating steadily clockwise.

Filename:tfigure-primitivewheel

To figure out the forces involved we draw a free-body diagram of the wheel. We neglect the wheel’s weight because it is generally much smaller than the forces it mediates. To make things more clear, the picture shows an unnaturally-large bearing hole $r$ .

Filename:tfigure-primitivewheelFBD

The force of the axle on the wheel has a normal component $N$ and a frictional component $F$ . The force of the ground on the wheel has a part holding the cart up $F_{y}$ and a part along the ground $F_{x}$ which will surely turn out to be negative for a cart moving to the right. If we take the wheel dimensions to be known, and also the vertical part of the ground reaction force $F_{y}$ (the weight born), we have as unknowns $N,F,\theta$ and $F_{x}$ . To find these we could use the friction equation for the sliding bearing contact

F=\mu N;

force balance

F_{x}\hat{\boldsymbol{\imath}}+F_{y}\hat{\boldsymbol{\jmath}}+N(-\sin\theta% \hat{\boldsymbol{\imath}}-\cos\theta\hat{\boldsymbol{\jmath}})+F(\cos\theta% \hat{\boldsymbol{\imath}}-\sin\theta\hat{\boldsymbol{\jmath}})={\accentset{% \rightharpoonup}{\boldsymbol{0}}},

which could be reduced to 2 scalar equations by taking components or dot products; and moment balance about C, which we calculate with forces and perpendicular distances as

Fr+F_{x}R=0.

Of key interest is finding the force resisting motion $F_{x}$ . With some mathematical manipulation we could solve the 4 scalar equations above for any of $F_{x},N,F,$ and $\theta$ in terms of $r,R,F_{y}$ , and $\mu$ . We follow a more intuitive approach instead.

As modeled, the wheel is a two-force body so the free-body diagram shows equal and opposite collinear forces at the two contact points.

Filename:tfigure-primitivewheelFBD2

The friction angle $\phi$ describes the friction between the axle and wheel (with $\tan\phi=\mu$ ). The angle $\alpha$ describes the effective friction of the wheel. This is not the friction angle for sliding between the wheel and ground which is assumed to be larger (if not, the wheel would skid and not roll), probably much larger. The specific resistance or the coefficient of rolling resistance or the specific cost of transport is $\mu_{\rm eff}=\tan\alpha$ . (If there was no wheel, and the cart or whatever was just dragged, the specific resistance would be the friction between the cart and ground $\mu_{\rm eff}=\mu$ .)

Although we can solve for $\alpha$ in terms of $\mu$ or $\phi$ let’s first consider two extreme cases: one is a frictionless bearing and the other is a bearing with infinite friction coefficient $\mu\to\infty$ and $\phi\to 90^{{}^{\circ}}$ .

Filename:tfigure-primitivelimits

In the case that the wheel bearing has no friction we satisfyingly see clearly that there is no ground resistance to motion. The case of infinite friction is perhaps surprising. Even with infinite friction we have that

\sin\alpha=\frac{r}{R}.

Thus if the axle has a diameter of $10\,\mathrm{cm}$ and the wheel of $1\,\mathrm{m}$ then $\sin\alpha$ is less than .1 no matter how bad the bearing material. For such small values we can make the approximation $\mu_{\rm eff}=\tan\alpha\approx\sin\alpha$ so that the effective coefficient of friction is .1 or less no matter what the bearing friction.

The genius of the wheel design is that it makes the effective friction less than $r/R$ no matter how bad the bearing friction.

Going back to the two-force body free-body diagram we can see that

	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\overbrace{r\sin\phi}^{d}$	$\displaystyle=$	$\displaystyle\overbrace{R\sin\alpha}^{d}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\sin\alpha$	$\displaystyle=$	$\displaystyle\frac{r}{R}\sin\phi.\qquad(*)$

From this formula we can extract the limiting cases discussed previously ( $\phi=0$ and $\phi=90^{{}^{\circ}}$ ). We can also plug in the small angle approximations ( $\sin\alpha\approx\tan\alpha$ and $\sin\phi\approx\tan\phi$ ) if the friction coefficient is low to get

\mu_{\rm eff}\approx\mu\frac{r}{R}.

The effective friction is the bearing friction attenuated by the radius ratio. Or, we can use the trig identity $\sin={\sqrt{1+\tan^{-2}}}^{-1}$ to solve the exact equation (*) for

\mu_{\rm eff}=\mu\frac{r}{R}\left(\frac{1}{\sqrt{1+\mu^{2}(1-r^{2}/R^{2})}}% \right),

where the term in parenthesis is always less than one and close to one if the sliding coefficient in the bearing is low.

Finally we combine the genius of the wheel with the genius of the rolling log and invent a wheel with rolling logs inside, a ball bearing wheel.

Filename:tfigure-ballbearing

Each ball is a two-force body and thus only transmits radial loads. It’s as if there were no friction on the bearing and we get a specific resistance of zero, $\mu_{\rm eff}=0$ . Of course real ball bearings are not perfectly smooth or perfectly rigid, so it’s good to keep $r/R$ small as a back up plan even with ball bearings.

By this means some wheels have effective friction coefficients as low as about .003. The force it takes to drag something on wheels can be as little as one three hundredth the weight.

Example: Rod pushed in a channel.

Filename:tfigure-twosolutions

A light rod is just long enough to make a $60^{{}^{\circ}}$ angle with the walls of a channel. One channel wall is frictionless and the other has $\mu=1$ . What is the force needed to keep it in equilibrium in the position shown? If we assume it is sliding we get the first free-body diagram. The forces shown can only be in equilibrium if all forces are zero. So a solution is that the rod slides in equilibrium with no force. If we assume that the rod is not sliding, the friction force on the lower wall can be at any angle between $\pm 45^{{}^{\circ}}$ . Thus we have equilibrium with the second FBD for arbitrary positive $F$ . This is a second set of solutions. A rod like this is said to be self locking in that it can hold an arbitrary large force $F$ without slipping. That we have found freely slipping solutions, with no force, and jammed solutions, with arbitrary force, corresponds physically to one being able to easily slide a rod like this down a slot and then also at a different instant, have the rod totally jamb. Some rock-climbing equipment depends on such self-locking and easy release.

Statically indeterminate problems

When there are two or more points of frictional contact and there is no slip nor impending slip, then static indeterminacy is likely.

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Figure 5.89: A chair with friction at the feet.

Example: Chair with friction

If we assume Coulomb friction at the chair feet, we know that

|F_{A}|\leq\mu N_{A}\qquad\mbox{and}\qquad|F_{B}|\leq\mu N_{B}

The equilibrium equations tell us (assuming for simplicity that $W$ acts in the middle of the chair):

F_{A}+F_{B}=0,N_{A}=N_{B}=W/2.

Putting these equations together we find that

-W/2\leq F_{A}\leq W/2\qquad\mbox{and}\qquad F_{B}=-F_{A}

and no more. That is, all we can tell is that both are within the friction limits and that the horizontal forces cancel each other.

If a free-body diagram shows two forces with a common line of action, like the friction forces $F_{A}$ and $F_{B}$ on the chair above, the laws of statics might only find their sum, but otherwise can’t untangle them.

Only if there is independent information, as would be the case if we knew the chair was sliding to the right (which it clearly isn’t in this static example), could we find the friction forces.

SAMPLE 5.12

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Figure 5.90:

A block on a ramp sliding down or up. Consider a block of mass $m=10\,\mathrm{kg}$ pushed up by the force $F$ on the ramp as shown in the figure. The coefficient of friction between the ramp and the block is $\mu=0.7$ .

1.

Let $\theta=60^{{}^{\circ}}$ and $\alpha=0^{{}^{\circ}}$ . Assuming that the block slides steadily downhill, find the tension in the string.
2.

Let $\theta=30^{{}^{\circ}}$ and $\alpha=30^{{}^{\circ}}$ . If the applied force $F=20\,{\mbox{N}}$ , find the force of friction on the block.
3.

Let $\theta=60^{{}^{\circ}}$ and $\alpha=30^{{}^{\circ}}$ . If the applied force $F=10\,{\mbox{N}}$ , find the force of friction on the block.
4.

For $\theta=60^{{}^{\circ}}$ and $\alpha=30^{{}^{\circ}}$ , what will be the required tension in the string to make the block just about slide up the slope? Express your answer in terms of the weight of the block.

Solution

Filename:sfig4-6-8a

Figure 5.91: Free-body diagram of the block. The friction force

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{f}

may be known or unknown depending on whether the block is sliding or not. Under downhill sliding,

F_{f}=\mu N

, whereas for no sliding

|F_{f}|\leq\mu N

The free-body diagram of the block is shown in fig. 5.91. We have assumed that the friction force acts upwards along the inclined plane. The direction of the friction force can be up or down depending on the direction of sliding. We will let the equilibrium equation tell us which way the friction force acts in a particular case. In fig. 5.91, we also use rotated unit vectors $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ , parallel and perpendicular to the inclined plane, respectively. This is just to make calculations easier. We can use these basis vectors in any orientation to suit our convenience.

The force balance equation for the static equilibrium of the block gives

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{F}}}+{{\accentset{\rightharpoonup}{\boldsymbol{N}% }}}+{{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{f}}+{{\accentset{% \rightharpoonup}{\boldsymbol{W}}}}={\accentset{\rightharpoonup}{\boldsymbol{0}% }}.$		(5.22)
		$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F(-\cos\alpha\hat{% \boldsymbol{\imath}}+\sin\alpha\hat{\boldsymbol{\jmath}})+N\hat{\boldsymbol{% \jmath}}-F_{f}\hat{\boldsymbol{\imath}}+mg(\sin\theta\hat{\boldsymbol{\imath}}% -\cos\theta\hat{\boldsymbol{\jmath}})={\accentset{\rightharpoonup}{\boldsymbol% {0}}}$		(5.23)

Now depending on what is given and what is unknown, we can manipulate this vector equation to find what we want.

1.

Block sliding down:

If the block slides down steadily or very slowly, we can use the static equilibrium equation written above with ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{f}=-\mu N\hat{\boldsymbol{% \imath}}$ (that is, the friction force is known. This is the case of sliding friction and the friction force is maximum possible). Substituting this value of ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{f}$ and separating out the $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ components of eqn. (5.23), we get

$\displaystyle-F\cos\alpha-\mu N+mg\sin\theta$ $\displaystyle=$ $\displaystyle 0$ (5.24)

$\displaystyle F\sin\alpha+N-mg\cos\theta$ $\displaystyle=$ $\displaystyle 0.$ (5.25)

Adding $\mu$ times eqn. (5.25) to eqn. (5.24) in order to get rid of $N$ , and rearranging terms, we get

$\displaystyle F(\cos\alpha-\mu\sin\alpha)$ $\displaystyle=$ $\displaystyle mg(\sin\theta-\mu\cos\theta)$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F$ $\displaystyle=$ $\displaystyle\frac{\sin\theta-\mu\cos\theta}{\cos\alpha-\mu\sin\alpha}\,mg.$ (5.26)

Substituting $\alpha=0^{{}^{\circ}},\,\theta=60^{{}^{\circ}}$ , and $\mu=0.7$ in eqn. (5.26), we get

$\displaystyle F$ $\displaystyle=$ $\displaystyle(\sin 60^{{}^{\circ}}-0.7\cdot\cos 60^{{}^{\circ}})\,mg=0.52\,mg.% =51\,{\mbox{N}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=-(51\,{\mbox{N}})\hat{% \boldsymbol{\imath}}$
2.

Block sliding or not sliding – not known:

Now, we are given that $F=20\,{\mbox{N}}$ , $\alpha=30^{{}^{\circ}}$ , and $\theta=30^{{}^{\circ}}$ . We do not know if the block is sliding or not. So, let us assume static equilibrium in the given configuration and solve for the friction force $F_{f}$ . Then, we will check if it satisfies friction law for static equilibrium ( $|F_{f}|\leq\mu N$ ).

Substituting ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{f}}=-F_{f}\hat{% \boldsymbol{\imath}}$ in eqn. (5.23) and separating out the $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ components of the equation, we get

Filename:sfig4-6-8b

Figure 5.92: Free-body diagram of the block. The friction force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{f}$ is not known because we do not know if the block is sliding or not.

$\displaystyle-F\cos\alpha-F_{f}+mg\sin\theta$ $\displaystyle=$ $\displaystyle 0$

$\displaystyle F\sin\alpha+N-mg\cos\theta$ $\displaystyle=$ $\displaystyle 0$

which are easily solved for $F$ and $N$ to give

$\displaystyle F_{f}$ $\displaystyle=$ $\displaystyle mg\sin\theta-F\cos\alpha$

$\displaystyle N$ $\displaystyle=$ $\displaystyle mg\cos\theta-F\sin\alpha.$

Substituting the given values of $F,\,\theta$ , and $\alpha$ , we get

$\displaystyle F_{f}$ $\displaystyle=$ $\displaystyle 10\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}\cdot\sin 30% ^{{}^{\circ}}-20\,{\mbox{N}}\cdot\cos 30^{{}^{\circ}}=31.73\,{\mbox{N}}$

$\displaystyle N$ $\displaystyle=$ $\displaystyle 10\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}\cdot\cos 30% ^{{}^{\circ}}-20\,{\mbox{N}}\cdot\sin 30^{{}^{\circ}}=74.96\,{\mbox{N}}.$

Now, the maximum possible value of friction force is $\mu N=0.7\cdot 74.96\,{\mbox{N}}=52.47\,{\mbox{N}}$ . Thus, $|F_{f}|<\mu N$ , and therefore, our assumption of static equilibrium is valid. This equilibrium requires that $F_{f}=31.73\,{\mbox{N}}$ .

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{f}}=-(31.73\,{\mbox% {N}})\hat{\boldsymbol{\imath}}$
3.

Block sliding or not sliding – not known, again:

Filename:sfig5-5-2

Figure 5.93: Free-body diagram of the block. The friction force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{f}$ is not known again because we do not know if the block is sliding or not.

In this case, $F=10\,{\mbox{N}}$ , $\alpha=30^{{}^{\circ}}$ , and $\theta=60^{{}^{\circ}}$ . Again, assuming static equilibrium, we do exactly the same calculations as above (in fact, use the same expressions) and substituting the given values, we get

$\displaystyle F_{f}$ $\displaystyle=$ $\displaystyle 10\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}\cdot\sin 60% ^{{}^{\circ}}-10\,{\mbox{N}}\cdot\cos 30^{{}^{\circ}}=76.3\,{\mbox{N}}$

$\displaystyle N$ $\displaystyle=$ $\displaystyle 10\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}\cdot\cos 60% ^{{}^{\circ}}-10\,{\mbox{N}}\cdot\sin 30^{{}^{\circ}}=44\,{\mbox{N}}.$

Here, $|F_{f}|>\mu N$ . Clearly, $F_{f}$ is not less than or equal to $\mu N$ , and therefore, our assumption of static equilibrium is not valid. In fact, the given parameters of the problem will make the block accelerate downhill — a problem of dynamics. However, the friction force remains constant, at its maximum ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{f}=\mu N=30.8\,{\mbox{N}}$ once the sliding starts, accelerating or not.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{f}=-30.8\,{\mbox{N}}\hat{% \boldsymbol{\imath}}$
4.

Block just about to slide upwards: If the block is about to slide upwards, then the friction force must act downwards as shown in fig. 5.94. We also know the magnitude, $F_{f}=\mu N$ because it is the case of impending slip. Now the force-balance equations in the $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ directions are:

Filename:sfig5-5-2a

Figure 5.94: Free-body diagram of the block about to slip upwards. The friction force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{f}$ is completely known because of impending slip, ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{f}=\mu N\hat{\boldsymbol{\imath}}$ .

$\displaystyle-F\cos\alpha+\mu N+mg\sin\theta$ $\displaystyle=$ $\displaystyle 0$

$\displaystyle F\sin\alpha+N-mg\cos\theta$ $\displaystyle=$ $\displaystyle 0$

Eliminating $N$ from the two equations, we get $F$ in terms of $m g$ and substituting $\alpha=30^{{}^{\circ}}$ , and $\theta=60^{{}^{\circ}}$ , we get the desired value:

$\displaystyle F$ $\displaystyle=$ $\displaystyle\frac{\sin\theta+\mu\cos\theta}{\cos\alpha+\mu\sin\alpha}\,mg$

$\displaystyle=$ $\displaystyle\frac{1.216}{1.216}\,mg=mg.$

Answer: $F=mg$

Does the answer make sense? Yes, it does. For the given $\theta$ and $\alpha$ , the string tension is vertical. If it balances the weight of the block, the normal force goes to zero and so does the friction force. The block is then ready to slide up if the tension increases by any tiny amount.

SAMPLE 5.13

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Figure 5.95:

How much friction does the cylinder need? A cylinder of mass $m$ sits between an incline and a vertical wall as shown in the figure. There is no friction between the wall and the cylinder but there is friction between the incline and the cylinder. Take the coefficient of friction to be $\mu$ and the angle of incline with the horizontal to be $\theta$ . Find the force of friction on the cylinder from the incline.

Solution

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Figure 5.96:

The free-body diagram of the cylinder is shown in fig. 5.96. We need to find the force of friction $F_{s}$ .

Note that the normal reaction of the vertical wall, $N$ , the force of gravity, $m g$ , and the normal reaction of the incline, $R$ , all pass through the center C of the cylinder. So, if we do moment balance about point C, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{C}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , none of these forces will appear in the equation since their moment about C is zero. Therefore, to find $F_{s}$ , we should use the moment-balance equation about point C. Noting that $F_{s}$ acts along the inclined plane, its normal distance (lever arm) from point C is simply $r$ , the radius of the cylinder, we have,

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{C}}={% \accentset{\rightharpoonup}{\boldsymbol{0}}}$	$\Rightarrow$	$\displaystyle{rF_{s}}(-\hat{\boldsymbol{k}})={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$
		$\Rightarrow$	$\displaystyle F_{s}=0.$

Thus the force of friction on the cylinder is zero! Note that $F_{s}$ is independent of $\theta$ , the angle of incline. Thus, irrespective of what the angle of incline is, in the static equilibrium condition, there is no force of friction on the cylinder.

Answer: $F_{s}=0$

Filename:pfigure-blue-99-1

Figure 5.97:

Note: The cylinder here is a three-force body since there are three forces acting on it — two contact forces (at A and B) and one gravity force. Therefore, for equilibrium, all the three forces must intersect at a single point. Now, lines of action of the gravity force and the normal reaction at B intersect at the center C of the cylinder. Therefore, the line of action of the contact force at A also must pass through the center. This is clearly not possible if the contact force is not normal to the incline (see the candidate contact forces marked by the dashed gray arrows in fig. 5.97. If there is any non-zero friction force at A, the contact force (the resultant of the normal reaction and the friction force) at A will be tipped away from the normal, thus making its line of action miss the center of the cylinder and, therefore, violate equilibrium condition.

SAMPLE 5.14

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Figure 5.98:

Will the ladder slip? A ladder of length $\ell=4\,\mathrm{m}$ rests against a wall at $\theta=60^{\circ}$ . Assume that there is no friction between the ladder and the vertical wall but there is friction between the ground and the ladder with $\mu=0.5$ . A person weighing $700\,{\mbox{N}}$ starts to climb up the ladder.

1.

Can the person make it to the top safely (without the ladder slipping)? If not, then find the distance $d$ along the ladder that the person can climb safely. Ignore the weight of the ladder in comparison to the weight of the person.
2.

Does the “no slip” distance $d$ depend on $\theta$ ? If yes, then find the angle $\theta$ which makes it safe for the person to reach the top.

Solution

Filename:pfigure-blue-89-1

Figure 5.99: The free-body diagram of the ladder indicates that it is a three-force body. Since the direction of the forces acting at points A and C are known (the normal, horizontal reaction at A and the vertical gravity force at C), it is easy to find the direction of the net ground reaction at B — it must pass through point D. The ground reaction

F

at B (shown by light grey arrow) can be decomposed into a normal reaction

N

and a horizontal reaction (the force of friction)

F_{s}

at B.

1.

The free-body diagram of the ladder is shown in fig. 5.99. There is only a normal reaction ⇀ R = R ˆ ı at A since there is no friction between the wall and the ladder. The force of friction at B is ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{s}}=-F_{s}\hat{% \boldsymbol{\imath}}$ where $F_{s}\leq\mu N$ . To determine how far the person can climb the ladder without the ladder slipping, we take the critical case of impending slip. In this case, $F_{s}=\mu N$ . Let the person be at point C, a distance $d$ along the ladder from point B. We need to find $d$ and check if $d<\ell$ (cannot make it to point A).

From moment balance about point B, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{B}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , we find

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/B}}% \times{\accentset{\rightharpoonup}{\boldsymbol{R}}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{W}}}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$

$\displaystyle-R\ell\sin\theta\hat{\boldsymbol{k}}+Wd\cos\theta\hat{\boldsymbol% {k}}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}R$ $\displaystyle=$ $\displaystyle W\frac{d\cos\theta}{\ell\sin\theta}.$ (5.27)

From force equilibrium, we get

$(R-\mu N)\hat{\boldsymbol{\imath}}+(N-W)\hat{\boldsymbol{\jmath}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}.$ (5.28)

Dotting eqn. (5.28) with ˆ ȷ and ˆ ı , respectively, we get

$\displaystyle N$ $\displaystyle=$ $\displaystyle W$

$\displaystyle R$ $\displaystyle=$ $\displaystyle\mu N=\mu W.$

Substituting this value of $R$ in eqn. (5.27) we get

$\displaystyle\mu W$ $\displaystyle=$ $\displaystyle W\frac{d\cos\theta}{\ell\sin\theta}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}d$ $\displaystyle=$ $\displaystyle\mu\ell\tan\theta$

$\displaystyle=$ $\displaystyle 0.5\cdot(4\,\mathrm{m})\cdot\tan 60^{\circ}=3.46\,\mathrm{m}.$

Thus, the ladder is about to slip when the person is at $d=3.46\,\mathrm{m}$ . But, $d<\ell$ , therefore, the person cannot make it to the top of the ladder safely.

Answer: $d=3.46\,\mathrm{m}$
2.

The “no slip” distance $d$ depends on the angle $\theta$ via the relationship in eqn. (1). The person can climb the ladder safely up to the top if

$\tan\theta=\frac{1}{\mu}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\theta=\tan^{% -1}(\mu^{-1})=63.43^{\circ}.$

Thus, any reasonable angle $\theta\geq 64^{\circ}$ will allow the person to climb up to the top safely.

Answer: $\theta\geq 64^{\circ}$

SAMPLE 5.15

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Figure 5.100:

Will it tip or will it slide? Whether or not a box of a given width and height will slide or tip over on an inclined plane depends on the slope of the plane and the coefficient of friction. For a given slope $\theta$ , find the relationship between the coefficient of friction $\mu$ and the aspect ratio of the box, $\gamma=b/h$ for impending tipping.

Solution Let us imagine that we put the box on a flat surface and then slowly start tilting the surface up with respect to the horizontal. At some slope, the box will either tip over or slide. Just before the instant the box starts to tip over or slide, it is in static equilibrium. The magnitude of the friction force at the contact points is $|F|\leq\mu N$ where $N$ is the magnitude of the normal force at the contact, and the equality holds only in the case of impending slip. That is, if the box is about to slip, then $F=\mu N$ at each contact point.

Filename:pfigure4-3Dpend

Figure 5.101:

The free-body diagram of the box is shown in fig. 5.101. Let us first write the equations of static equilibrium assuming there is no impending slip.

The force balance in the $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ directions (see fig. 5.104) gives

	$\displaystyle F_{A}+F_{B}$	$\displaystyle=$	$\displaystyle mg\sin\theta$		(5.30)
	$\displaystyle N_{A}+N_{B}$	$\displaystyle=$	$\displaystyle mg\cos\theta.$		(5.31)

The moment equilibrium about the center-of-mass, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{C}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , in the $\hat{\boldsymbol{k}}$ direction gives

N_{B}\cdot\frac{b}{2}-N_{A}\frac{b}{2}-(F_{A}+F_{B})\frac{h}{2}=0.

(5.32)

Substituting $F_{A}+F_{B}=mg\sin\theta$ from eqn. (5.30) in eqn. (5.32), and solving eqns. (5.31) and (5.32) simultaneously, we get

N_{A}=\frac{1}{2}mg\left(\cos\theta-\frac{h}{b}\sin\theta\right),\mbox{ and}% \quad N_{B}=\frac{1}{2}mg\left(\cos\theta+\frac{h}{b}\sin\theta\right).

If the box were to tip over (about point B), the support forces at A will go to zero (because of loss of contact). Thus, for impending tipping,

N_{A}=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\cos\theta-\frac{h}{b}\sin% \theta=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\tan\theta=\frac{b}{h}=\gamma.

Thus, the condition for impending tipping is

\tan\theta=\gamma.

(5.33)

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Figure 5.102: The impending tipping condition

\tan\theta=\gamma\equiv b/h

, provided

\tan\theta\leq\mu

This condition, however, does not guarantee that the box will tip over. In fact, it may start sliding before it tips over. We need to check if sliding condition is met before eqn. (5.33) is satisfied. In other words, we need to check the value of friction forces and make sure that $|F_{A}+F_{B}|\leq\mu(N_{A}+N_{B})$ . Thus, for no slipping,

F_{A}+F_{B}\leq\mu(N_{A}+N_{B})\hbox to36.135pt{\hfil$\Rightarrow$\hfil}mg\sin% \theta\leq\mu mg\cos\theta\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\tan\theta% \leq\mu.

Using this condition (with equality) in eqn. (5.33), we get the critical condition for tipping:

\gamma=\mu.

Answer: $\gamma=\tan\theta\leq\mu$

You may know this condition geometrically as the line of action of the weight of the box must pass through B and beyond for tipping over (see fig. 5.102).

SAMPLE 5.16

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Figure 5.103:

How big does the friction force get? Consider the box on the inclined plane of Sample 5.99 again. The box has aspect ratio $\gamma=b/h$ . The coefficient of friction is $\mu$ . Imagine that the angle $\theta$ of the inclined plane can be varied. How does the force of friction on the box vary with $\theta$ ? How does the maximum value of this force depend on $\mu$ ?

Solution

Filename:pg84-3

Figure 5.104:

If we imagine the inclined plane to be not inclined ( $\theta=0$ ) but horizontal and the box to be just sitting there, the force of friction on the box has to be zero. As we tilt the plane up ( $\theta>0$ ), the friction force starts increasing. It increases up to the point of impending slip unless the box tips over before that. Assuming that the aspect ratio of the box prevents it from tipping (see Sample 5.99), we can determine the maximum value up to which the friction force rises before the box starts slipping.

From Sample 5.99, we know that the total friction force $F_{s}=F_{A}+F_{B}=mg\sin\theta$ . Thus the normalized friction force (as a fraction of the weight of the block), $F_{s}/mg$ is

\frac{F_{s}}{mg}=\sin\theta.

Filename:pfigure-blue-36-1

Figure 5.105: The maximum net friction force on the box as a function of the friction coefficient

\mu

. Note that the relationship is almost linear for most practical values of

\mu

Thus the total friction force varies as sine of the ramp angle. However, this variation is valid only upto the maximum value of the friction force ( $\mu N$ ) when the block starts sliding. The critical angle at which this maximum is attained is $\theta_{\rm slip}=\tan^{-1}\mu\equiv\phi$ (friction angle). Thus,

\left.\frac{F_{s}}{mg}\right|_{\rm max}=\sin\phi.

Figure 5.105 shows how the maximum normalized friction force varies with $\mu$ . Note that for lower values of $\mu$ (which covers most practical values of $\mu$ ), the relationship is almost linear. Thus, $|F_{s}/mg|\approx\mu$ for $\mu\leq 0.5$ .

Answer: $\frac{F_{s}}{mg}=\sin\theta,\quad\left.\frac{F_{s}}{mg}\right|_{\rm max}=\sin\phi$

What happens to the friction force after it attains the maximum value $F_{s}=mg\sin\phi$ ? For a given ramp angle, the friction force remains constant and the box slides.

SAMPLE 5.17

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Figure 5.106:

A spool of mass $m=2\,\mathrm{kg}$ rests on an incline as shown in the figure. The inner radius of the spool is $r=200\,{\rm mm}$ and the outer radius is $R=500\,{\rm mm}$ . The coefficient of friction between the spool and the incline is $\mu=0.4$ , and the angle of incline $\theta=60^{\circ}$ .

1.

Which way does the force of friction act, up or down the incline?
2.

What is the required horizontal pull $T$ to balance the spool on the incline?
3.

Is the spool about to slip?

Solution

1.

The free-body diagram of the spool is shown in fig. 5.107.

Note that the spool is a 3-force body. Therefore, in static equilibrium all the three forces — the force of gravity $m g$ , the horizontal pull $T$ , and the incline reaction $F$ — must intersect at a point. Since $T$ and $m g$ intersect at the top of the inner drum (point B), the reaction force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ of the incline must be along the direction AB. Now the incline reaction ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ is the vector sum of two forces — the normal (to the incline) reaction $N$ and the friction force $F_{s}$ (along the incline). The normal reaction force $N$ passes though the center C of the spool. Therefore, the force of friction $F_{s}$ must point up along the incline to make the resultant ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ point along AB.

Filename:pfigure4-rpi

Figure 5.107:

Answer: Up the incline
2.

We need to find the tension $T$ in the string. From the free-body diagram, we see that the force equilibrium will involve ${\accentset{\rightharpoonup}{\boldsymbol{T}}}$ along with another unknown force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ , the reaction of the incline. On the other hand, if we do moment balance about point A, we can get rid of ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ and get one scalar equation involving $T$ and $m g$ , giving $T$ in terms of $m g$ . So, writing the moment equilibrium equation about point A,

$\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}},$

we get

${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}\times(-mg\hat{% \boldsymbol{\jmath}})+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm% {B/A}}\times(T\hat{\boldsymbol{\imath}})={\accentset{\rightharpoonup}{% \boldsymbol{0}}}.$ (5.34)

These cross products can be easily evaluated by using the scalar form of the moment of a force—the product of force and the lever arm. Thus the moment of $m g$ is $mg\cdot R\sin\theta$ and the moment of $T$ is $-T\cdot(r+R\cos\theta)$ about point A in the $\hat{\boldsymbol{k}}$ direction.

Thus the scalar form of the moment-balance equation gives

$\displaystyle mgR\sin\theta$ $\displaystyle=$ $\displaystyle T(R\cos\theta+r)$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T$ $\displaystyle=$ $\displaystyle mg\,\frac{\sin\theta}{\cos\theta+r/R}$

$\displaystyle=$ $\displaystyle 2\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}\cdot\frac{% \frac{\sqrt{3}}{2}}{\frac{1}{2}+\frac{.2\,\mathrm{m}}{.5\,\mathrm{m}}}$

$\displaystyle=$ $\displaystyle 18.88\,{\mbox{N}}.$

Answer: $T=18.88\,{\mbox{N}}$

Alternatively,
We can also evaluate the net moment on the spool, given by eqn. (5.34), using direct cross products of vectors in the equation. We can use mixed basis vectors ( $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ , $\hat{\boldsymbol{\lambda}}$ , and $\hat{\boldsymbol{n}}$ ) as shown in fig. 5.107. Since,

${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}=R\hat{% \boldsymbol{n}}\quad\mbox{ and }\quad{\accentset{\rightharpoonup}{\boldsymbol{% r}}}_{\!\mathrm{B/A}}=R\hat{\boldsymbol{n}}+r\hat{\boldsymbol{\jmath}},$

we have,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}% \times(-mg\hat{\boldsymbol{\jmath}})$ $\displaystyle=$ $\displaystyle-mgR(\hat{\boldsymbol{n}}\times\hat{\boldsymbol{\jmath}})$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}% \times T\hat{\boldsymbol{\imath}}$ $\displaystyle=$ $\displaystyle T[R(\hat{\boldsymbol{n}}\times\hat{\boldsymbol{\imath}})+r(\hat{% \boldsymbol{\jmath}}\times\hat{\boldsymbol{\imath}})].$

Now, from the geometry of the basis vectors (see fig. 5.107), we have,

$\hat{\boldsymbol{n}}\times\hat{\boldsymbol{\jmath}}=-\sin\theta\hat{% \boldsymbol{k}}\quad\mbox{and}\quad\hat{\boldsymbol{n}}\times\hat{\boldsymbol{% \imath}}=-\cos\theta\hat{\boldsymbol{k}}.$

Therefore,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}% \times(-mg\hat{\boldsymbol{\jmath}})$ $\displaystyle=$ $\displaystyle mgR\sin\theta\hat{\boldsymbol{k}},$

$\displaystyle\hbox to36.135pt{and\hfil}{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{B/A}}\times T\hat{\boldsymbol{\imath}}$ $\displaystyle=$ $\displaystyle-TR\cos\theta\hat{\boldsymbol{k}}-r\hat{\boldsymbol{k}}.$

Hence, eqn. (5.34) becomes

$mgR\sin\theta\hat{\boldsymbol{k}}-TR\cos\theta\hat{\boldsymbol{k}}-r\hat{% \boldsymbol{k}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.$

Dotting both sides of this equation with $\hat{\boldsymbol{k}}$ , we get the scalar equation

$mgR\sin\theta=T(R\cos\theta+r)$

which is the same equation as obtained above using moment lever arms.
3.

To find if the spool is about to slip, we need to find the force of friction $|F_{s}|$ and see if it satisfies the condition of impending slip: $F_{s}=\mu N$ .

The force balance on the spool, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ gives

$T\hat{\boldsymbol{\imath}}-mg\hat{\boldsymbol{\jmath}}+F_{s}\hat{\boldsymbol{% \lambda}}+N\hat{\boldsymbol{n}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}$ (5.35)

where ˆ λ and ˆ n are unit vectors along the incline and normal to the incline, respectively. Dotting eqn. (5.35) with ˆ λ we get

$\displaystyle F_{s}$ $\displaystyle=$ $\displaystyle-T(\underbrace{\hat{\boldsymbol{\imath}}\cdot\hat{\boldsymbol{% \lambda}}}_{\cos\theta})+mg(\underbrace{\hat{\boldsymbol{\jmath}}\cdot\hat{% \boldsymbol{\lambda}}}_{\sin\theta})$

$\displaystyle=$ $\displaystyle-T\cos\theta+mg\sin\theta$

$\displaystyle=$ $\displaystyle-18.88\,{\mbox{N}}(1/2)+19.62\,{\mbox{N}}(\sqrt{3}/2)$

$\displaystyle=$ $\displaystyle 7.55\,{\mbox{N}}.$

Similarly, we compute the normal force $N$ by dotting eqn. (5.35) with ˆ n :

$\displaystyle N$ $\displaystyle=$ $\displaystyle-T(\hat{\boldsymbol{\imath}}\cdot\hat{\boldsymbol{n}})+mg(\hat{% \boldsymbol{\jmath}}\cdot\hat{\boldsymbol{n}})$

$\displaystyle=$ $\displaystyle T\sin\theta+mg\cos\theta$

$\displaystyle=$ $\displaystyle 18.88\,{\mbox{N}}(\sqrt{3}/2)+19.62\,{\mbox{N}}(1/2)$

$\displaystyle=$ $\displaystyle 26.16\,{\mbox{N}}.$

Now we find that $\mu N=0.4(26.16\,{\mbox{N}})=10.46\,{\mbox{N}}$ which is greater than $F_{s}=7.55\,{\mbox{N}}$ . Thus $F_{s}<\mu N$ , and therefore, the spool is not about to slip.

Answer: Not about to slip

Problems for 5.3 Friction and equilibrium

Preparatory Problems

5.3.1 For the block shown in the figure, what do you know about $F$ if

(a)

the block is sliding to the right
(b)

the block is sliding to the left
(c)

the block is not sliding.

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Figure 5.108:

5.3.2 A block weighing 500 N is dragged slowly on the ground as shown in the figure. Find the tension in the string.

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Figure 5.109:

5.3.3 Find the tension in the cable assuming the car is dragged at constant speed.

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Figure 5.110:

5.3.4 Consider the tow truck dragging the car in Problem 5.109 again. In order to ensure safety, you would like to minimize the tension in the rope attached to the car. Assume that the angle shown at point B is $\theta$ .

(a)

What value of $\theta$ minimizes the tension in the rope?
(b)

What is the corresponding value of $T$ ?
(c)

What is the force of the ground on the car?

More-Involved Problems

5.3.5 A $30,000\,{\mbox{N}}$ stone cube one meter on a side was dragged up a $20^{\circ}$ ramp by 100 of a Pharaoh’s slaves by a rope parallel to the slope. The coefficient of friction was $\mu=0.2$ . Assume all the ground contact is at the front and back edges of the cube.

(a)

Find the dragging force.
(b)

Find the force on the front and back edges of the cube.

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Figure 5.111:

5.3.6 The $20\,{\rm lbf}$ uniform rectangular sign is suspended from the strut ABCD by two wires. The strut is supported by cable DE and a pin at A.

(a)

Find tension DE.
(b)

Suppose the workers who hung the sign forgot to pin the strut to the wall at point A. What is the least value of $\mu$ between the strut and wall for the system to maintain equilibrium.

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Figure 5.112:

5.3.7 A horizontal force $F$ is applied to slide the bead on the rod shown in the figure. Find the value of $F$ that is required to initiate sliding up the rod. Why is $F$ so big or small?

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Figure 5.113:

5.3.8 A $130$ pound person climbs a $120$ pound ladder that is $30\,\mathrm{ft}$ long. The ladder leans against a frictionless wall and makes an angle of $53^{\circ}$ with the ground.

(a)

Find the force of the ground on the ladder when the person is one third of the way up the ladder.
(b)

When the person gets two thirds of the way up, the bottom of the ladder starts to slip. What is $\mu$ between the ladder and ground?

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Figure 5.114:

5.3.9 A uniform $200\,{\mbox{N}}$ , $10\,\mathrm{m}$ ladder leans between a frictionless ground and a wall. It is kept from sliding away from the wall by a horizontal cable $2\,\mathrm{m}$ above the ground. Find

(a)

The tension in the cable.
(b)

The force of the ground on the ladder.
(c)

The force of the wall on the ladder.

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Figure 5.115:

5.3.10 A uniform ladder of length $\ell$ and weight $W$ rests against a frictionless slanted wall. What is the minimum $\mu$ between ladder and ground that is needed to hold the ladder in position?

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Figure 5.116:

5.3.11 A uniform ladder with weight $W$ and length $\ell$ leans against a frictionless vertical wall and makes an angle $\theta$ with the ground. In terms of the given quantities, find the values of $\mu$ at the ground for which the ladder will not slip.

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Figure 5.117:

5.3.12 A uniform ladder with weight $W$ and length $\ell$ leans against a frictional vertical wall and is supported by the frictional ground. The same coefficient of friction $\mu$ applies to the wall and to the ground. In terms of the given quantities, find the values of $\theta$ between the ladder and ground for which the ladder can be in equilibrium without slipping. Answer: $\theta\geq\tan^{-1}\left((1-\mu^{2})/2\mu\right)$

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Figure 5.118:

5.3.13 A $2\,\mathrm{m}$ square $500\,{\mbox{N}}$ 4-leg table is pushed across a floor by a horizontal force at its top surface and normal to one edge. Assume the table is $0.8\,\mathrm{m}$ high, that its center of mass is $0.6\,\mathrm{m}$ high and that all four legs slide on the floor with friction coefficient $\mu=0.3$ . Which legs carry the most load and what is the magnitude of the force from the ground on one of those legs?

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Figure 5.119:

5.3.14 An $80\,{\mbox{N}}$ chair is pulled steadily to the right by a rope. The coefficient of friction between the ground and floor is $\mu=0.25$ .

(a)

What is the force needed to pull the chair?
(b)

What is the highest point on the chair that the rope can be tied without the chair tipping over?

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Figure 5.120:

5.3.15 A candidate rock-climbing device consists of a roller (radius 2 cm) frictionlessly pinned at A to diagonal-member AC. The length of AC from point A to the wall-contact point at C is $L_{AC}=15cm$ . The climber ( $m=60\,\mathrm{kg}$ ) hangs from a rope connected to AC by a pin at B. B is on the line AC and located as shown in the figure. If needed, assume $g=10\mbox{N/kg}$ . What is the minimum coefficient of friction $\mu$ at C that is needed to hold up the climber? Answer: For this device to hold, $\mu\geq 1$ . (Demanding $\mu\geq 1$ is large for a practical device because typical rock friction has $\mu\approx 0.5$ . The too-large number follows from the simplified geometry and numbers chosen for a homework problem.)

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Figure 5.121:

5.3.16 A uniform $W=50\,{\mbox{N}}$ block with width $a$ and height $h$ is held against a wall with a horizontal force of $F$ acting on the left side half way up the block. The block is prevented from sliding down the wall by friction. There is no glue (no tension between wall and block).

(a)

Assuming friction is high enough to prevent slip, what is the minimum $F$ to keep the block from tipping away from the wall?
(b)

For twice that $F$ what is the minimum friction to keep the block from sliding down the wall?
(c)

For $a=h$ and $F=3W$ the resultant of all the wall normal and contact forces is a single force that acts on the right side of the block at what position $y$ above the bottom of the block?

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Figure 5.122:

5.3.17 In the figure shown, what is force $F$ required to push the block along the floor? This problem has no solution. Explain why (using free-body diagrams and mechanics equations).

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Figure 5.123:

5.3.18 Consider the situation shown in the figure. Give your answers to the following questions in terms of some or all of $W,\theta,\beta,g$ , and $\phi$ or $\mu$ . Assume all values of $\beta$ and $0\leq\theta\leq\pi/2$ , $0\leq\mu$ , $g>0$ , $W>0$ .

(a)

Assume the block slides steadily uphill. Find $F$ . For what values of $\theta,\beta$ , and $\mu$ does no such $F$ exist (allow $F<0$ )?
(b)

Assume the block slides downhill. What is $F$ ? For what values of $\theta,\beta$ , and $\mu$ does no such solution exist?
(c)

assume the block is not sliding. What are the possible values of F? For what values of $\theta,\beta$ , and $\mu$ does such a solution exist?
(d)

For what values of $\theta,\beta$ , and $\mu$ can you have the block slide up, slide down, or lock (that is, no incipient slip) depending on the value of $F$ ?

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Figure 5.124:

5.3.19 A car is being towed. Unfortunately all the wheels are locked and skidding with friction coefficient $\mu$ . The tow cable AB has a slope of $1/3$ .

(a)

In terms of some or all of $e, b, c, d, m, g$ & $\mu$ , find the tension in the tow cable AB. Answer: $T_{AB}=\sqrt{10}\mu mg/(3+\mu)$
(b)

Instead of an angle with slope $1/3$ , what should the cable angle be to minimize the tension. Answer: Minimum tension if rope slope is $\mu$ (instead of $1/3$ )

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Figure 5.125:

5.3.20 A weight $M$ is steadily raised by pulling with a force $F$ on a rope going over a negligible-mass pulley on an unlubricated journal bearing (no ball bearings). For an ideal frictionless pulley $F=Mg$ . Here, however, we have a friction coefficient between the bearing and its axle which is $\mu=\tan\phi$ .

[Hint: Finding the location of the contact point D is part of the problem.]

(a)

Find $F$ in terms of $M, g, R, r$ and $\mu$ (or $\phi$ or $\sin\phi$ or $\cos\phi$ — whichever is most convenient. For example $\cos(\tan^{-1}(\mu))$ is more simply expressed as $\cos\phi$ ), and
(b)

Evaluate $F$ in the special case that $M=100\,\mathrm{kg},g=10\,\mathrm{m}/\mathrm{s}^{2},r=1\,\mathrm{cm},R=2\,% \mathrm{cm},$ and $\mu=\sqrt{3}/3$ (so $\phi=\pi/6,\sin\phi=1/2,\cos\phi=\sqrt{3}/2$ ).
(c)

Referring back to the general case, for fixed $r, R, M$ , and $g$ what happens to $F$ as $\mu\rightarrow\infty$ (does it go to $\infty$ )?

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Figure 5.126:

5.3.21 A reel of mass $M$ and outer radius $R$ is connected by a horizontal string from point $P$ across a pulley to a hanging object of mass $m$ . The inner cylinder of the reel has radius $r=\frac{1}{2}R$ . The slope has angle $\theta$ . There is no slip between the reel and the slope. There is gravity.

(a)

Find the ratio of the masses so that the system is at rest. Answer: $\frac{m}{M}=\frac{R\sin\theta}{R\cos\theta+r}=\frac{2\sin\theta}{1+2\cos\theta}$ .
(b)

Find the corresponding tension in the string, in terms of $M$ , $g$ , $R$ , and $\theta$ . Answer: $T=mg=2Mg\frac{\sin\theta}{1+2\cos\theta}$ .
(c)

Find the corresponding force on the reel at its point of contact with the slope, point $C$ , in terms of $M$ , $g$ , $R$ , and $\theta$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}=Mg\left[-\frac{2\sin\theta}{% 2\cos\theta+1}\hat{\boldsymbol{\imath}}^{\prime}+\hat{\boldsymbol{\jmath}}^{% \prime}\right]$ (where $\hat{\boldsymbol{\imath}}^{\prime}$ and $\hat{\boldsymbol{\jmath}}^{\prime}$ are aligned with the horizontal and vertical directions)
Harder

Draw a careful sketch and find a point where the lines of action of the gravity force and string tension intersect. For the reel to be in static equilibrium, the line of action of the reaction force at $C$ must pass through this point. Using this information, what must the tangent of the angle $\phi$ of the reaction force at $C$ be, measured with respect to the normal to the slope? Does this answer agree with that you would obtain from your answer in part(c)? Answer: $\tan\phi=\frac{\sin\theta}{2+\cos\theta}$ . Needs somewhat involved trigonometry, geometry, and algebra.
(d)

What is the relationship between the angle $\phi$ of the reaction at $C$ , measured with respect to the normal to the ground, and the mass ratio required for static equilibrium of the reel? Answer: $\tan\psi=\frac{m}{M}=\frac{2\sin\theta}{1+2\cos\theta}$ .
(e)

What is the minimum coefficient of friction $\mu$ at C needed to prevent slip.

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Figure 5.127:

5.3.22 This problem is similar to problem 5.126. A reel of mass $M$ and outer radius $R$ is connected by an inextensible string from point $P$ across a pulley to a hanging object of mass $m$ . The inner cylinder of the reel has radius $r=\frac{1}{2}R$ . The slope has angle $\theta$ . There is no slip between the reel and the slope. There is gravity. In terms of $M$ , $g$ , $R$ , and $\theta$ , find:

(a)

the ratio of the masses so that the system is at rest, Answer: $\frac{m}{M}=\frac{R\sin\theta}{R\cos\theta r}=\frac{2\sin\theta}{2\cos\theta-1}$ .
(b)

the corresponding tension in the string, and Answer: $T=mg=2Mg\frac{\sin\theta}{2\cos\theta-1}$ .
(c)

the corresponding force on the reel at its point of contact with the slope, point $C$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}=\frac{Mg}{1-2\cos\theta}% \left[\sin\theta\hat{\boldsymbol{\imath}}+(\cos\theta-2)\hat{\boldsymbol{% \jmath}}\right]$ .
(d)

What is the minimum coefficient of friction $\mu$ at C needed to prevent slip.

Check that for $\theta=0$ , your solution gives $\frac{m}{M}=0$ and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}=Mg\hat{\boldsymbol{\jmath}}$ and for $\theta=\frac{\pi}{2}$ , it gives $\frac{m}{M}=-2$ and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}=Mg(\hat{\boldsymbol{\imath}}% -2\hat{\boldsymbol{\jmath}})$ .The negative mass ratio is impossible since mass cannot be negative and the negative normal force is impossible unless the wall or the reel or both can ‘suck’ or they can ‘stick’ to each other (that is, provide some sort of suction, adhesion, or magnetic attraction).

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Figure 5.128:

5.3.23 Assume a massless pulley is round and has outer radius $R_{2}$ . It slides on a shaft that has radius $R_{i}$ . Assume there is friction between the shaft and the pulley with coefficient of friction $\mu$ , and friction angle $\phi$ defined by $\mu=\tan(\phi)$ . Assume the two ends of the line that are wrapped around the pulley are parallel.

(a)

What is the relation between the two tensions when the pulley is turning? You may assume that the bearing shaft touches the hole in the pulley at only one point. Answer: $\frac{F_{1}}{F_{2}}=\frac{R_{o}+R_{i}\sin\phi}{R_{o}-R_{i}\sin\phi}$ .
(b)

Plug in some reasonable numbers for $R_{i},R_{o}$ and $\mu$ (or $\phi$ ) to see one reason why wheels (say pulleys) are such a good idea even when the bearings are not all that well lubricated. Answer: For $R_{o}=3R_{i}$ and $\mu=0.2$ , $\frac{F_{1}}{F_{2}}\approx 1.14$ .

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Figure 5.129:

5.3.24 The so-called pipe-clamp has a bracket ABC which loosely fits around the slide-shaft (the‘pipe’). When not clamped there is no big force at C and the bracket freely slides on the shaft. However the bracket frictionally locks once the load $F$ at C gets large. Neglecting gravity, find the minimum coefficient of friction $\mu$ at A and B for which this clamp holds well (which it does).

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Figure 5.130:

5.3.25 Find the minimum coefficient of friction $\mu$ needed for a front wheel drive car to go up hill. Answer in terms of some or all of $a, b, h, m, g$ and $\theta$ .

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Figure 5.131:

5.3.26 Solve Problem 5.130 for a rear wheel drive car.

5.3.27 Solve Problem 5.130 for a four wheel drive car.

5.4 Internal forces

The vague concept of ‘forces inside’ a structure is superficially in conflict with the subject of mechanics. Why? Because mechanics equations only concern the forces on an object shown in a free-body diagram;‘internal forces’ have no place on a free-body diagram and thus no place in mechanics.

Example: Pulling on the ends of a rope; nothing internal

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Figure 5.132: a) Two people pulling on a rope that is likely to break in the middle, b) A free-body diagram of the rope.

Consider two people pulling apart the frayed rope of fig. 5.132a. A free-body diagram of the rope is shown in fig. 5.132b. The laws of mechanics use the external forces on an isolated system. These are the forces that show on a free-body diagram. For the rope, these are the forces at the ends. The free-body diagram does not include internal forces. Thus nothing about the ‘internal forces’ at the fraying part of the rope shows up in the mechanics equations describing the rope.

Mechanics has nothing to say about so called ‘internal forces’ and thus nothing to say about the rope breaking in the middle. ‘Internal forces’ are meaningless in mechanics. The section title describes a non-existent subject.

Something’s wrong. The problem is somewhat one of language: ‘internal forces’ are not really internal and they are not really forces!

‘Internal forces’ represent external forces on a smaller body

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Figure 5.133: a) a free-body diagram of the right part of the rope, b) the same free-body diagram, with the force distribution at the cut replaced with an equivalent force couple system, c) further simplified by using the laws of mechanics, and d) a free-body diagram of the left portion of the rope.

On page ‣ Chapter 0 What is mechanics? we advertised mechanics as being useful for predicting when things will break. And our intuitions strongly tell us that there is something about the forces in the rope that make it break. Yet mechanics equations are based on the forces that show on free-body diagrams. And free-body diagrams only show external forces. How can we use mechanics based on external forces to describe the ‘forces’ inside a body? We use an idea whose simplicity hides its incredible utility:

You cut the body, and what was inside is now on the outside of a smaller body.

In the case of the rope, we cut it in the middle. Then we fool the rope into thinking it wasn’t cut using forces (remember, ‘forces are the measure of mechanical interaction’), one force, say, at each fiber that is cut. Then we get the free-body diagram of fig. 5.133a. We can simplify this to the free-body diagram of fig. 5.133b because we know that every force system is equivalent to a force and couple at any point, in this case the middle of the rope. If we apply the equilibrium conditions to this cut rope, we see that

\begin{array}[]{ccccccc}\mbox{Sum of vertical forces is zero}&\hbox to36.135pt% {\hfil$\Rightarrow$\hfil}&F_{y}&=&0\\ \mbox{Sum of horizontal forces is zero}&\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}&F_{x}&=&-T\\ \mbox{Sum of moments about the cut is zero}&\hbox to36.135pt{\hfil$\Rightarrow% $\hfil}&M&=&0.\end{array}

Thus we get the simpler free-body diagram of fig. 5.133c as you probably already guessed without using the equilibrium equations explicitly.

Tension

We have just discovered the concept of ‘tension in a rope’, also sometimes called the ‘axial force’. The tension is the pulling force on a free-body diagram of the cut rope. If we had used the same cut for a free-body diagram of the left half of the rope we would see the free-body diagram of fig. 5.133d. Either by the principle of action and reaction, or by the equilibrium equations for the left half of the rope, you see also a tension $T$ . The force vector is the opposite of the force vector on the right half of the rope. So it doesn’t make sense to talk about the tension force vector in the rope since different (opposite) force vectors manifest themselves on the two sides of the cut ( $-T\hat{\boldsymbol{\imath}}$ on the left end of the right half and $T\hat{\boldsymbol{\imath}}$ on the right end of the left half). Instead we talk about the scalar tension $T$ which expresses the force vector at the cut as

{\accentset{\rightharpoonup}{\boldsymbol{F}}}=T\hat{\boldsymbol{\lambda}}

where $\hat{\boldsymbol{\lambda}}$ is a unit vector pointing out from the free-body diagram cut. Because $\hat{\boldsymbol{\lambda}}$ switches direction depending on which half rope you are looking at, the same scalar $T$ works for both pieces.

The tension in a rope, cable, or bar is the amount of force pulling out on a free-body diagram of the cut rope, cable, or bar. Tension is a scalar.

Internal ‘forces’ are not force vectors

Note our abuse of language: force is a vector, tension is an ‘internal force’ and tension is a scalar. What we call ‘internal forces’ are not really forces. We can’t talk about the internal force vector at a point in the string because there are two different vectors for each cut, one for left half of string and one for the right. An ‘internal force’ isn’t a force vector. Rather it is a quantity from which we can find a force vector once we have made a cut and picked which side of the cut we care about. We use this confusing language because of its firm place in the engineering workplace.

The common phrase internal force means ‘a scalar with dimensions of force from which you can find the force on one side of a free-body diagram cut’.

^†^†margin: Calling tension a scalar is a deception for pedagogical purposes. The best representation of ‘internal forces’ is with tensors which are too mathematically advanced for this book. But it is fun to notice that the concept of a tensor, something prominent in Einstein’s theory of general relativity for example, has its origin in tension, our object of study here. Note the non-coincidental similarity of the words tensor and tension. What is a tensor? Loosely, a tensor is a quantity that helps you find a vector (the force at a cut) once you are told another vector (the unit vector pointing outwards from the cut). [Aside for hyper-experts: The relation between the tension tensor and tension scalar can be expressed by the dyadic representation

\underline{\underline{\boldsymbol{T}}}=T\hat{\boldsymbol{e}}_{1}\hat{% \boldsymbol{e}}_{1}

Summarizing:

•

Internal forces are not internal. Rather they describe the forces on the boundary of a smaller system that has a free-body diagram cut that is inside the system of previous interest.
•

Internal forces are not force vectors. Rather they are scalars from which you can find the force vector acting on one side of a free-body diagram cut.

What is the strength of a structural piece?

Getting back to the question of whether or not the rope will break, we can now characterize the rope by the tension it can carry. A $10kN$ cable can carry a tension of $10,000\,{\mbox{N}}$ all along its length. This means a free-body diagram of the rope, cut anywhere along its length, could show forces up to but not bigger than $10,000\,{\mbox{N}}$ . If the rope is frayed it may break at, say, a tension of $2,000\,{\mbox{N}}$ , meaning a free-body diagram with a cut at the fray can only show forces up to $2,000\,{\mbox{N}}$ .

Note that tension is not always positive. A negative tension (negative pulling out from the ends) is also called a positive compression (positive pushing in at the ends). For ropes we don’t see much negative tension, the rope bends with just a hint of compression. But for metal and wood bars, and bones, compression is as important as tension.

Shear force and bending moment

To characterize the strength of more than just 2-force bodies we need to generalize the concept of tension. The main idea, which was emphasized in Chapter 3, is this:

You can make a free-body diagram cut anywhere on any body no matter how it is loaded.

As for tension, we define internal forces in terms of the forces (and moments) that show up on a free-body diagram cut. Again we consider things (bars) that are rather longer than they are wide or thick because

•

Long narrow pieces are commonly used in construction of buildings, machines, plants and animals.
•

Internal forces in long narrow things are easier to understand than in bulkier objects.

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Figure 5.134: a) A piece of a structure, loads not shown; b) a partial free-body diagram of the right part of the bar; c) a partial free-body diagram of the left part of the bar.

For now we limit ourselves to 2D statics. At an arbitrary cut we can find the force and moment on the remaining piece in the same manner as in Section 5.2. And we could look at the $x$ and $y$ components of the force. Fine. The problem is that the force and moment we find do not just depend on the cut, but on which body we look at. On the right side of the cut a force and moment act. On the left side of the cut, the opposite force and moment act on the other object. Another problem with $x y$ components is that they don’t necessarily line up with the natural directions for the structural part.

So, for the purposes of thinking about internal forces we break the force into two components (see fig. 5.134) lined up with the part. And we measure the internal forces with scalars that are the same for both sides of the cut:

•

The tension $T$ is the scalar part of the force directed along the bar assumed positive when pulling away from the free-body diagram cut.
•

The shear force $V$ is the force perpendicular to the bar (tangent to the free-body diagram cut). Our sign convention is that shear is positive if it tends to rotate the cut object clockwise. An equivalent statement of the sign convention is that shear is positive if down on cuts at the right of a bar and positive if up on a cut on the left of bar (and to the right on top and to the left on the bottom).

Since we are just doing 2D problems now, the moment is always in the out-of-plane (typically $\hat{\boldsymbol{k}}$ ) direction.

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Figure 5.135: The smiling beam sign convention for bending moment. For a horizontal beam, moments which tend to make the beam smile (curve up) are called positive.

•

The bending moment $M$ is the scalar part of the bending moment. The sign convention is that for a smiling beam (fig. 5.135): A clockwise ( $-\hat{\boldsymbol{k}}$ ) couple is positive on a left cut and a counterclockwise ( $\hat{\boldsymbol{k}}$ ) couple is positive on a right cut^†^†margin: Note that neither $V$ nor $T$ changes if you rotate your paper until the picture is upside down. However, the definition for the sign convention for $M$ has the disadvantage that the bending moment changes sign if you turn your paper upside down. (This ambiguity can only be avoided if one picks a favored side or end of the beam). .

The tension $T$ , shear $V$ , and bending moment $M$ on fig. 5.134 follow these sign conventions.

Filename:tfigure-intforce

Example: Internal forces in a bent rod

The internal forces at B can be found by making a free-body diagram of a portion of the structure with a cut at B.

\begin{array}[]{ccccccc}\mbox{Sum of vertical forces is zero}&\hbox to36.135pt% {\hfil$\Rightarrow$\hfil}&V&=&(100/\sqrt{2})\,{\mbox{N}}\\ \mbox{Sum of horizontal forces is zero}&\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}&T&=&(100/\sqrt{2})\,{\mbox{N}}\\ \mbox{Sum of moments about the cut at B is zero}&\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}&M&=&-100\sqrt{2}\,{\mbox{N}}\,\mathrm{m}.\end{array}

You may have noticed that we did get ahead of ourselves and used the concept of tension in a rope or rod as a source of loading with known direction on a particle and rigid body. We will use the concept of tension extensively in our analysis of trusses.

Calculating how internal forces vary from point to point in a structure is picked up in Section 8 on page 8.

SAMPLE 5.18

Filename:p-f96-p2-3

Figure 5.136:

A structure is made up of two bars – a thick bent bar ABC and a thin rod CE. Point C is halfway between B and D, $\ell=0.8\,\mathrm{m}$ and $\theta=60^{{}^{\circ}}$ . Bar ABC is pulled up by a force $F=500\,{\mbox{N}}$ at point A.

1.

Find the internal forces in the bar ABC just to the left of point B.
2.

Find the force in bar CE at the section s-s shown in the figure.

Solution

Filename:s97p3-2

Figure 5.137:

We cut the bar ABC at point B. The free-body diagram of the left part AB is shown in fig. 5.137. The internal forces acting at the cut section are tension $T$ , shear force $V$ and the bending moment $M$ . From force balance of part AB in $x$ and $y$ directions, we have

T=0,\quad\mbox{and}\quad V=F=500\,{\mbox{N}}.

From the moment balance about point B, we have

M-F\ell/4=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}M=F\ell/4=100\,{\rm N}% \mathord{\cdot}{\rm m}.

Answer: $T=0,\quad V=500\,{\mbox{N}},\quad M=100\,{\rm N}\mathord{\cdot}{\rm m}$

Filename:pfigure-dynbalancers

Figure 5.138:

For finding the tension in rod CE at the given section, we cut the rod at s-s and draw the free-body diagram of the structure along with the upper part of the rod attached at point C. The tension in bar CE is $T$ and the reaction of the support at pin D is $R$ . We need to find $T$ .

We can write the moment-balance equation about point D, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{D}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , so that the unknown force $R$ (that we are not interested in) disappears from the equation:

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/D}}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/D}}\times{\accentset{% \rightharpoonup}{\boldsymbol{T}}}

\displaystyle=

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.

The moments of $F$ and $T$ about point D can be easily evaluated using the scalar formula ‘force times the lever arm’ (see fig. 5.139). Thus, the moment-balance equation in $\hat{\boldsymbol{k}}$ direction is:

	$\displaystyle-F\ell(1/4+\cos\theta)+T\frac{\ell}{2}\sin 2\theta$	$\displaystyle=$	$\displaystyle 0$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T=\frac{2(1/4+\cos\theta% )}{\sin 2\theta}F.$

Filename:pfigure-s94h7p5

Figure 5.139: Moments of

F

and

T

about point D can be evaluated by multiplying the forces with their respective lever arms

\ell/4+\ell\cos\theta

and

\ell/2\sin 2\theta

, respectively.

Substituting the given values, $F=500\,{\mbox{N}}$ and $\theta=60^{{}^{\circ}}$ , we get

T=866\,{\mbox{N}}.

Answer: $T=866\,{\mbox{N}}$

Note: Evaluation of the moment equation about point D using vectors and cross products is as follows. Since ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/D}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/B}}+{\accentset{\rightharpoonup% }{\boldsymbol{r}}}_{\!\mathrm{B/D}}$ $=-\frac{\ell}{4}\hat{\boldsymbol{\imath}}+\ell(-\cos\theta\hat{\boldsymbol{% \imath}}+\sin\theta\hat{\boldsymbol{\jmath}})$ , ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/D}}=\frac{\ell}{2}(% -\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{\jmath}})$ , ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=F\hat{\boldsymbol{\jmath}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{T}}}=T(-\cos\theta\hat{\boldsymbol{% \imath}}-\sin\theta\hat{\boldsymbol{\jmath}})$ ,

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/D}}\times{% \accentset{\rightharpoonup}{\boldsymbol{F}}}=-F\left(\frac{\ell}{4}+\ell\cos% \theta\right)\hat{\boldsymbol{k}},\mbox{ and }{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{C/D}}\times{\accentset{\rightharpoonup}{% \boldsymbol{T}}}=T\ell\cos\theta\sin\theta\hat{\boldsymbol{k}}.

Therefore, the moment-balance equation is

-F\ell(1/4+\cos\theta)\hat{\boldsymbol{k}}+T\frac{\ell}{2}\sin 2\theta\hat{% \boldsymbol{k}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.

SAMPLE 5.19

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Figure 5.140:

A ladder of length $2d=4\,\mathrm{m}$ rests against a wall as shown. A person of weight $W=700\,{\mbox{N}}$ stands at C. Assume that the ladder does not slip. Neglecting the weight of the ladder, find the internal forces in the ladder at sections $a$ - $a$ and $b$ - $b$ , at mid points of AC and AB, respectively. (See Sample 5.97.)

Solution

Filename:pfigure-blue-137-2

Figure 5.141:

To find the internal forces at the indicated sections, we need to cut the ladder at those sections, one at a time, draw the free-body diagram of each part and carry out the force and moment-balance equations. A little anticipation shows that we will need the support reactions at A and B in our calculations. So, let us first determine the support reactions. The free-body diagram of the ladder is shown in fig. 5.141. The moment balance about point B in $\hat{\boldsymbol{k}}$ direction gives

\displaystyle-R(2d\sin\theta)+W(d\cos\theta)=0\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}R=\frac{W}{2}\frac{\cos\theta}{\sin\theta}.

The force balance, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ , gives

R\hat{\boldsymbol{\imath}}-W\hat{\boldsymbol{\jmath}}+{\accentset{% \rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}{\boldsymbol{0}}% }\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{\rightharpoonup}{% \boldsymbol{F}}}=-R\hat{\boldsymbol{\imath}}+W\hat{\boldsymbol{\jmath}}.

Substituting the given values of $\theta(60^{{}^{\circ}})$ and $W(700\,{\mbox{N}})$ , we get,

{\accentset{\rightharpoonup}{\boldsymbol{R}}}=(202\,{\mbox{N}})\hat{% \boldsymbol{\imath}},\hbox to36.135pt{\hfil and\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{F}}}=(-202\hat{\boldsymbol{\imath}}+700\hat{% \boldsymbol{\jmath}})\,{\mbox{N}}.

Section $a$ - $a$ : Now, we cut the ladder at $a$ - $a$ and draw the free-body diagram of the upper part of the ladder as shown in fig. 5.142. The force balance for this part gives

$\displaystyle T\hat{\boldsymbol{\lambda}}-V\hat{\boldsymbol{n}}+R\hat{% \boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T$	$\displaystyle=$	$\displaystyle-R(\hat{\boldsymbol{\imath}}\cdot\hat{\boldsymbol{\lambda}})=-R\cos\theta$
$\displaystyle\mbox{and}\quad V$	$\displaystyle=$	$\displaystyle R(\hat{\boldsymbol{\imath}}\cdot\hat{\boldsymbol{n}})=R\sin\theta.$

Substituting the numerical values of $R$ and $\theta$ , we get $T=-101\,{\mbox{N}}$ and $V=175\,{\mbox{N}}$ . Now, the moment-balance equation about $a$ (the cut) gives

M-R(d/2)\sin\theta=0\hbox to36.135pt{\hfil$\Rightarrow$\hfil}M=(1/2)Rd\sin\theta

which, with numerical values, gives $M=175\,{\rm N}\mathord{\cdot}{\rm m}$ .

Filename:pfigure-blue-78-2

Figure 5.142: Free-body diagram of the upper part of the ladder cut at section

a

a

. Note that

\hat{\boldsymbol{\lambda}}=\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{% \boldsymbol{\jmath}}

and

\hat{\boldsymbol{n}}=\sin\theta\hat{\boldsymbol{\imath}}+\cos\theta\hat{% \boldsymbol{\jmath}}

Answer: $T=-101\,{\mbox{N}},\quad V=175\,{\mbox{N}},\quad M=175\,{\rm N}\mathord{\cdot}% {\rm m}$

Section $b$ - $b$ : Now we consider the internal forces at section $b$ - $b$ . We cut the ladder at the given section. We can consider the free-body diagram of the upper part or the lower part of the ladder to find the internal forces. Considering the upper part, (see fig. 5.143) we get, from force balance,

Filename:pfigure-blue-130-2

Figure 5.143:

T\hat{\boldsymbol{\lambda}}-V\hat{\boldsymbol{n}}+R\hat{\boldsymbol{\imath}}-W% \hat{\boldsymbol{\jmath}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}

which, as the analysis above, gives

	$\displaystyle T$	$\displaystyle=$	$\displaystyle-R(\hat{\boldsymbol{\imath}}\cdot\hat{\boldsymbol{\lambda}})+W(% \hat{\boldsymbol{\jmath}}\cdot\hat{\boldsymbol{\lambda}})=-R\cos\theta+W(-\sin% \theta)=-707\,{\mbox{N}}$
	$\displaystyle V$	$\displaystyle=$	$\displaystyle R(\hat{\boldsymbol{\imath}}\cdot\hat{\boldsymbol{n}})-W(\hat{% \boldsymbol{\jmath}}\cdot\hat{\boldsymbol{n}})=R\sin\theta-W\cos\theta=-175\,{% \mbox{N}}.$

Similarly, the scalar moment-balance equation about point $b$ gives

M-R\frac{3d}{2}\sin\theta+W\frac{d}{2}\cos\theta=0\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}M=175\,{\rm N}\mathord{\cdot}{\rm m}.

Answer: $T=-707\,{\mbox{N}},\quad V=-175\,{\mbox{N}},\quad M=175\,{\rm N}\mathord{\cdot% }{\rm m}$

Problems for 5.4 Internal forces

Preparatory Problems

5.4.1 For the bar shown which ones of the following statements are true? Answer: None are true. The tension is $100\,{\mbox{N}}$ .

(a)

The two forces cancel so the tension is zero.
(b)

The two forces add so the tension is $200\,{\mbox{N}}$ .
(c)

The tension is $100\,{\mbox{N}}\hat{\boldsymbol{\imath}}$ .
(d)

The tension is $-100\,{\mbox{N}}\hat{\boldsymbol{\imath}}$ .
(e)

The tension is $100\,{\mbox{N}}\hat{\boldsymbol{\imath}}$ on the right end and $-100\,{\mbox{N}}\hat{\boldsymbol{\imath}}$ on the left end.

Filename:pfigure-s94h10p4

Figure 5.144:

5.4.2 What letters and case (upper or lower) are used in this book for tension, shear force, and bending moment?

5.4.3 Mechanics depends on free-body diagrams. And free-body diagrams only show the external forces on an object. So how can mechanical sense be made of the concept of “internal” force?

5.4.4 A string is conceptually cut in half by making free-body diagrams of the left and right halves of the string. At the cut on the left half of the string acts the force $50\,{\mbox{N}}\hat{\boldsymbol{\imath}}$ . At the cut on the right half of the string acts the force $-50\,{\mbox{N}}\hat{\boldsymbol{\imath}}$ . With two different forces acting on the two halves how can one define a single ‘tension’?

5.4.5 Define as precisely as you can:

(a)

Shear force
(b)

Bending moment

5.4.6 Find the tension, shear force and bending moment at C for each of the structures below. Neglect gravity. Assume dimensions as needed.

Filename:pfigure-blue-81-2

Figure 5.145:

5.4.7 Find the tension, shear force and bending moment at C for each of the structures below. There is no gravity. Assume dimensions if needed.

Filename:pg147-2

Figure 5.146:

5.4.8 Find the tension, shear and bending moment at section C for each of the structures below. And also at D, if marked. Assume reasonable dimensions as needed.

Filename:pfigure-s94h9p1

Figure 5.147:

5.4.9 The tension in the bow-saw blade BC is $250\,{\mbox{N}}$ . Find the tension, shear, and bending moment at A.

Filename:pfigure-blue-151-4

Figure 5.148:

More-Involved Problems

5.4.10 Find the tension, shear and bending moment at section C for each of the structures below. And also at D, if marked. Include gravity, assume all bars are uniform with density of (100 N/m). Assume reasonable dimensions as needed.

Filename:pfigure4-rpf

Figure 5.149:

5.4.11 Find the tension, shear and bending moment at section C for each of the structures below. And also at D, if marked. Neglect gravity. Assume reasonable dimensions as needed.

Filename:pfigure-slipperymoney

Figure 5.150:

5.5 3D statics

The structures and machines we study are most-often adequately modeled as 2D. In those cases a 2D analysis gives about the same answer as a 3D analysis would have given. However, sometimes a 2D model is inadequate, and a 3D analysis is needed. Here we use 3D statics to find various unknown aspects of forces acting on one part. By learning the 3D approach you can get a better sense of when to use a 2D model (which is most of the time for most engineers).

3D statics is conceptually the same as 2D: draw a free-body diagram and use the force and moment-balance equations. However, the geometry can be more of a challenge, the moment balance equation becomes a full vector equation (instead of just having one non-zero component, it has three)^†^†margin: For 2D problems we used the phrase ‘moment about a point’ to be short for ‘moment about an axis in the $z$ direction that passes through the point. In 3D moment about a point is a 3-component vector. ,

and the number of scalar equations from one free-body diagram increases from 3 to 6. In 3D, issues related to static-indeterminacy arise more often and more subtly.

The statics-of-a-3D-object recipe

Our recipe here:

1)

Draw a free-body diagram (FBD) of the part of interest.
Use knowledge of the contact conditions (see Chapter 3) to draw known and unknown aspects of the forces appropriately (see fig. 2.64 on page 2.64) [hint: use of the form $F\hat{\boldsymbol{\lambda}}$ is often appropriate];
2)

Write equilibrium equations in terms of the forces (and couples) shown on the FBD;
3)

Solve the equilibrium equations for unknowns.

The brute-force approach to statically determinate problems

A problem is statically determinate when all as-yet-unknown forces can be found using the equilibrium equations. In 3D statics, this generally means that the two vector equilibrium equations

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}\qquad\mbox{and}\qquad\sum{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathrm{i/C}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}

(where C is any point that you like) make up 6 independent scalar equations which you can solve for 6 unknown aspects of the applied forces (say the magnitudes of 6 forces whose directions are known a priori)^†^†margin: Most elementary text-book problems are statically determinate. Unfortunately most real-world problems, when you first model them, are not statically determinate. .

Alternative equation sets

In 2D single-part statics, we noted various alternatives to using vector force balance and moment about one point (see page 5.2). Similarly, here there are also an infinite number of true equilibrium equations, for example,

•

$\left(\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}}\right% )\cdot\hat{\boldsymbol{\lambda}}=0$ where $\hat{\boldsymbol{\lambda}}$ is a vector in any direction you please; and
•

$\left(\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/C}}\right)% \cdot\hat{\boldsymbol{\lambda}}=0$ . This is moment balance about an axis through C in the $\hat{\boldsymbol{\lambda}}$ direction.

From these there are various ways to extract 6 independent scalar equations, including:

•

Cartesian components of force balance and moment balance about any point C: $\sum F_{x}=0,\sum F_{y}=0,\sum F_{z}=0,\sum M_{Cx}=0,\sum M_{Cy}=0,$ and $\sum M_{Cz}=0$ . This always works, although it does not necessarily minimize algebra.
•

Force balance in any 3 non-coplanar directions and moment balance about point C resolved in any three non-coplanar directions.
•

Moment balance about 6 independent axes. There seems to be no simple description of independent axes but that they give independent equilibrium equations ^†^†margin: One can test the sufficiency of the equations by this check: if a force at the origin and a couple are the only forces applied to a system, do the equations demand that they must both be zero? If so, then you have an independent set of equations. .

Practically speaking, six moment-about-an-axis equations are likely to be independent if not too many axes are parallel with each other, not too many are coplanar, and not too many intersect at one point.

In any case, force balance contributes at most 3 independent equations and moment balance can contribute up to 6 (thus rendering force balance a non-essential tool).

Solving 6 equations in 6 unknowns, or even setting up such for computer solution, is relatively time-consuming and error-prone. Thus one looks for shortcuts when one can, namely:

Useful shortcuts:

•

Use moment balance about an axis that intersects, or is parallel to, as many unknown force lines-of-action as possible (thus those forces do not show up in that equilibrium equation);
•

Use force balance in a direction orthogonal to as many of the unknown forces as possible (so those forces don’t show up in that equation).

Special loadings

Two- and three-force bodies

The concepts of two-force (page 5.2) and three-force (page 5.2) bodies are identical in 3D.

•

If there are only two forces applied to a body in equilibrium, they must be equal and opposite and acting along the line connecting the points of application. The full set of six equations tells you no more.
•

If there are only three forces applied to an object, they must all be in the plane of the points of application and the three forces must have lines of action that intersect at one point. The three equations of force balance are an additional restriction on these three forces.

There are other special loadings where the equilibrium equations offer less than 6 independent equations:

•

2D. If all of the forces have lines of action in one plane, then there are only three independent scalar equations and thus one can solve for 3 unknowns. For example, if all the forces lie in the $x y$ plane, then automatically $\sum F_{iz}=0$ , $\sum M_{ix/C}=0$ , and $\sum M_{iy/C}=0$ .
•

Concurrent forces. If all the lines of action intersect in one point, say D, then $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/D}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ is automatically satisfied and only the 3 equations of force balance are independent.
•

If all the forces are parallel in, say, the $\hat{\boldsymbol{k}}$ direction, then force balance in the $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ directions as well as moment balance about any axis in the $\hat{\boldsymbol{k}}$ direction, is automatically satisfied and there are only three independent equilibrium equations (say $\sum F_{z}=0$ , $\sum M_{x}=0$ and $\sum M_{y}=0$ .

What does it mean for a problem to be ‘2D’?

The world we live in is three-dimensional: all the objects we wish to study mechanically are three-dimensional, and if they are in equilibrium, they satisfy the three-dimensional equilibrium equations. How then can an engineer justify doing 2D mechanics? There are a variety of overlapping justifications.

•

The 2D equilibrium equations are a subset of the 3D equations. In both 2D and 3D, $\sum F_{x}=0$ , $\sum F_{y}=0$ , and $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{/0}}\cdot\hat{% \boldsymbol{k}}=0$ . So, if when doing 2D mechanics, one just neglects the $z$ component of any applied forces and the $x$ and $y$ components of any applied couples, one is doing correct 3D mechanics, just not all of 3D mechanics. If the forces or conditions of interest to you are contained in the 2D equilibrium equations, then 2D mechanics is really 3D mechanics, ignoring equations you don’t need.
•

If the $x y$ plane is a plane of symmetry for the object and of any applied loading, then the 3D equilibrium equations not covered by the 2D equations are automatically satisfied. For a car, say, the assumption of symmetry implies that the forces in the $z$ direction will automatically add to zero, and the moments about the $x$ and $y$ axis will automatically be zero.
•

If the object is thin and there are constraint forces holding it near the $x y$ plane, and these constraint forces are not of interest, then 2D statics is also appropriate. This last case is caricatured by all the poor mechanical objects you have drawn so. They are conceptually constrained to lie in your flat paper by invisible slippery glass in front of and behind the paper.

“Internal forces” in 3D

At a free-body-diagram cut on a long narrow structural piece in 2D, we have two force components (tension and shear) and one scalar moment. In 3D, such a cut shows a force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ and a moment ${\accentset{\rightharpoonup}{\boldsymbol{M}}}$ each with three components. If one picks a coordinate system with the $x$ axis aligned with the bar at the cut, the concept of tension remains the same. Tension is the force component along the bar.

T=F_{x}={\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot\hat{\boldsymbol{% \imath}}.

The two other force components, $F_{y}$ and $F_{z}$ , are two components of shear. The net shear force is a vector in the plane orthogonal to $\hat{\boldsymbol{\imath}}$ .

The new concept, often called torsion is the component of ${\accentset{\rightharpoonup}{\boldsymbol{M}}}$ along the axis:

\mbox{torsion}=M_{x}={\accentset{\rightharpoonup}{\boldsymbol{M}}}\cdot\hat{% \boldsymbol{\imath}}

Torsion is the part of the moment that twists the shaft.

The remaining part of the ${\accentset{\rightharpoonup}{\boldsymbol{M}}}$ , in the $y z$ plane, is the bending moment. It has two components $M_{y}$ and $M_{z}$ .

The preponderance of statically indeterminate problems

As noted in box 5.1 on page 5.1, non-uniqueness of solutions is common in the real world. Especially in 3D, real world problems are, at first blush, rarely statically determinate. The statics equations are relevant and provide useful information, but are not sufficient for finding all unknowns of interest. Finding the forces then depends on knowing the deformation properties of the structures as well as details of their initial state.

Example: Four-leg furniture

Filename:pfigure-blue-129-1

Figure 5.151: A symmetric 4 leg table on a flat floor. Statics is insufficient to find the 4 forces of the ground on the table. The full weight could be carried by either diagonal pair of legs.

Take the table, chair or bed you are now interacting with. It probably has 4 legs. To keep it simple, imagine the legs are on a slippery (negligible-friction) floor and the table is symmetric (left-right and front-back). What are the forces of the floor on the legs? The most we can get from the statics equations (force and moment balance) are that

F_{1}=F_{3},\quad F_{2}=F_{4},\qquad\mbox{and}\qquad F_{1}+F_{2}=W/2.

If we insist that there is no glue between the floor and table then $F_{1}\geq 0,F_{2}\geq 0,F_{3}\geq 0,F_{4}\geq 0$ . But we still can’t find the reactions. Here is the variety of solutions:

\begin{array}[]{llcl}&F_{1}=F_{3}=W/2&\qquad\mbox{and}&F_{2}=F_{4}=0\\ \mbox{or}&F_{1}=F_{3}=0&\qquad\mbox{and}&F_{2}=F_{4}=W/2\\ \mbox{or}&F_{1}=F_{3}=W/4&\qquad\mbox{and}&F_{2}=F_{4}=W/4\\ \mbox{or}&F_{1}=F_{3}=C&\qquad\mbox{and}&F_{2}=F_{4}=W/2-C\\ &&&(\mbox{with}\ C\ \mbox{anything in the interval}\quad 0\leq C\leq W/2).\end% {array}

It takes more than just statics to find the forces. One has to know the exact initial shape of the table and floor, and how the table and floor ‘give way’ in response to loads.

The lack of static determinacy of a table is not merely an academic curiosity. If you measured the forces of the floor on your table legs, they each could well differ noticeably from $W/4$ , with, probably, two bigger and two smaller. Once friction is taken into account, the situation is near hopeless.

Example: Statically determinate stool

Is it even possible to make a stool in 3 dimensions that is statically determinate? Here’s one way. Give it three legs. One leg can have a point frictional contact (3 reaction components), one leg can have a wheel (2 reaction components) and one can be frictionless (like with a castered wheel, 1 reaction component). $3+2+1=6$ .

In general, it is hard to hold an object in place in three dimensions in a statically determinate manner. Here are some other ways (besides the unusual stool above):

•

with six rods that have ball-and-socket joints at both the object-end and at the ground-end. The rods need to have a variety of orientations and attachment points (this idea is used in a ‘Stewart Platform’).
•

With one ball-and-socket joint and three rods.
•

A 3-leg stool with three wheels (at the contact points. For each wheel one can draw a line in the plane in the direction normal to rolling. These three lines must not intersect at a point).
•

With one hinge and one two-force-member rod.
•

With one axially sliding hinge and two rods.
•

With a single welded connection.

Given that many things are held in place in a manner that seems statically indeterminate, what can one do in practice? A common approach is to remove reaction components that you think are relatively unimportant. Some examples:

•

A door held by two hinges. That’s 10 reaction components. Usually one replaces, in the analysis, the hinges with ball-and-socket joints. That makes 6 unknown reaction components but is still statically indeterminate no matter what the loading (the force along the line connecting the joints cannot be decomposed into parts acting at each joint). So one joint is allowed to slide along the nominal hinge axis.
•

4-leg furniture. Counting friction, there are 12 reaction components. If side loads are not an issue, then we can assume-away friction. Thus we have only 4 reaction components for 3 equations (see table example above). We can get a unique solution by assuming the forces share the symmetry of the table (thus $F_{1}=F_{2}$ ).

Given this complex state of affairs in 3D, it is easy to see why engineers often resort to the more-easily-made determinate 2D world for their models and analyses.

Box 5.8 Statically determinate ways to hold an object in 3D

An advanced aside.

It takes some thought to find ways to hold something in place in 3D that are statically determinate. If you hold it securely enough so that it can’t move you will often do it in a way that there can be reaction forces even when there are no loads on the structure. These ‘locked-in’ forces are sometimes called ‘pre-stresses’, ‘pre-loads’ or ‘self-stresses’. The values of these locked-in forces will depend on the construction. You can’t find them from the equations of statics. Hence the indeterminacy.

Below we show some ways to hold things in a statically determinate way. Static determinacy is great for homework problems. Whether it is good or bad in structural design is not a simple question.

The big tradeoff. Before showing how to achieve determinacy, a warning.

A statically determinate structure has no redundancy.

A redundant structure can have locked-in forces.

And locked-in forces can contribute to structural failure. Two parallel rods holding something in place can break because they fight each other, even when there is no external load on the thing.

Counting rules. Counting dominates the discussion. Here’s how to do that counting. The rule of thumb for determinacy is:

\underbrace{\textrm{Number of equations}}_{n_{eq}}=\underbrace{\textrm{number % of unknowns}}_{n_{unkn}}

For one object in 3 dimensions we have 6 independent equilibrium equations, for example 3 components of force balance and 3 components of moment balance about a given reference point. Thus $n_{eq}=6$ . To achieve determinacy we thus want $n_{unkn}=6$ . That is, in a free-body diagram of the object we also want 6 unknown reaction components.

Attachment schemes. For each attachment point we can select an attachment type from the table of common connections on page Back tables. Here’s a compact list with the number (#) of unknown reaction components:

3D Connection	#	what
weld	6	3 moment and 3 force comps
keyed slot	5	3 moment & 2 force comps
hinge	5	3 moment & 2 force comps
linear bearing	4	2 moment & 2 force comps
ball &socket	3	3 force components
no-slip point contact	3	3 force comps
skate or wheel	2	normal force and side force
sloppy linear bearing	2	2 force components
rod or string	1	tension along 2-force member
frictionless point contact	1	normal force

Mix and match. At least to satisfy the counting rules, you can hold something in place in a statically determinate manner by selecting any number of connections from the table above just make the number of unknown components add to 6.

Here are some options:

Scheme	$\sum=6$
One weld	6=6
Hinge+rod	5+1=6
Ball & socket +
sloppy linear bearing + rod	3+2 + 1 =6
6 rods	1+1+1+1+1+1=6
Ball & socket + 3 rods	3 +1+1+1=6
No-slip point contact + wheel
+ frictionless point contact	3 + 2+1 =6
3 wheels or skates	2+2+2 = 6
Etc	n + m + … = 6

Filename:tfigure-statdethold3D

A statically determinate stool has one normal leg (with friction), one wheel and one frictionless leg (shown as a ball). A box is held in a statically determinate way with a ball & socket and three bars.

Bad luck. Although the rule

n_{eq}=n_{unkn}

is a good starting point, it is not enough. In the bad cases, two things happen together:

1)

The object is not held in place (some loads can not be equilibrated), and
2)

The constraints can fight each other.

Then, in the language of linear algebra, the $6\times 6$ matrix for the linear equilibrium equations is singular. So, for some loadings no solution exists (the object is not held in place). And if a solution exists, it is not unique (the homogeneous equations have non-trivial solutions, there can be locked-in forces).

Bad examples. Some ways of holding where 6 reaction forces lead to indeterminacy include:

•

A hinge + rod with the rod co-planar with the hinge.
•

6 rods where some 4 or more of them have lines of action that intersect in one point.
•

A ball & socket with 3 rods where one or more of the rods has a line of action that goes through the ball & socket.
•

A ball & socket with two rods that are coplanar with the ball.
•

Three wheels on a plane where the normal-lines to the rolling direction all intersect in a single point (the wheels roll on a common circle).

SAMPLE 5.20

Filename:sfig4-intern-cant3D

Figure 5.152:

3-D moment at the support: A ’T’-shaped cantilever beam is loaded as shown in the figure. Find all the support reactions at A.

^†^†margin:

Solution The free-body diagram of the beam is shown in Fig. 5.153. Note that the forces acting on the beam can produce in-plane as well as out-of-plane moments. Therefore, we show the unknown reactions ${\accentset{\rightharpoonup}{\boldsymbol{R}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}$ as general 3-D vectors at A.

Filename:sfig4-intern-cant3D-a

Figure 5.153: Free-body diagram of the cantilever.

The moment equilibrium about point A, $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , gives

{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}\times({\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}+{\accentset{\rightharpoonup% }{\boldsymbol{F}}}_{\!\!\mathrm{2}})+{\accentset{\rightharpoonup}{\boldsymbol{% r}}}_{\!\mathrm{D/A}}\times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{3}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}$	$\displaystyle=$	$\displaystyle-({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}+% {\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/B}})\times({% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}+{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}})-({\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}+{\accentset{\rightharpoonup% }{\boldsymbol{r}}}_{\!\mathrm{D/B}})\times{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{\!\!\mathrm{3}}$
	$\displaystyle=$	$\displaystyle-(\ell\hat{\boldsymbol{\imath}}+a\hat{\boldsymbol{\jmath}})\times% (-F_{1}\hat{\boldsymbol{k}}-F_{2}\hat{\boldsymbol{\imath}})-(\ell\hat{% \boldsymbol{\imath}}-a\hat{\boldsymbol{\jmath}})\times F_{3}\hat{\boldsymbol{% \imath}}.$



	$\displaystyle=$	$\displaystyle-(\ell\hat{\boldsymbol{\imath}}+a\hat{\boldsymbol{\jmath}})\times% (-F_{1}\hat{\boldsymbol{k}}-F\hat{\boldsymbol{\imath}})-(\ell\hat{\boldsymbol{% \imath}}-a\hat{\boldsymbol{\jmath}})\times F\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle-F_{1}\ell\hat{\boldsymbol{\jmath}}+F_{1}a\hat{\boldsymbol{\imath% }}-2Fa\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle-30\,{\rm lbf}\cdot 3\,\mathrm{ft}\hat{\boldsymbol{\jmath}}+30\,{% \rm lbf}\cdot 1\,\mathrm{ft}\hat{\boldsymbol{\imath}}-2(15\,{\rm lbf}\cdot 1\,% \mathrm{ft})\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle(30\hat{\boldsymbol{\imath}}-90\hat{\boldsymbol{\jmath}}-30\hat{% \boldsymbol{k}})\,{\rm lbf}\cdot\,\mathrm{ft}.$



$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{R}}}$	$\displaystyle=$	$\displaystyle-{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}-{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}-{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{3}}$
	$\displaystyle=$	$\displaystyle-{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}-{% \accentset{\rightharpoonup}{\boldsymbol{F}}}+{\accentset{\rightharpoonup}{% \boldsymbol{F}}}$
	$\displaystyle=$	$\displaystyle-(-F_{1}\hat{\boldsymbol{k}})=F_{1}\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle 30\,{\rm lbf}\hat{\boldsymbol{k}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{A}}}=30\,{\rm lbf}\hat{\boldsymbol{k}% },\qquad\mbox{and}\qquad{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm% {A}}=(30\hat{\boldsymbol{\imath}}-90\hat{\boldsymbol{\jmath}}-30\hat{% \boldsymbol{k}})\,{\rm lbf}\cdot\,\mathrm{ft}$

SAMPLE 5.21

Filename:sfig4-3d-plate

Figure 5.154:

An unsolvable problem? A $0.6\,\mathrm{m}\times 0.4\,\mathrm{m}$ uniform rectangular plate of mass $m=4\,\mathrm{kg}$ is held horizontally with two strings BE and CF and linear hinges at A and D as shown in the figure. The plate is loaded uniformly with books of total mass $6\,\mathrm{kg}$ . If the maximum tension the strings can hold is $100\,{\mbox{N}}$ , how much more load can be applied to the plate?

Solution

Filename:sfig4-3d-plate-a

Figure 5.155:

The free-body diagram of the plate is shown in fig. 5.155. Note that we model the hinges at A and D with no resistance in the $y$ -direction. Since the plate has uniformly distributed load (including its own weight), we replace the distributed load with an equivalent concentrated load ${\accentset{\rightharpoonup}{\boldsymbol{W}}}$ acting vertically through point G.

The various forces acting on the plate are

{\accentset{\rightharpoonup}{\boldsymbol{W}}}=-W\hat{\boldsymbol{k}},\;{% \accentset{\rightharpoonup}{\boldsymbol{T}}}_{\mathrm{1}}=T_{1}\hat{% \boldsymbol{\lambda}}_{\mathrm{BE}},\;{\accentset{\rightharpoonup}{\boldsymbol% {T}}}_{\mathrm{2}}=T_{2}\hat{\boldsymbol{\lambda}}_{\mathrm{CF}},\;{\accentset% {\rightharpoonup}{\boldsymbol{A}}}=A_{x}\hat{\boldsymbol{\imath}}+A_{z}\hat{% \boldsymbol{k}},\;{\accentset{\rightharpoonup}{\boldsymbol{D}}}=D_{x}\hat{% \boldsymbol{\imath}}+D_{z}\hat{\boldsymbol{k}}.

Here, $\hat{\boldsymbol{\lambda}}_{\mathrm{BE}}=\hat{\boldsymbol{\lambda}}_{\mathrm{% CF}}=-\cos\theta\hat{\boldsymbol{\imath}}+\sin\theta\hat{\boldsymbol{k}}=\hat{% \boldsymbol{\lambda}}({\rm let})$ . Now, we apply moment equilibrium about point A, i.e., $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ .

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}\times{\accentset{% \rightharpoonup}{\boldsymbol{T}}}_{\mathrm{1}}+{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{C}}\times{\accentset{\rightharpoonup}{\boldsymbol{% T}}}_{\mathrm{2}}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G}}% \times{\accentset{\rightharpoonup}{\boldsymbol{W}}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D}}\times{\accentset{% \rightharpoonup}{\boldsymbol{D}}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}

(5.36)

where,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B}}% \times{\accentset{\rightharpoonup}{\boldsymbol{T}}}_{\mathrm{1}}$	$\displaystyle=$	$\displaystyle a\hat{\boldsymbol{\imath}}\times T_{1}\hat{\boldsymbol{\lambda}}% =-aT_{1}\sin\theta\hat{\boldsymbol{\jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}% \times{\accentset{\rightharpoonup}{\boldsymbol{T}}}_{\mathrm{2}}$	$\displaystyle=$	$\displaystyle(a\hat{\boldsymbol{\imath}}+b\hat{\boldsymbol{\jmath}})\times T_{% 2}\hat{\boldsymbol{\lambda}}=T_{2}b\sin\theta\hat{\boldsymbol{\imath}}-T_{2}a% \sin\theta\hat{\boldsymbol{\jmath}}+T_{2}b\cos\theta\hat{\boldsymbol{k}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G}}% \times{\accentset{\rightharpoonup}{\boldsymbol{W}}}$	$\displaystyle=$	$\displaystyle\frac{1}{2}(a\hat{\boldsymbol{\imath}}+b\hat{\boldsymbol{\jmath}}% )\times(-W\hat{\boldsymbol{k}})=-\frac{Wa}{2}\hat{\boldsymbol{\imath}}+\frac{% Wa}{2}\hat{\boldsymbol{\jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D}}% \times{\accentset{\rightharpoonup}{\boldsymbol{D}}}$	$\displaystyle=$	$\displaystyle b\hat{\boldsymbol{\jmath}}\times(D_{x}\hat{\boldsymbol{\imath}}+% D_{z}\hat{\boldsymbol{k}})=D_{z}b\hat{\boldsymbol{\imath}}-D_{x}b\hat{% \boldsymbol{k}}.$

Substituting these products in eqn. (5.36) and dotting with $\hat{\boldsymbol{\imath}},\;\hat{\boldsymbol{\jmath}}$ and $\hat{\boldsymbol{k}}$ , we get

$\displaystyle T_{2}\sin\theta+D_{z}$	$\displaystyle=$	$\displaystyle\frac{W}{2}$	(5.37)
$\displaystyle T_{2}\cos\theta-D_{x}$	$\displaystyle=$	$\displaystyle 0$	(5.38)
$\displaystyle(T_{1}+T_{2})\sin\theta$	$\displaystyle=$	$\displaystyle\frac{W}{2}.$	(5.39)

The force equilibrium, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ , gives

{\accentset{\rightharpoonup}{\boldsymbol{A}}}+{\accentset{\rightharpoonup}{% \boldsymbol{D}}}+{\accentset{\rightharpoonup}{\boldsymbol{T}}}_{\mathrm{1}}+{% \accentset{\rightharpoonup}{\boldsymbol{T}}}_{\mathrm{2}}+{\accentset{% \rightharpoonup}{\boldsymbol{W}}}={\accentset{\rightharpoonup}{\boldsymbol{0}}}.

Again, substituting the forces in their component form and dotting with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{k}}$ (there are no $\hat{\boldsymbol{\jmath}}$ components), we get

$\displaystyle A_{x}+D_{x}-(T_{1}+T_{2})\cos\theta$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}A_{x}-T_{1}\cos\theta$	$\displaystyle=$	$\displaystyle 0$	(5.40)
$\displaystyle A_{z}+D_{z}+(T_{1}+T_{2})\sin\theta$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}A_{z}+T_{1}\sin\theta$	$\displaystyle=$	$\displaystyle\frac{W}{2}.$	(5.41)

These are all the equations that we can get. Now, note that we have five independent equations (eqns. (5.37) to (5.41)) but six unknowns. Thus we cannot solve for the unknowns uniquely. This is an indeterminate structure! No matter which point we use for our moment equilibrium equation, we will always have one more unknown than the number of independent equations. We can, however, solve the problem with an extra assumption (see comments below) — the structure is symmetric about the axis passing through G and parallel to $x$ -axis. From this symmetry we conclude that $T_{1}=T_{2}$ . Then, from eqn. (5.40) we have

2T\sin\theta=\frac{W}{2}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T=\frac{W}{4% \sin\theta}.

We can now find the maximum load that the plate can take subject to the maximum allowable tension in the strings.

$\displaystyle W$	$\displaystyle=$	$\displaystyle 4T\sin\theta$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}W_{\rm max}$	$\displaystyle=$	$\displaystyle 4T_{\rm max}\sin\theta$
	$\displaystyle=$	$\displaystyle 4(100\,{\mbox{N}})\cdot\frac{1}{2}=200\,{\mbox{N}}.$

The total load as given is $(6+4)\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}=98.1\,{\mbox{N}}% \approx 100\,{\mbox{N}}$ . Thus we can double the load before the strings reach their break-points. Now the reactions at D and A follow from eqns. (5.37), (5.38), (5.40), and (5.41).

	$\displaystyle D_{z}=A_{z}=\frac{W}{2}-T\sin\theta$	$\displaystyle=$	$\displaystyle\frac{W}{2}$
	$\displaystyle D_{x}=A_{x}=T\cos\theta$	$\displaystyle=$	$\displaystyle\frac{W}{4}\cot\theta.$

Answer: $W_{\rm max}=200\,{\mbox{N}}$

Comments:

1.

We got only five independent equations (instead of the usual 6) because the force equilibrium in the $y$ -direction gives a zero identity (0 = 0). There are no forces in the $y$ -direction. The structure seems to be free in the $y$ -direction — if you push a little, it will move. This freedom to move in the $y$ -direction arose because we chose to model the hinges at A and D as allowing sliding, keeping in mind only the vertical loading. The actual hinges used on a bookshelf will not allow movement in the $y$ -direction either. If we model the hinges as ball and socket joints, we introduce two more unknowns, one at each joint, and get just one more scalar equation. Thus we are back to square one. There is no way to determine $A_{y}$ and $D_{y}$ from equilibrium equations alone.
2.

The assumption of symmetry and the consequent assumption of equality of the two string tensions is, mathematically, an extra independent equation based on deformations (strength of materials). At this point, you may not know any strength of material calculations or deformation theory, but your intuition is likely to lead you to make the same assumption. Note, however, that this assumption is sensitive to accuracy in fabrication of the structure. If the strings were slightly different in length, the angles were slightly off, or the wall was not perfectly vertical, the symmetry argument would not hold and the two tensions would not be the same.

Most real problems are like this — indeterminate. Our modelling, if good, makes them determinate, solvable and useful.

Problems for 5.5 Advanced statics

Preparatory Problems

5.5.1 In 2D, the force balance and moment-balance equations for equilibrium of a body give three independent scalar equations that can be used to solve for three unknowns. How many independent scalar equations can you get from force and moment balance in 3D? Write down a set of such equations.

5.5.2 In 3D, how many independent scalar equations can you write for equilibrium of a particle?

5.5.3 How is moment-balance equation about an axis different from moment balance about a point? Illustrate your answer with an example.

5.5.4 How many independent scalar equations of equilibrium can you get by writing moment-balance equations about different lines or axes in 3D?

More-Involved Problems

5.5.5 Assume identical uniform rigid blocks with weight $W=1\,{\mbox{N}}$ , height $h=1\,\mathrm{cm}$ , and length $\ell=10\,\mathrm{cm}$ are put one on top of the other. Assume there is no glue so blocks can only push against each other.

(a)

For two blocks what is the biggest overhang $a$ so that the top block does not tip over?
(b)

For three blocks what is the biggest total overhang $=2a$ (the same overhang $a$ at each layer) so that the top block doesn’t tip, nor does the middle block?

Filename:pfigureSoodak4-43

Figure 5.156:

5.5.6 See simpler problem 5.156. Stacking identical rigid blocks, one on top of the other, one wants to get the biggest overhang possible without the tower toppling. Each block has, say, $W=1\,{\mbox{N}}$ , height $h=1\,\mathrm{cm}$ , and length $\ell=10\,\mathrm{cm}$ .

(a)

For three blocks find the biggest $a_{1}$ and $a_{2}$ so there is no toppling. [First put the top block as far to the right as you can, $a_{1}$ , for no toppling. Then put that pair as far to the right as possible for no toppling over the bottom block.] The total overhang is $a_{1}+a_{2}$ .
(b)

For 4 blocks find the largest possible overhang $a_{1}+a_{2}+a_{3}$ by placing the tower of three above as far to the right as possible relative to the bottom block. [Note that you place the center of mass of the top 3 blocks over the right edge of the fourth bottom block].
(c)

For $n$ blocks what is the biggest possible overhang ( $=a_{1}+a_{2}+a_{3}+\dots+a_{n-1}$ )?
(d)

Using blocks with length $\ell=10\,\mathrm{cm}$ how many blocks $n$ are needed to get an overhang of $1\,\mathrm{m}$ ? $2\,\mathrm{m}$ ?

Filename:pfigureSoodak4-45

Figure 5.157:

5.5.7 Uniform plate ADEH with mass $m$ is connected to the ground with a ball and socket joint at A. It is also held by three massless bars (IE, CH and BH) that have ball and socket joints at each end, one end at the rigid ground (at I, C and B) and one end on the plate (at E and H).

In terms of some or all of $m, g,$ and $L$ find

(a)

the reaction at A (the force of the ground on the plate),
(b)

$T_{\rm IE}$ ,
(c)

$T_{\rm CH}$ ,
(d)

$T_{\rm BH}$ .

Filename:pfigure-01-f1

Figure 5.158:

5.5.8 An 80 kg square table has one quarter cut away. The remaining 60 kg are supported on 3 massless legs on a level floor. Use $g=10\mbox{N/kg}$ . What is the load carried by leg AB? (State your assumptions clearly.) Answer: Assuming no side-loads from floor the support from leg AB is $250\,{\mbox{N}}$ , $T_{AB}=-250\,{\mbox{N}}$ .

Filename:S03q4p7table

Figure 5.159:

5.5.9 Uniform plate ADEH with mass $m$ is connected to the ground with a ball and socket joint at A. It is also held by three massless bars (CE, CH and BH) that have ball and socket joints at each end, one end at the rigid ground (at C and B) and one end on the plate (at E and H). In terms of some or all of $m, g,$ and $L$ find the reaction at A (the force of the ground on the plate) and the three bar tensions $T_{CE}$ , $T_{CH}$ and $T_{BH}$ . Answer: $T_{CE}=mg/2,T_{CH}=T_{BH}=-0,A_{z}=mg/2,A_{y}=A_{x}=0$

Filename:F01final1-plate

Figure 5.160:

5.5.10 A massless triangular plate rests against a frictionless wall at point $D$ and is rigidly attached to a massless rod supported by two ideal bearings fixed to the floor. A ball of mass $m$ is fixed to the centroid of the plate. There is gravity and the system is at rest. What is the reaction at point $D$ on the plate?

Filename:ch3-1b

Figure 5.161:

5.5.11 A uniform equilateral triangular plate with weight $W=1000N$ and sides $\ell=2\,\mathrm{m}$ rests against a slippery plane S. Point C is negligibly above the $x y$ plane. The bottom edge of the triangle has ball-and-socket joints at A and B, with the line AB on the $x y$ plane making an angle of $15^{{}^{\circ}}$ with the $x$ direction.

(a)

Find the reaction at C
(b)

Find all you can about the reactions at A and B.

Filename:pfigure-restingtriangle

Figure 5.162:

5.5.12 A uniform $5\,\mathrm{kg}$ shelf is supported at one corner with a ball and socket joint and the other three corners with strings. At the instant of interest the shelf is at rest. Gravity acts in the $-\hat{\boldsymbol{k}}$ direction. The shelf is in the $x y$ plane.

(a)

Draw a FBD of the shelf.
(b)

Challenge: without doing any calculations on paper can you find one of the reaction force components or the tension in any of the cables? Give yourself a few minutes of staring to try to find this force. If you can’t, then come back to this question after you have done all the calculations.
(c)

Write down the equation of force equilibrium.
(d)

Write down the moment-balance equation using the center of mass as a reference point.
(e)

By taking components, turn (b) and (c) into six scalar equations in six unknowns.
(f)

Solve these equations by hand or on the computer.
(g)

Instead of using a system of equations try to find a single equation which can be solved for $T_{EH}$ . Solve it and compare to your result from before. Answer: $T_{EH}=0$ as you can find a number of ways.
(h)

Challenge: For how many of the reactions can you find one equation which will tell you that particular reaction without knowing any of the other reactions? [Hint, try moment balance about an appropriate axis as well as force balance in an appropriate direction. It is possible to find five of the six unknown reaction components this way.] Must these solutions agree with (f)? Do they?

Filename:pfigure-s94h2p10-a

Figure 5.163:

5.5.13 The sign is held up by 6 rods. Find the tension in bars

(a)

BH Answer: Use axis EC.
(b)

EB Answer: Use axis AH.
(c)

AE Answer: Use $\hat{\boldsymbol{\jmath}}$ axis through B.
(d)

IA Answer: Use axis DE.
(e)

JD Answer: Use axis EH.
(f)

EC Answer: Can’t do in one shot.

[One game you can play is to see how many of the tensions you can find without knowing any of the others. Another approach is to set up and solve 6 equations in 6 unknowns.]

Filename:pfigure-crookedsign

Figure 5.164:

5.5.14 The 100 kg, 2 m square, uniform sign KHNA is held up by 6 bars.
Structure and geometry clarifications: The sign is held vertically, 1m in front of, and orthogonal-to a vertical wall. Each bar holding the sign has a ball-and-socket joint both where it attaches to the sign and where it attaches to the wall. The points L, M, J, I, K, P and H lie in the same horizontal plane that includes the top edge of the sign. The points M, O, and C lie on a vertical line that is coplanar with the sign. Points B, O, D, A, and N lie in a horizontal plane shared with the bottom edge of the sign. The center of mass of the sign is at G. $g=10\mbox{N/kg}$ .

(a)

Find the “bar force” in bar AC.
[hint: $\Sigma F_{z}=\left\{\Sigma{\accentset{\rightharpoonup}{\boldsymbol{F}}}\right% \}\cdot\hat{\boldsymbol{k}}=0$ ]. Answer: $T_{AC}=-\sqrt{2}mg=-1000\sqrt{2}\,{\mbox{N}}\approx-1410\,{\mbox{N}}$ (the bar is in compression)
(b)

Find the “bar force” in bar IP.
[hint: $\Sigma M_{AK}=\left\{\Sigma{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/A}% \right\}\cdot\hat{\boldsymbol{k}}=0$ ]. Answer: $T_{IP}=0$
(c)

Find the “bar force” in bar KL. Answer: $T_{KL}=\sqrt{2}mg/6=\left(1000\sqrt{2}/6\right)N\approx 408\,{\mbox{N}}$ (the bar is in tension)

Filename:F02p1p2sign

Figure 5.165:

5.5.15 Below is a highly schematic picture of a tricycle. The wheels are at C, B and A. The person-trike system has center of mass at G directly over the rear axle. The wheels at C and A are good free-turning, high friction wheels. The wheel at B is in a small ditch and can’t move. Assume no slip and that $F,m,g,w,\ell,$ and $h$ are given.

(a)

Of the 9 possible reaction components at A, B, and C, which do you know are zero a priori.
(b)

Find all the reaction components (the full reaction force) at A.
(c)

Find the vertical component of the reaction at C.
(d)

Find the $x$ and $z$ reaction components at B.
(e)

Find the sum of the $y$ components of the reactions at B and C.
(f)

Can you find the $y$ component of the reaction at C? Why or why not?

Filename:pfigure-trikerock

Figure 5.166:

5.5.16 A 3-wheeled robot with mass $m$ is parked on a hill with slope $\theta$ . The ideal massless robot wheels are free to roll but not to slip sideways. The robot steering mechanism has turned the wheels so that wheels at A and C are free to roll in the $\hat{\boldsymbol{\jmath}}$ direction and the wheel at B is free to roll in the $\hat{\boldsymbol{\imath}}$ direction. The center of mass of the robot at G is $h$ above (normal to the slope) the trailer bed and symmetrically above the axle connecting wheels A and B. The wheels A and B are a distance $b$ apart. The length of the robot is $\ell$ .

Find the force vector ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{A}$ of the ground on the robot at A in terms of some or all of $m,g,\ell,\theta,b,h,\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},$ and $\hat{\boldsymbol{k}}$ . Answer: Hint: With reference to a free-body diagram of the robot, use moment balance about axis BC.

Filename:pfigure-threewheelparked

Figure 5.167:

$\displaystyle\hat{\boldsymbol{\lambda}}_{\mathrm{AB}}$	$\displaystyle=$	$\displaystyle\frac{-4\hat{\boldsymbol{\imath}}+3\hat{\boldsymbol{\jmath}}+12% \hat{\boldsymbol{k}}}{\sqrt{4^{2}+3^{2}+12^{2}}}=-\frac{4}{13}\hat{\boldsymbol% {\imath}}+\frac{3}{13}\hat{\boldsymbol{\jmath}}+\frac{12}{13}\hat{\boldsymbol{% k}}$
$\displaystyle\hat{\boldsymbol{\lambda}}_{\mathrm{AC}}$	$\displaystyle=$	$\displaystyle-\hat{\boldsymbol{\jmath}}$
$\displaystyle\hat{\boldsymbol{\lambda}}_{\mathrm{AD}}$	$\displaystyle=$	$\displaystyle\frac{12\hat{\boldsymbol{\imath}}+5\hat{\boldsymbol{k}}}{\sqrt{12% ^{2}+5^{2}}}=\frac{12}{13}\hat{\boldsymbol{\imath}}+\frac{5}{13}\hat{% \boldsymbol{k}}.$

$\displaystyle-\frac{4}{13}T_{\rm AB}+\frac{12}{13}T_{\rm AD}$	$\displaystyle=$	$\displaystyle 0$
$\displaystyle\frac{3}{13}T_{\rm AB}-T_{\rm AC}$	$\displaystyle=$	$\displaystyle 0$	(5.11)
$\displaystyle\frac{12}{13}T_{\rm AB}+\frac{5}{13}T_{\rm AD}$	$\displaystyle=$	$\displaystyle mg.$

	$\displaystyle-F\cos\alpha-\mu N+mg\sin\theta$	$\displaystyle=$	$\displaystyle 0$		(5.24)
	$\displaystyle F\sin\alpha+N-mg\cos\theta$	$\displaystyle=$	$\displaystyle 0.$		(5.25)

	$\displaystyle F(\cos\alpha-\mu\sin\alpha)$	$\displaystyle=$	$\displaystyle mg(\sin\theta-\mu\cos\theta)$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}F$	$\displaystyle=$	$\displaystyle\frac{\sin\theta-\mu\cos\theta}{\cos\alpha-\mu\sin\alpha}\,mg.$		(5.26)

	$\displaystyle-F\cos\alpha-F_{f}+mg\sin\theta$	$\displaystyle=$	$\displaystyle 0$
	$\displaystyle F\sin\alpha+N-mg\cos\theta$	$\displaystyle=$	$\displaystyle 0$

	$\displaystyle F_{f}$	$\displaystyle=$	$\displaystyle mg\sin\theta-F\cos\alpha$
	$\displaystyle N$	$\displaystyle=$	$\displaystyle mg\cos\theta-F\sin\alpha.$

	$\displaystyle F_{f}$	$\displaystyle=$	$\displaystyle 10\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}\cdot\sin 30% ^{{}^{\circ}}-20\,{\mbox{N}}\cdot\cos 30^{{}^{\circ}}=31.73\,{\mbox{N}}$
	$\displaystyle N$	$\displaystyle=$	$\displaystyle 10\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}\cdot\cos 30% ^{{}^{\circ}}-20\,{\mbox{N}}\cdot\sin 30^{{}^{\circ}}=74.96\,{\mbox{N}}.$

	$\displaystyle-F\cos\alpha+\mu N+mg\sin\theta$	$\displaystyle=$	$\displaystyle 0$
	$\displaystyle F\sin\alpha+N-mg\cos\theta$	$\displaystyle=$	$\displaystyle 0$

	$\displaystyle F$	$\displaystyle=$	$\displaystyle\frac{\sin\theta+\mu\cos\theta}{\cos\alpha+\mu\sin\alpha}\,mg$
		$\displaystyle=$	$\displaystyle\frac{1.216}{1.216}\,mg=mg.$

$\displaystyle\mu W$	$\displaystyle=$	$\displaystyle W\frac{d\cos\theta}{\ell\sin\theta}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}d$	$\displaystyle=$	$\displaystyle\mu\ell\tan\theta$
	$\displaystyle=$	$\displaystyle 0.5\cdot(4\,\mathrm{m})\cdot\tan 60^{\circ}=3.46\,\mathrm{m}.$

$\displaystyle mgR\sin\theta$	$\displaystyle=$	$\displaystyle T(R\cos\theta+r)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}T$	$\displaystyle=$	$\displaystyle mg\,\frac{\sin\theta}{\cos\theta+r/R}$
	$\displaystyle=$	$\displaystyle 2\,\mathrm{kg}\cdot 9.81\,\mathrm{m}/\mathrm{s}^{2}\cdot\frac{% \frac{\sqrt{3}}{2}}{\frac{1}{2}+\frac{.2\,\mathrm{m}}{.5\,\mathrm{m}}}$
	$\displaystyle=$	$\displaystyle 18.88\,{\mbox{N}}.$

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}% \times(-mg\hat{\boldsymbol{\jmath}})$	$\displaystyle=$	$\displaystyle-mgR(\hat{\boldsymbol{n}}\times\hat{\boldsymbol{\jmath}})$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/A}}% \times T\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle T[R(\hat{\boldsymbol{n}}\times\hat{\boldsymbol{\imath}})+r(\hat{% \boldsymbol{\jmath}}\times\hat{\boldsymbol{\imath}})].$

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/A}}% \times(-mg\hat{\boldsymbol{\jmath}})$	$\displaystyle=$	$\displaystyle mgR\sin\theta\hat{\boldsymbol{k}},$
	$\displaystyle\hbox to36.135pt{and\hfil}{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{B/A}}\times T\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle-TR\cos\theta\hat{\boldsymbol{k}}-r\hat{\boldsymbol{k}}.$

$\displaystyle F_{s}$	$\displaystyle=$	$\displaystyle-T(\underbrace{\hat{\boldsymbol{\imath}}\cdot\hat{\boldsymbol{% \lambda}}}_{\cos\theta})+mg(\underbrace{\hat{\boldsymbol{\jmath}}\cdot\hat{% \boldsymbol{\lambda}}}_{\sin\theta})$
	$\displaystyle=$	$\displaystyle-T\cos\theta+mg\sin\theta$
	$\displaystyle=$	$\displaystyle-18.88\,{\mbox{N}}(1/2)+19.62\,{\mbox{N}}(\sqrt{3}/2)$
	$\displaystyle=$	$\displaystyle 7.55\,{\mbox{N}}.$

$\displaystyle N$	$\displaystyle=$	$\displaystyle-T(\hat{\boldsymbol{\imath}}\cdot\hat{\boldsymbol{n}})+mg(\hat{% \boldsymbol{\jmath}}\cdot\hat{\boldsymbol{n}})$
	$\displaystyle=$	$\displaystyle T\sin\theta+mg\cos\theta$
	$\displaystyle=$	$\displaystyle 18.88\,{\mbox{N}}(\sqrt{3}/2)+19.62\,{\mbox{N}}(1/2)$
	$\displaystyle=$	$\displaystyle 26.16\,{\mbox{N}}.$