Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap1 Vectors: position, force and moment 3 Computation: Algebra, ODEs, etc.

Chapter 2 Free-Body Diagrams

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2.0Introductionstudy PDF
2.1Equivalent force setsstudy PDF + solutions
2.2Center of mass and gravitystudy PDF + solutions
2.3Interactions, forces & partial FBDsstudy PDF + solutions
2.4Contact forces: Sliding, rolling, and collisionsstudy PDF

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A free-body diagram is a sketch of the system to which you will apply the laws of mechanics. The diagram shows all of the non-negligible external forces and couples which act on the system. The diagram tells what material is in the system and also what is known, and what is not known, about the forces. Mechanics reasoning depends on free-body diagrams. So, we give tips about how to avoid common, and often calamitous, FBD errors.

We also discuss two key related ideas:

1) Equivalent force systems; and
2) Center of Gravity (or, Center of Mass)

On a free-body diagram, systems of forces are often replaced with ‘equivalent’ forces, a special important case of which is a weight force at the center of gravity. The center of gravity is also the Center of Mass, a key concept in Dynamics.

First, a language issue. Traditionally in mechanics, like in Newton’s and Shakespeare’s times, a system, or object, was often called a body (as in ‘heavenly bodies’). When such a system is conceptually isolated it is free, as in free from its surroundings. So, translating to modern language,

A free-body diagram (FBD) is a sketch of an isolated system and of the external forces, from outside the system, which act on it.

The laws of mechanics use the forces shown on a free-body diagram, and no other forces. Thus, as we can’t repeat too much (but will attempt to), drawing good free-body diagrams is essential for both statics and dynamics. The skills for drawing these diagrams are presented in the following sections.

Some basic mechanics assumptions

Isolated systems. One way to understand something is to isolate it, see how it behaves on its own, and see how it responds to various stimuli. Then, when the thing is not isolated, you still think of it as isolated, but think of the effects of all its surroundings as stimuli. Reversing the point of view, we can also see the system’s behavior as causing stimulus to other things around it, which themselves can be thought of as isolated and stimulating back, and so on.

This reductionist approach is used throughout the physical and social sciences. A tobacco plant is understood in terms of its response to light, heat flow, the chemical environment, insects, and viruses. The economy of Singapore is understood in terms of the flow of money and goods in and out of Singapore. And social behavior is regarded as being a result of individuals reacting to the sights, sounds, smells, and touch of other individuals and thus causing sights, sounds, smell and touch that the others react to in turn, etc.The concept of an isolated system is central in most of the reductionist sciences

^†^†margin: Closed systems in thermodynamics.The isolated system approach to understanding is made most clear in thermodynamics courses. A system, usually a fluid, is isolated with rigid walls that allow no heat, motion or material to pass. Then, bit by bit, as the subject is developed, the response of the system to certain interactions across the boundaries is allowed. Eventually, enough interactions are understood that the system can be viewed as isolated even when in a useful context. The gas expanding in a refrigerator follows the same rules of heat-flow and work as when it was expanded in its ‘isolated’ container.

The “free-body” is a closed system. As in elementary thermodynamics, we only consider here so-called closed systems. A (closed) system, in mechanics, is a fixed collection of material. You can draw an imaginary boundary around a system, then in your mind paint all the atoms inside the boundary red, and then define the system as being the red atoms, no matter whether they later cross the original spatial boundary markers or not^†^†margin: Open Systems. The mechanics of open systems, where material crosses the system boundaries, is important in fluid mechanics. Such open fluid systems are first seen in some elementary dynamics problems (like rockets), where material is allowed to cross the system boundaries. But the equations governing these open systems are deduced from careful application of the more fundamental governing mechanics equations of closed systems. So we have to master the mechanics of closed systems first. .

Mechanics, as used by engineers, depends on bits of matter as being durable and non-ephemeral. Violating the theory of relativity:

A given bit of matter exists forever, has the same mass forever. A given bit of matter is either in a given system, or out of that system, for all time.

Mechanics is based on the notion that any part of a system (i.e., a subsystem) is itself a system . And we assume that all interactions between systems or subsystems have certain simple rules, most basically:

Force is the measure of mechanical interaction.

Thus a person can be moved by forces, but not by the sight of a tree falling towards them or the attractive smell of a flower. Sights and smells may cause, by rules that fall outside of mechanics, forces (on the feet, etc.) that move a person. When a person moves towards a flower or away from a falling tree she is moved by the force of the ground on her feet not by sight, sound, smell or fear. Finally,

The principle of “action and reaction”: what one system does to another, the other does back to the first.

When a person accelerates away from a falling tree because of the force on the ground, her feet push equally hard on the ground the other direction. Of course, when you think of this you think the other way around. To start running, you make a quick action, pushing on the ground with your feet. The reaction force accelerates you. But, what causes what is not the issue. Rather, if system A is pushed by B then B is pushed back equally by A. In kindergarten talk, ‘it doesn’t matter who started it.’

The rules above, which we call the zeroth^†^†margin: Why do we awkwardly number the most basic laws as zero? Because they are really more of an underlying assumption, a background concept, than a law. As a law they are a little imprecise since force has not yet been defined. You could take the latter two of these zeroth laws as an implicit and partial definition of force. The phrase “zeroth law” means “important implicit assumption”. The third of the zeroth laws is usually called “Newton’s third law.”

laws of mechanics, imply that all the mechanical effects of the outside world on a system can be represented by a sketch of the system with arrows showing the forces of interaction. If we want to know how the system, in turn, affects some part of its surroundings, we draw the opposite arrows on a sketch of that outside part.

Every single atomic force? OK, it is not practical to show every force from every atom that acts on a system. We cluster the forces, add them up and replace them with a single ‘equivalent’ force. ‘Equivalent forces’ are described in the next section. A special case of an equivalent force is the replacement of the billions of gravity forces acting on all the atoms in a system with a single force acting at the center of gravity, as shown in the next section.

Draw good free-body diagrams. The final two sections of this chapter tell you how to draw free-body diagrams. Draw a good free-body diagram and then well-use the vector skills of Chapter 1 and you have near-guaranteed success at doing mechanics problems.

2.1 Equivalent force sets

The definition of the word equivalent

^†^†margin: Other phrases used to describe the equivalent force sets include:
* statically equivalent,
* mechanically equivalent, and
* equipollent.

when applied to force sets in mechanics, is:

Two force systems are said to be equivalent if they have the same sum (the same resultant, the same net force) and the same resultant moment (net moment) about any one point C.

So that we don’t get confused about the system of interest (a collection of material), and the force system (a collection of forces), we will sometimes call the force system a force set.

Why use equivalent force sets?

First, the replacement of one set with an equivalent set is often used to help simplify or solve mechanics problems.

Second, the concept of equivalent force sets allows us to define a couple, a concept we will use throughout the book.

And finally, most often one does not want to know the complete details of all the forces acting on a system. When you think of the force of the ground on your bare foot you do not think of the thousands of little forces at each micro-asperity or the billions and billions of molecular interactions between the wood (say) and your skin. Instead you think of some kind of equivalent force.

In what way are two ‘equivalent’ force systems (or sets) actually equivalent? Well, because all that the equations of mechanics know about forces is their net force and net moment, you have a criterion. You replace the actual force set with a simpler force set, possibly just a single well-placed force, that has the same total force and same total moment with respect to a reference point C^†^†margin: A special point C? No, any point will do. As we will show, equivalent for C implies equivalent for D, E and G etc.. (see box 2.1 on page 2.1

We have already discussed two important cases of equivalent force sets.

1.

On page 1.1 we stated the mechanics assumption that a set of forces applied at one point is equivalent to a single resultant force, their sum, applied at that point. Thus when doing a mechanics analysis you can replace a collection of forces at a point with their sum. If you think of your whole foot as a ‘point’, this justifies the replacement of the billions of little atomic ground contact forces with a single force.
2.

On page 1.49 we discovered that a force applied at a different point is equivalent to the same force applied at a point displaced in the direction of the force. You can thus harmlessly slide the point of force application along the line of the force and have its mechanically equivalent effect: One force applied at a point is equivalent to the same force at a different point that is displaced in the direction of the force.

More generally, we can compare two sets of forces. The first set consists of

{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{1},{\accentset{% \rightharpoonup}{\boldsymbol{F}}}^{(1)}_{2},{\accentset{\rightharpoonup}{% \boldsymbol{F}}}^{(1)}_{3},\textit{etc.}

applied at positions

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{1/C}^{(1)},{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{2/C}^{(1)},{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{3/C}^{(1)},\textit{etc.}

In short hand, these forces are

${\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i}$ applied at positions ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i/C}^{(1)}$ ,

where each value of $i$ describes a different force ( $i=7$ refers to the seventh force in the set). The second set of forces consists of

${\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(2)}_{j}$ applied at positions ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{j/C}^{(2)}$ ,

where each value of $j$ describes a different force in the second set.

Now, we compare the net (resultant) force and net moment of the two sets.

Filename:tfigure8-rel-ang-vel

Figure 2.1: The force set

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{A}},{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{C}}

is equivalent to the force set

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{0}},{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{B}}

If,

{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{\rm tot}={\accentset{% \rightharpoonup}{\boldsymbol{F}}}^{(2)}_{\rm tot}\qquad\mbox{and}\qquad{% \accentset{\rightharpoonup}{\boldsymbol{M}}}^{(1)}_{\rm C}={\accentset{% \rightharpoonup}{\boldsymbol{M}}}^{(2)}_{\rm C},

(2.1)

then the two sets are equivalent.

Here we have defined the net forces and net moments by

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{\mbox{tot}}$	$\displaystyle=$	$\displaystyle\displaystyle\sum_{\mbox{\tiny all forces $i$}}{\accentset{% \rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i},\par$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}^{(1)}_{\rm C}$	$\displaystyle=$	$\displaystyle\displaystyle\sum_{\mbox{\tiny all forces $i$}}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{i/C}^{(1)}\times{\accentset{\rightharpoonup% }{\boldsymbol{F}}}^{(1)}_{i},$		(2.2)
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(2)}_{\rm tot}$	$\displaystyle=$	$\displaystyle\displaystyle\sum_{\mbox{\tiny all forces $j$}}{\accentset{% \rightharpoonup}{\boldsymbol{F}}}^{(2)}_{j},\quad\mbox{and}\$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}^{(2)}_{\rm C}$	$\displaystyle=$	$\displaystyle\displaystyle\sum_{\mbox{\tiny all forces $j$}}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{j/C}^{(2)}\times{\accentset{\rightharpoonup% }{\boldsymbol{F}}}^{(2)}_{j}.$		(2.3)

If you find the $\sum$ (sum) symbol intimidating see box 2.1 on page 2.1.

Example:

Consider force set (1) with forces ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{A}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{C}}$ and force set (2) with forces ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{0}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{B}}$ as shown in fig. 2.1. Are the sets equivalent? First check the sum of forces.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{\rm tot}$	$\displaystyle\mathrel{\mathop{=}\limits^{?}}$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(2)}_{\rm tot}$
$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i}$	$\displaystyle\mathrel{\mathop{=}\limits^{?}}$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(2)}_{j}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{A}}+{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{C}}$	$\displaystyle\mathrel{\mathop{=}\limits^{?}}$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{0}}+{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{B}}$
$\displaystyle 1\,{\mbox{N}}\hat{\boldsymbol{\imath}}+2\,{\mbox{N}}\hat{% \boldsymbol{\jmath}}$	$\displaystyle\mathrel{\mathop{=}\limits^{\surd}}$	$\displaystyle(1\,{\mbox{N}}\hat{\boldsymbol{\imath}}+1\,{\mbox{N}}\hat{% \boldsymbol{\jmath}})+1\,{\mbox{N}}\hat{\boldsymbol{\jmath}}$

Then check the sum of moments about C.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}^{(1)}_{\rm C}$	$\displaystyle\mathrel{\mathop{=}\limits^{?}}$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}^{(2)}_{\rm C}$
$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i/C}^{(1)}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i}$	$\displaystyle\mathrel{\mathop{=}\limits^{?}}$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{j/C}^{(2)}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(2)}_{j}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/C}}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{A}}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/C}}\times{\accentset% {\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{C}}$	$\displaystyle\mathrel{\mathop{=}\limits^{?}}$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0/C}}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{0}}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{B/C}}\times{\accentset% {\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{B}}$
$\displaystyle(-1\,\mathrm{m}\hat{\boldsymbol{\imath}}+1\,\mathrm{m}\hat{% \boldsymbol{\jmath}})\times 1\,{\mbox{N}}\hat{\boldsymbol{\imath}}+{\accentset% {\rightharpoonup}{\boldsymbol{0}}}\times 2\,{\mbox{N}}\hat{\boldsymbol{\jmath}}$	$\displaystyle\mathrel{\mathop{=}\limits^{?}}$	$\displaystyle(-1\,\mathrm{m}\hat{\boldsymbol{\imath}})\times(1\,{\mbox{N}}\hat% {\boldsymbol{\imath}}+1\,{\mbox{N}}\hat{\boldsymbol{\jmath}})+1\,\mathrm{m}% \hat{\boldsymbol{\jmath}}\times 1\,{\mbox{N}}\hat{\boldsymbol{\jmath}}$
$\displaystyle-1\,\mathrm{m}\,{\mbox{N}}\hat{\boldsymbol{k}}$	$\displaystyle\mathrel{\mathop{=}\limits^{\surd}}$	$\displaystyle-1\,\mathrm{m}\,{\mbox{N}}\hat{\boldsymbol{k}}$

So, the two force sets are equivalent.

What is so special about the point C in the example above? Nothing.

If two force sets are equivalent with respect to some point C, they are equivalent with respect to any point, as justified in box 2.1 on page 2.1.

Filename:tfigure8-ang-vel-ex

Figure 2.2: Frictionless wheel bearing. All the bearing forces are equivalent to a single force acting at the center of the wheel.

For example, both of the force sets in the example above have the same moment of $2\,{\mbox{N}}\,\mathrm{m}\hat{\boldsymbol{k}}$ about the point A. See box 2.1 for the proof of the general case.

Example: Frictionless wheel bearing

If the contact of an axle with a bearing housing is perfectly frictionless then each of the contact forces has no moment about the center of the wheel (fig. 2.2). Thus the whole force set is equivalent to a single force at the center of the wheel.

Box 2.1 $\displaystyle\sum$ means add

In mechanics, we often need to add up lots of things: all the forces on a body, all the moments they cause, all the mass of a system, etc. One notation for adding up all 14 forces on some body is

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}+{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}+{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{3}}+{\accentset{\rightharpoonup% }{\boldsymbol{F}}}_{\!\!\mathrm{4}}+{\accentset{\rightharpoonup}{\boldsymbol{F% }}}_{\!\!\mathrm{5}}+{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{6}}+{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{7}}$
			$\displaystyle+{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{8}}+{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{9}}+{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{10}}+{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{11}}+{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{12}}+{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{13}}+{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{14}}.$

which is a bit long, so we might abbreviate it as

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}

\displaystyle=

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}+{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}+\dots+{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{14}}.

But this is definition by pattern recognition. A more explicit statement would be

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}

\displaystyle=

\displaystyle\mbox{The sum of all 14 forces}{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{\!\!\mathrm{i}}\ \mbox{where $i=1\dots 14$}

which is too space consuming. This kind of summing is so important that mathematicians use up a whole letter of the greek alphabet as a short hand for ‘the sum of all’. They use the capital greek ’S’ (for Sum) called sigma which looks like this:

\sum.

When you read $\sum$ aloud you don’t say ‘S’ or‘sigma’ but rather ‘the sum of.’ The $\sum$ (sum) notation may remind you of infinite series, and convergence thereof. We will rarely be concerned with infinite sums in this book, and never with convergence issues. So, panic on those grounds is unjustified. We just want to write easily about adding things. For example, we use the $\sum$ (sum) to write the sum of 14 forces ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}}$ explicitly and concisely as

\sum_{i=1}^{14}{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}},

and say ‘the sum of F sub i where i goes from one to fourteen’. Sometimes we don’t know, say, how many forces are being added. We just want to add all of them so we write (a little informally)

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}}\quad\mbox{% meaning}\quad{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}+{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}+\mbox{{etc.}},

where the subscript $i$ lets us know that the forces are numbered.

Rather than panic when you see something mathy looking like $\displaystyle\sum_{i=1}^{14}$ , just relax and think: oh, we want to add up a bunch of things all of which look like the next thing written. In general,

\sum{\mbox{(thing)}}_{i}\quad\mbox{translates to}\quad{\mbox{(thing)}}_{1}+{% \mbox{(thing)}}_{2}+{\mbox{(thing)}}_{3}+\mbox{{etc.}}

no matter how intimidating the ‘thing’ is. In time, you can skip the translation into English and will enjoy the concise notation.

Box 2.2 Two force sets that are equivalent for one reference point are equivalent for all reference points.

Consider two sets of forces ${\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i}$ and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(2)}_{j}$ with corresponding points of application ${\rm P}^{(1)}_{i}$ and ${\rm P}^{(2)}_{j}$ at positions relative to the origin of ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i}^{(1)}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{j}^{(2)}$ . To simplify the discussion let’s define the net forces of the two systems as

{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{\rm tot}\equiv\sum{% \accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i}\quad\mbox{and}\quad{% \accentset{\rightharpoonup}{\boldsymbol{F}}}^{(2)}_{\rm tot}\equiv\sum{% \accentset{\rightharpoonup}{\boldsymbol{F}}}^{(2)}_{j},

and the net moments about the origin as

{\accentset{\rightharpoonup}{\boldsymbol{M}}}^{(1)}_{0}\equiv\sum{\accentset{% \rightharpoonup}{\boldsymbol{r}}}^{(1)}_{i}\times{\accentset{\rightharpoonup}{% \boldsymbol{F}}}^{(1)}_{i}\quad\mbox{and}\quad{\accentset{\rightharpoonup}{% \boldsymbol{M}}}^{(2)}_{0}\equiv\sum{\accentset{\rightharpoonup}{\boldsymbol{r% }}}^{(2)}_{j}\times{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(2)}_{j}.

Using point 0 as a reference, the statement that the two sets are equivalent is then ${\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{\rm tot}={\accentset{% \rightharpoonup}{\boldsymbol{F}}}^{(2)}_{\rm tot}$ and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}^{(1)}_{0}={\accentset{% \rightharpoonup}{\boldsymbol{M}}}^{(2)}_{0}$ . Now consider point C with position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/0}}=-{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0/C}}$ . What is the net moment of force set (1) about point C?

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}^{(1)}_{\rm C}$	$\displaystyle\equiv$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}^{(1)}_{i/\rm C}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i}$
	$\displaystyle=$	$\displaystyle\sum\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}^{(1)}_{i}% -{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}\right)\times{% \accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i}$
	$\displaystyle=$	$\displaystyle\sum\left({\accentset{\rightharpoonup}{\boldsymbol{r}}}^{(1)}_{i}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i}-{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i}\right)$
	$\displaystyle=$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}^{(1)}_{i}\times% {\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i}-\sum{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i}$
	$\displaystyle=$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}^{(1)}_{i}\times% {\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i}-{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}\times\left(\sum{\accentset{% \rightharpoonup}{\boldsymbol{F}}}^{(1)}_{i}\right)$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}^{(1)}_{0}-{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C}}\times{\accentset{% \rightharpoonup}{\boldsymbol{F}}}^{(1)}_{\rm tot}.$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}^{(1)}_{0}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0/C}}\times{\accentset% {\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{\rm tot}.$

[ Note. The calculation above uses the ‘move’ (A math operation is kind of analogous to a chess move) of factoring a constant vector out of a sum. This move, of using the distributive rule backwards (undistributing), is used often in mechanics theory. ]

Similarly, for force set (2):

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}^{(2)}_{\rm C}

\displaystyle=

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}^{(2)}_{0}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{0/C}}\times{\accentset% {\rightharpoonup}{\boldsymbol{F}}}^{(2)}_{\rm tot}.

If the two force sets are equivalent for reference point 0 then ${\accentset{\rightharpoonup}{\boldsymbol{F}}}^{(1)}_{\rm tot}={\accentset{% \rightharpoonup}{\boldsymbol{F}}}^{(2)}_{\rm tot}$ and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}^{(1)}_{0}={\accentset{% \rightharpoonup}{\boldsymbol{M}}}^{(2)}_{0}$ and the expressions above imply that ${\accentset{\rightharpoonup}{\boldsymbol{M}}}^{(1)}_{\rm C}={\accentset{% \rightharpoonup}{\boldsymbol{M}}}^{(2)}_{\rm C}$ . Because we specified nothing special about the point C, the systems are equivalent for any reference point. Thus, to demonstrate equivalence we need to use a reference point, but once equivalence is demonstrated we need not name the point since the equivalence holds for all points.

By the same reasoning we find that once we know the net force and net moment of a force system ( ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{tot}}$ ) relative to some point C (call it ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{C}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{C}$}}}$ ), we know the net moment relative to point D as

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

Note that if the net force is ${\accentset{\rightharpoonup}{\boldsymbol{0}}}$ (and the force set is then called a couple) that ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{D}}={\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathrm{C}}$ so the net moment is the same for all reference points.

Couples

Filename:tfigure8-ang-vel

Figure 2.3: One couple. The forces add to zero. The net moment they cause does not.

Consider a pair of equal and opposite forces that are not collinear. Such a pair is called a couple.

The net moment caused by a couple is the size of the force times the perpendicular distance between the two lines of action and doesn’t depend on the reference point. In fact, any force system that has ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{\rm tot}}={% \accentset{\rightharpoonup}{\boldsymbol{0}}}$ causes the same moment about all different reference points (as shown at the end of box 2.1).

So, in modern usage, any force system with any number of forces and with ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{\rm tot}}={% \accentset{\rightharpoonup}{\boldsymbol{0}}}$ is called a couple. A couple is described by its net moment.

^†^†margin: People who have been in difficult long-term relationships don’t need a mechanics textbook to learn that a couple is a pair of equal and opposite forces that push each other around.

A couple is any force system that has a total force of ${\accentset{\rightharpoonup}{\boldsymbol{0}}}$ . It is described by the net moment ${\accentset{\rightharpoonup}{\boldsymbol{M}}}$ that it causes.

We then think of the couple ${\accentset{\rightharpoonup}{\boldsymbol{M}}}$ as representing an equivalent force system that contributes ${\accentset{\rightharpoonup}{\boldsymbol{0}}}$ to the net force and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}$ to the net moment with respect to every reference point.

The concept of a couple (also called an applied moment or an applied torque) is especially useful for representing the net effect of a complicated collection of forces that causes some turning. The complicated set of electromagnetic forces turning a motor shaft can be replaced by a single couple.

Every set of forces is equivalent to a force and a couple

Given any point C, we can calculate the net moment of a system of forces relative to C. We then can replace the sum of forces with a single force at C and the net moment with a couple at C and we have an equivalent force system.

A force system is equivalent to a force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{\rm tot}$ acting at C and a couple $M$ equal to the net moment of the forces about C, i.e., ${\accentset{\rightharpoonup}{\boldsymbol{M}}}={\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{\mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{C}$}}{% \lower 1.0pt\hbox{$\textstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptstyle% \mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle\mathrm{C}$}}}$ .

If instead we want a force system at D, we could recalculate the net moment about D, or just use the translation formula (see box 2.1 on page 2.1).

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{tot}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{tot}},% \quad\mbox{and}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{D}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{C}}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/D}}\times{\accentset% {\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{tot}}.$

The total force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}$ stays the same and the moment at D is the moment at C plus the moment caused by ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}$ acting at position C relative to D. The net effect of the forces of the ground on a tree, for example, is of a force and a couple acting on the base of the tree.

Box 2.3 The tidiest representation of a force system: A “wrench”

This is an advanced aside.

Any force system can be represented by an equivalent force and a couple at any point. But force systems can be reduced to simpler forms. That this is so is of more theoretical than practical import. We state the results here without proof.

In 2D one of these two things is always true:

•

The system is equivalent to a couple, or most often
•

There is a line parallel to the force along which the system can be described by an equivalent force with no couple.

In 3D one of these three things is true:

•

The system is equivalent to a couple (applied anywhere), or
•

The system is equivalent to a force (applied on a given line parallel to the force), or most often
•

There is a line for which the system can be reduced to a force and a couple where the force, couple, and line are all parallel. The representation of the system of forces as a force and a parallel moment is called a wrench.

Equivalent does not mean equivalent for all purposes

We have perhaps oversimplified.

Imagine you stayed up late studying and overslept. Your roommate was not so diligent; woke up on time and went to wake you by gently shaking you. Having read this chapter so far and no further, and being rather literal, your roommate gets down on the floor and presses on the linoleum underneath your bed applying a force that is equivalent to pressing on you. Obviously this is not equivalent in the ordinary sense of the word. It isn’t even equivalent in all of its mechanics effects. One force moves you even if you don’t wake up, and the other doesn’t.

Some force systems that are ‘equivalent’ but different do have different mechanical effects. In what sense are two force systems that have the same net force and the same net moment really equivalent?

‘Equivalent’ force systems are equivalent in their contributions to the equations of mechanics (equations 0-II on the inside cover) for any system to which they are both applied.

But full mechanical analysis of a situation requires looking at the mechanics equations of many subsystems. In the mechanics equations for each subsystem, two ‘equivalent’ force systems are equivalent if they are both applied to that subsystem.

Filename:tfigure8-ang-accel

Figure 2.4: It feels different if someone presses on you or presses on the floor underneath you with an ‘equivalent’ force. The equivalence of ‘equivalent’ force systems depends on them both being applied to the same system.

For the analysis of the subsystem that is you sleeping, the force of your roommate’s hand on the floor isn’t applied to you, so it doesn’t show up in the mechanics equations for you, and doesn’t have the same effect as a force on you.

In summary, repeating:

Two force systems are called mechanically ‘equivalent’, with regard to a particular system on which they both act, if they have the same net force and also, the same net moment about some common point.

Two force sets can be equivalent, but not have all of the same mechanical effects, if they are not equivalent for all of the relevant subsystems. That is, you can only replace a force set with an equivalent force set after you have chosen the system of study^†^†margin: Or, at least you know that both sets act, or not, on all subsystems you will study .

SAMPLE 2.1

Filename:tfigure8-ang-accel-ex

Figure 2.5:

Equivalent force on a particle: Four forces ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}=2\,{\mbox{N}}% \hat{\boldsymbol{\imath}}-1\,{\mbox{N}}\hat{\boldsymbol{\jmath}},{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}=-5\,{\mbox{N}}\hat{% \boldsymbol{\jmath}},{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{3}}=3\,{\mbox{N}}\hat{\boldsymbol{\imath}}+12\,{\mbox{N}}\hat{% \boldsymbol{\jmath}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{4}}=1\,{\mbox{N}}% \hat{\boldsymbol{\imath}}$ act on a particle. Find the equivalent force on the particle.

Solution The equivalent force on the particle is the net force, i.e., the vector sum of all forces acting on the particle. Thus,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}+{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}+{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{3}}+{\accentset{\rightharpoonup% }{\boldsymbol{F}}}_{\!\!\mathrm{4}}$
	$\displaystyle=$	$\displaystyle(2\,{\mbox{N}}\hat{\boldsymbol{\imath}}-1\,{\mbox{N}}\hat{% \boldsymbol{\jmath}})+(-5\,{\mbox{N}}\hat{\boldsymbol{\jmath}})+(3\,{\mbox{N}}% \hat{\boldsymbol{\imath}}+12\,{\mbox{N}}\hat{\boldsymbol{\jmath}})+(1\,{\mbox{% N}}\hat{\boldsymbol{\imath}})$
	$\displaystyle=$	$\displaystyle 6\,{\mbox{N}}\hat{\boldsymbol{\imath}}+6\,{\mbox{N}}\hat{% \boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}=6\,{\mbox{N}}% (\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$

Note that there is no net couple since all the four forces act at the same point. This is always true for particles. Thus, the equivalent force-couple system for particles consists of only the net force.

SAMPLE 2.2

Filename:tfigure8-ang-accel1

Figure 2.6:

Equivalent force with no net moment: In the figure shown, $F_{1}=50\,{\mbox{N}},F_{2}=10\,{\mbox{N}},F_{3}=30\,{\mbox{N}}$ , and $\theta=60^{\circ}$ . Find the equivalent force-couple system about point D of the structure.

Solution From the given geometry, we see that the three forces ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}},{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}},\qquad\mbox{and}\qquad{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{3}}$ pass through point D. Thus they are concurrent forces. Since point D is on the line of action of these forces, we can simply slide the three forces to point D without altering their mechanical effect on the structure. Then the equivalent force-couple system at point D consists of only the net force, ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}$ , with no couple (the three forces passing through point D produce no moment about D). This is true for all concurrent forces. Thus,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}+{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}+{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{3}}$
	$\displaystyle=$	$\displaystyle F_{1}(\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat{% \boldsymbol{\jmath}})-F_{2}\hat{\boldsymbol{\jmath}}+F_{3}\hat{\boldsymbol{% \imath}}$
	$\displaystyle=$	$\displaystyle(F_{1}\cos\theta+F_{3})\hat{\boldsymbol{\imath}}-(F_{1}\sin\theta% +F_{2})\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle(50\,{\mbox{N}}\cdot\frac{1}{2}+30\,{\mbox{N}})\hat{\boldsymbol{% \imath}}-(50\,{\mbox{N}}\cdot\frac{\sqrt{3}}{2}+10\,{\mbox{N}})\hat{% \boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle 55\,{\mbox{N}}\hat{\boldsymbol{\imath}}-53.3\,{\mbox{N}}\hat{% \boldsymbol{\jmath}},$
$\displaystyle\hbox to36.135pt{and\hfil}{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{\mathrm{D}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{0}}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}=55\,{\mbox{N}% }\hat{\boldsymbol{\imath}}-53.3\,{\mbox{N}}\hat{\boldsymbol{\jmath}},{% \accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{D}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$

Graphically, the solution is shown in Fig. 2.7

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Figure 2.7:

SAMPLE 2.3

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Figure 2.8:

An equivalent force-couple system: Three forces $F_{1}=100\,{\mbox{N}},F_{2}=50\,{\mbox{N}}$ , and $F_{3}=30\,{\mbox{N}}$ act on a structure as shown in the figure where $\alpha=30^{\circ},\,\theta=60^{\circ},\ell=1\,\mathrm{m}$ and $h=0.5\,\mathrm{m}$ . Find the equivalent force-couple system about point D.

Solution The net force is the sum of all applied forces, i.e.,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}+{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}+{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{3}}$
	$\displaystyle=$	$\displaystyle F_{1}(-\sin\alpha\hat{\boldsymbol{\imath}}-\cos\alpha\hat{% \boldsymbol{\jmath}})+F_{2}(\cos\theta\hat{\boldsymbol{\imath}}-\sin\theta\hat% {\boldsymbol{\jmath}})+F_{3}\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle(-F_{1}\sin\alpha+F_{2}\cos\theta)\hat{\boldsymbol{\imath}}+(-F_{% 1}\cos\alpha-F_{2}\sin\theta+F_{3})\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle(-100\,{\mbox{N}}\cdot\frac{1}{2}+50\,{\mbox{N}}\cdot\frac{1}{2})% \hat{\boldsymbol{\imath}}+(-100\,{\mbox{N}}\cdot\frac{\sqrt{3}}{2}-50\,{\mbox{% N}}\cdot\frac{\sqrt{3}}{2}+30\,{\mbox{N}})\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle-25\,{\mbox{N}}\hat{\boldsymbol{\imath}}-99.9\,{\mbox{N}}\hat{% \boldsymbol{\jmath}}.$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{D}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/D}}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}$
	$\displaystyle=$	$\displaystyle h\hat{\boldsymbol{\jmath}}\times F_{2}(\cos\theta\hat{% \boldsymbol{\imath}}-\sin\theta\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle-F_{2}h\cos\theta\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle-50\,{\mbox{N}}\cdot 0.5\,\mathrm{m}\frac{1}{2}\hat{\boldsymbol{k% }}=-12.5\,{\rm N}\mathord{\cdot}{\rm m}\hat{\boldsymbol{k}}.$

Filename:sfig8-2-3b

Figure 2.9:

The equivalent force-couple system is shown in Fig. 2.9

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}=-25\,{\mbox{N% }}\hat{\boldsymbol{\imath}}-99.9\,{\mbox{N}}\hat{\boldsymbol{\jmath}}\qquad% \mbox{and}\qquad{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{D}}=-12% .5\,{\rm N}\mathord{\cdot}{\rm m}\hat{\boldsymbol{k}}$

SAMPLE 2.4

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Figure 2.10:

Translating a force-couple system: The net force and couple acting about point O on the ’L’ shaped bar shown in the figure are 100 N and 20 $\,{\rm N}\mathord{\cdot}{\rm m}$ , respectively. Find the net force and moment about point G.

Solution

Filename:sfig8-3-1a

Figure 2.11:

The net force on a structure is the same about any point since it is just the vector sum of all the forces acting on the structure and is independent of their point of application. Therefore,

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}={\accentset{% \rightharpoonup}{\boldsymbol{F}}}=-100\,{\mbox{N}}\hat{\boldsymbol{\jmath}}.

The net moment about a point, however, depends on the location of points of application of the forces with respect to that point. Thus,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{G}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{O}}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{O/G}}\times{\accentset% {\rightharpoonup}{\boldsymbol{F}}}$
	$\displaystyle=$	$\displaystyle M\hat{\boldsymbol{k}}+(-\ell\hat{\boldsymbol{\imath}}+h\hat{% \boldsymbol{\jmath}})\times(-F\hat{\boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle(M+F\ell)\hat{\boldsymbol{k}}$
	$\displaystyle=$	$\displaystyle(20\,{\rm N}\mathord{\cdot}{\rm m}+100\,{\mbox{N}}\cdot 1\,% \mathrm{m})\hat{\boldsymbol{k}}=120\,{\rm N}\mathord{\cdot}{\rm m}\hat{% \boldsymbol{k}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}=-100\,{\mbox{% N}}\hat{\boldsymbol{\jmath}},\qquad\mbox{and}\qquad{\accentset{\rightharpoonup% }{\boldsymbol{M}}}_{\mathrm{G}}=120\,{\rm N}\mathord{\cdot}{\rm m}\hat{% \boldsymbol{k}}$

SAMPLE 2.5 Checking equivalence of force-couple systems: In the figure shown below, which of the force-couple systems shown in (b), (c), and (d) are equivalent to the force system shown in (a)?

Filename:sfig8-5-wiper

Figure 2.12:

Solution The equivalence of force-couple systems require that (i) the net force be the same, and (ii) the net moment about any reference point be the same. For the given systems, let us choose point B as our reference point for comparing their equivalence. For the force system shown in Fig. 2.12(a), we have,

Filename:sfig8-5-wiper-a

Figure 2.13:

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}+{% \accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}=-10\,{\mbox{N}}% \hat{\boldsymbol{\jmath}}-10\,{\mbox{N}}\hat{\boldsymbol{\jmath}}=-20\,{\mbox{% N}}\hat{\boldsymbol{\jmath}}$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{B_{net}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/B}}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}=1\,% \mathrm{m}\hat{\boldsymbol{\imath}}\times(-10\,{\mbox{N}}\hat{\boldsymbol{% \jmath}})=-10\,{\rm N}\mathord{\cdot}{\rm m}\hat{\boldsymbol{k}}.$

Now, we can compare the systems shown in (b), (c), and (d) against the computed equivalent force-couple system, ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}\qquad\mbox{% and}\qquad{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{D}}$ .

•

Figure (b) shows exactly the system we calculated. Therefore, it represents an equivalent force-couple system.

•

Figure (c): Let us calculate the net force and moment about point B for this system.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{C}}% \mathrel{\mathop{=}\limits^{\surd}}-20\,{\mbox{N}}\hat{\boldsymbol{\jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{B}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{C}}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{C/B}}\times{\accentset% {\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{C}}$
	$\displaystyle=$	$\displaystyle-10\,{\rm N}\mathord{\cdot}{\rm m}\hat{\boldsymbol{k}}+1\,\mathrm% {m}\hat{\boldsymbol{\imath}}\times(-20\,{\mbox{N}}\hat{\boldsymbol{\jmath}})=-% 30\,{\rm N}\mathord{\cdot}{\rm m}\hat{\boldsymbol{k}}\neq{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathrm{B_{net}}}.$

Thus, the given force-couple system in this case is not equivalent to the force system in (a).

•

Figure (d): Again, we compute the net force and the net couple about point B:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{D}}% \mathrel{\mathop{=}\limits^{\surd}}-20\,{\mbox{N}}\hat{\boldsymbol{\jmath}}$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{B}}$ $\displaystyle=$ $\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D/B}}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{D}}$

$\displaystyle=$ $\displaystyle 0.5\,\mathrm{m}\hat{\boldsymbol{\imath}}\times(-20\,{\mbox{N}}% \hat{\boldsymbol{\jmath}})=-10\,{\rm N}\mathord{\cdot}{\rm m}\hat{\boldsymbol{% k}}\mathrel{\mathop{=}\limits^{\surd}}{\accentset{\rightharpoonup}{\boldsymbol% {M}}}_{\mathrm{B_{net}}}.$

Thus, the given force-couple system (with zero couple) at D is equivalent to the force system in (a).

Answer: (b) and (d) are equivalent to (a); (c) is not.

SAMPLE 2.6

Filename:sfig8-5-2disks

Figure 2.14:

Equivalent force with no couple: For a body, an equivalent force-couple system at point A consists of a force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=20\,{\mbox{N}}\hat{\boldsymbol{% \imath}}+15\,{\mbox{N}}\hat{\boldsymbol{\jmath}}$ and a couple ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}=10\,{\rm N}\mathord% {\cdot}{\rm m}\hat{\boldsymbol{k}}$ . Find a point on the body such that the equivalent force-couple system at that point consists of only a force (zero couple).

Solution The net force in the two equivalent force-couple systems has to be the same. Therefore, for the new system, ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}={\accentset{% \rightharpoonup}{\boldsymbol{F}}}=20\,{\mbox{N}}\hat{\boldsymbol{\imath}}+15\,% {\mbox{N}}\hat{\boldsymbol{\jmath}}$ .

Filename:sfig8-4-1

Figure 2.15:

Let B be the point at which the equivalent force-couple system consists of only the net force, with zero couple. We need to find the location of point B. Let A be the origin of a $x y$ coordinate system in which the coordinates of B are $(x,y)$ . Then, the moment about point B is,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{B}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{A/B}}\times{\accentset% {\rightharpoonup}{\boldsymbol{F}}}$
	$\displaystyle=$	$\displaystyle M_{A}\hat{\boldsymbol{k}}+(-x\hat{\boldsymbol{\imath}}-y\hat{% \boldsymbol{\jmath}})\times(F_{x}\hat{\boldsymbol{\imath}}+F_{y}\hat{% \boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle M_{A}\hat{\boldsymbol{k}}+(-F_{y}x+F_{x}y)\hat{\boldsymbol{k}}.$

$\displaystyle F_{y}x-F_{x}y$	$\displaystyle=$	$\displaystyle M_{A}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}y$	$\displaystyle=$	$\displaystyle\frac{F_{y}}{F_{x}}x-\frac{M_{A}}{F_{x}}$
	$\displaystyle=$	$\displaystyle\frac{15\,{\mbox{N}}}{20\,{\mbox{N}}}x-\frac{10\,{\rm N}\mathord{% \cdot}{\rm m}}{20\,{\mbox{N}}}$
	$\displaystyle=$	$\displaystyle 0.75x-0.5\,\mathrm{m}.$

This is the equation of a line. Thus, we can select any point on this line and apply the force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=20\,{\mbox{N}}\hat{\boldsymbol{% \imath}}+15\,{\mbox{N}}\hat{\boldsymbol{\jmath}}$ with zero couple as an equivalent force-couple system.

Answer: Any point on the line $y=0.75x-0.5\,\mathrm{m}.$

Filename:sfig8-4-1a

Figure 2.16:

So, how or why does it work? The line we obtained is shown in gray in Fig. 2.15. Note that this line has the same slope as that of the given force vector (slope $=0.75=F_{y}/F_{x}$ ) and the offset is such that shifting the force ⇀ F to this line counter balances the given couple at A. To see this clearly, let us select three points C, D, and E on the line as shown in Fig. 2.16. From the equation of the line, we find the coordinates of C(0,-.5m), D(.24m,.32m) and E(.67m,0). Now imagine moving the force ⇀ F to C, D, or E. In each case, it must produce the same moment ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{A}}$ about point A. Let us do a quick check.

• ⇀ F

at point C: The moment about point A is due to the horizontal component $F_{x}=20\,{\mbox{N}}$ , since $F_{y}$ passes through point A. The moment is $F_{x}\cdot AC=20\,{\mbox{N}}\cdot 0.5\,\mathrm{m}=10\,{\rm N}\mathord{\cdot}{% \rm m}$ , same as $M_{A}$ . The direction is counterclockwise as required.
• ⇀ F

at point D: The moment about point A is $|{{\accentset{\rightharpoonup}{\boldsymbol{F}}}}|\cdot AD=25\,{\mbox{N}}\cdot 0% .4\,\mathrm{m}=10\,{\rm N}\mathord{\cdot}{\rm m}$ , same as $M_{A}$ . The direction is counterclockwise as required.
• ⇀ F

at point E: The moment about point A is due to the vertical component $F_{y}$ , since $F_{x}$ passes through point A. The moment is $F_{y}\cdot AE=15\,{\mbox{N}}\cdot 0.67\,\mathrm{m}=10\,{\rm N}\mathord{\cdot}{% \rm m}$ , same as $M_{A}$ . The direction here too is counterclockwise as required.

Once we check the calculation for one point on the line, we should not have to do any more checks since we know that sliding the force along its line of action (line CB) produces no couple and thus preserves the equivalence.

Problems for 2.1 Equivalent force systems and couples

2.1.1 Find the net force on the particle shown in the figure.

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Figure 2.17

2.1.2 Replace the forces acting on the particle of mass $m$ shown in the figure by a single equivalent force.

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Figure 2.18

2.1.3 Find the net force on the pulley due to the belt tensions shown in the figure.

Filename:efig1-2-28

Figure 2.19

2.1.4 The net force on A from the two cables is a force that points down and has magnitude of $125\,{\mbox{N}}$ . Find the tension in cable AB. Answer: $T_{AB}=75\,{\mbox{N}}$

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Figure 2.20

2.1.5 Replace the forces shown on the rectangular plate by a single equivalent force. Where should this equivalent force act on the plate and why?

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Figure 2.21

2.1.6 Three forces act on a Z-section ABCDE as shown in the figure. Point C lies in the middle of the vertical section BD. Find any equivalent force-couple system and describe where to apply it.

Filename:Danef94s1q2

Figure 2.22

2.1.7 Find a force-couple system at D that is equivalent to the single force at C shown. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{D}=(15\sqrt{3}\hat{\boldsymbol{% \imath}}-15\hat{\boldsymbol{\jmath}})\,{\mbox{N}},{\accentset{\rightharpoonup}% {\boldsymbol{M}}}_{D}=30(1+\sqrt{3})\hat{\boldsymbol{k}}\,{\rm N}\mathord{% \cdot}{\rm m}$

Filename:pfigure-blue-123-1

Figure 2.23

2.1.8 The three forces acting on the circular plate are shown. They act at the corners of a square which is concentric with the plate. Find an equivalent force-couple system acting at point C.

Filename:pfigure-blue-119-2

Figure 2.24

2.1.9 The forces and the moment acting on point C of the frame ABC shown in the figure are $C_{x}=48\,{\mbox{N}},\;C_{y}=40\,{\mbox{N}}$ , and $M_{c}=20\,{\rm N}\mathord{\cdot}{\rm m}$ . Find an equivalent force couple system at point B.

Filename:pfigure-s95q14

Figure 2.25

2.1.10 The force system ( ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{1},{\accentset{\rightharpoonup}% {\boldsymbol{F}}}_{2}$ ) is equivalent to a force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=10\hat{\boldsymbol{\imath}}\,{% \mbox{N}}$ at the origin and a couple $M\hat{\boldsymbol{k}}$ . Find $M$ . Answer: $M=30\left(\sqrt{3}-3\right)\,{\rm N}\mathord{\cdot}{\rm m}\hat{\boldsymbol{k}}$

Filename:Danef94s3q2

Figure 2.26

2.1.11 Find an equivalent force-couple system for the forces acting on the beam shown in the figure, if the equivalent system is to act at

(a)

point B,
(b)

point D.

Filename:bikefork1-ang-accel

Figure 2.27

2.1.12 ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{1}$ acts at A and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{2}=7\,{\mbox{N}}\hat{% \boldsymbol{\imath}}$ acts at an unknown location. Together they are equivalent to a force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{B}$ and moment ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{B}=48\,{\rm N}\mathord{\cdot}{% \rm m}\hat{\boldsymbol{k}}$ at B. Together they are also equivalent to a force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}$ and moment ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{C}=75\,{\rm N}\mathord{\cdot}{% \rm m}\hat{\boldsymbol{k}}$ at C.

(a)

Find ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}=(-9\hat{\boldsymbol{\imath}}% -12\hat{\boldsymbol{\jmath}})\,{\mbox{N}}$
(b)

Find the line of action of ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{2}$ . Answer: Through B and also through a point $3\,\mathrm{m}$ above A.

Filename:bikefork-ang-accel

Figure 2.28

2.1.13 The figure shows three different force-couple systems acting on a square plate. Identify which force-couple systems are equivalent.

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Figure 2.29

2.1.14 The force and moment acting at point C of a machine part are shown in the figure where $M_{c}$ is not known. It is found that if the given force-couple system is replaced by a single horizontal force of magnitude $10\,{\mbox{N}}$ acting at point A then the net effect on the machine part is the same. What is the magnitude of the moment $M_{c}$ ?

Filename:bikefork-alt

Figure 2.30

2.1.15 2D. Assume a force system is equivalent to a force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}\neq{\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ and couple ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{1}=M_{1}\hat{\boldsymbol{k}}$ acting at point ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{1}$ .

(a)

Find some point ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}$ , and force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}$ so that ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}$ acting at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}$ is equivalent to ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{1}}$ acting at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}+{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{\!\!\mathrm{1}}\times\hat{\boldsymbol{k}}M_{1}/|{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}|^{2}$ , ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}$ .
(b)

Find all possible wrenches (combinations of point location, force and moment) equivalent to the system with ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{1}}$ acting at the point with position vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}+{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{\!\!\mathrm{1}}\times\hat{\boldsymbol{k}}M_{1}/|{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}|^{2}+c{\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}$ where $c$ is any real number, ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}$ .
(c)

Describe the situation in the special case when ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathrm{1}}$ applied at any point in the plane.

2.1.16 3D. Assume a force system is equivalent to a force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}$ and couple ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{1}}$ acting at point with position vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}$ .

(a)

Find a point P with position vector ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}$ , so that an equivalent force system ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{2}}$ acting at P has ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}$ is parallel to ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{2}}$ . (Finding such a point, force and moment is called “reducing the force system to a wrench”). Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}+{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{\!\!\mathrm{1}}\times{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{\mathrm{1}}/|{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{% \!\!\mathrm{1}}|^{2}$ , ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathrm{1}}\cdot{\accentset{\rightharpoonup% }{\boldsymbol{F}}}_{\!\!\mathrm{1}}{\accentset{\rightharpoonup}{\boldsymbol{F}% }}_{\!\!\mathrm{1}}/|{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{1}}|^{2}$ . If ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ then ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathrm{1}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}$ is any point at all in space.
(b)

Find all possible wrenches (combinations of point location, force and moment) equivalent to the system with ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{1}}$ acting at ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}+{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{\!\!\mathrm{1}}\times{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{\mathrm{1}}/|{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{% \!\!\mathrm{1}}|^{2}+c{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{1}}$ where $c$ is any real number, ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{2}}={\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{\mathrm{1}}\cdot{\accentset{\rightharpoonup% }{\boldsymbol{F}}}_{\!\!\mathrm{1}}{\accentset{\rightharpoonup}{\boldsymbol{F}% }}_{\!\!\mathrm{1}}/|{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{1}}|^{2}$ . See above for the special case of ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ .

Note, one special case with a slightly different result than the other cases is if ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{1}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , so it should be treated separately.

2.2 Center of mass and gravity

The center-of-mass of a system is the point at the position ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}$ defined by

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\frac{\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{i}}m_{i}}{m_{\mathrm{tot}}}$	for discrete systems
		$\displaystyle=$	$\displaystyle\frac{\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}dm}{m_{% \mathrm{tot}}}$	for continuous systems

where $m_{\mathrm{tot}}=\sum m_{i}$ for discrete systems and $m_{\mathrm{tot}}=\int dm$ for continuous systems (see boxes 2.1 and 2.2 on pages 2.1 and 2.2 for a discussion of the $\sum$ and $\int$ sum notations).

For every system, and at every instant in time, this is a unique location in space that is the average position of the system’s mass. The center of mass is commonly designated by

•

cm
•

c.o.m.,
•

COM,
•

G,
•

c.g., or
•

with a circle with diagonally-opposite quadrants shaded

One of the important (if routine) tasks of many real engineers is to find the center-of-mass of a complex machine.

What is the center of mass good for? Just knowing the location of the center-of-mass of a car, for example, is enough to estimate whether it can be tipped over by maneuvers on level ground. The center-of-mass of a boat must be low enough for the boat to be stable. Any propulsive force on a space craft must be directed towards the center-of-mass in order to not induce rotations. Tracking the trajectory of the center-of-mass of an exploding plane can determine whether or not it was hit by a massive object. Any rotating piece of machinery must have its center-of-mass on the axis of rotation if it is not to cause much vibration.

Filename:pfigure-blue-128-1

Figure 2.31: The center of mass of the ink making up the letter C is approximately at G (marked with a circle with the diagonally-opposite quadrants shaded). The various notations for center of mass include: G and COM.

Also, many calculations in mechanics are greatly simplified by making use of a system’s center-of-mass. In particular,

the whole complicated distribution of near-earth gravity forces on a body is equivalent to a single force at the body’s center-of-mass.

Many of the important quantities in dynamics are similarly simplified using the center-of-mass.

The following rearranged definition of center of mass, might be easier to remember, and is sometimes more directly useful:

{m_{\mathrm{tot}}}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}% }={\sum m_{i}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i}}}% \hbox to36.135pt{\hfil or\hfil}{m_{\mathrm{tot}}}{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{cm}}={\int{\accentset{\rightharpoonup}{\boldsymbol% {r}}}dm}.

That is, the center of mass is the position that, when multiplied by the total mass, gives the same result as the sum of all the mass bits each multiplied by their individual positions.

For theoretical purposes, we rarely need to evaluate these sums and integrals. And, for simple problems there are sometimes shortcuts that reduce the calculation to a matter of observation. However, for complex machines, one or both of the formulas 2.2 must be evaluated in detail

^†^†margin: This routine work is generally done with CAD (computer aided design) software. But an engineer still needs to know the basic calculation skills in order to make sense of, and roughly check, the computer calculations and in order to think of appropriate design changes.

Box 2.4 Like $\displaystyle\sum$ , the symbol $\displaystyle\int$ also means add

We often add things up in mechanics. For example, the total mass of some particles is

m_{\mathrm{tot}}=m_{1}+m_{2}+m_{3}+\dots=\sum m_{i}

or more specifically the mass of 137 particles is, say, $\displaystyle m_{\mathrm{tot}}=\sum_{i=1}^{137}m_{i}$ . And the total mass of a bicycle is:

m_{\rm bike}={\sum_{i=1}^{100,000,000,000,000,000,000,000}}\!\!\!\!\!\!\!\!\!% \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\ m_{i}

where $m_{i}$ are the masses of each of the $10^{23}$ (or so) atoms of metal, rubber, plastic, cotton, and paint. But atoms are so small and there are so many of them. Instead we often think of a bike as built of macroscopic parts. The total mass of the bike is then the sum of the masses of the tires, the tubes, the wheel rims, the spokes and nipples, the ball bearings, the chain pins, and so on. And we would write:

m_{\rm bike}=\sum_{i=1}^{2,000}m_{i}

where now the $m_{i}$ are the masses of the 2,000 or so bike parts. This sum is more manageable, but still too detailed for some purposes.

An approach that avoids attending to atoms or ball bearings, is to think of sending the bike to a big shredding machine that cuts it up into very small bits. Now we write

m_{\rm bike}=\sum m_{i}

where the $m_{i}$ are the masses of the very small bits. We don’t fuss over whether one bit is a piece of ball bearing or fragment of cotton from the tire walls. We just chop the bike into bits and add up the contribution of each bit. If you take the letter S, as in SUM, and distort it and you get a big old fashioned German ‘S’ used in calculus as the integral sign

Filename:tablefigure-SisInt

Evolution.
From left to right, the letter S is distorted into the integral sign $\int$ .

So, instead of writing, like Lagrange would have (if he knew about bicycles and cared about their mass), $m_{\rm bike}=$ S $d m$ , we write

m_{\rm bike}=\int dm.

This is the $\int\!$ um of all the teeny bits of mass $d m$ . More formally, we mean the value of that sum in the limit that all the bits are infinitesimal (not minding the technical fine point that it’s hard to chop atoms into infinitesimal pieces).

The mass is one of many things we would like to add up. Many of the other things also involve mass. For example, to find the center-of-mass we add up the positions ‘weighted’ by mass.

\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}\,dm\quad\mbox{which means}% \sum_{\lim m_{i}\to 0}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm% {i}}m_{i}.

That is, to find the center of mass, you take your object of interest and chop it into a billion pieces. Then you re-assemble it. For each piece, you make the vector which is the position vector of the piece multiplied by (‘weighted by’) its mass. Then you add up the billion vectors. Well, really you chop the thing into a trillion trillion $\dots$ pieces, but a billion gives the idea.

Example: Center of mass of two point masses

Intuitively, the center-of-mass of the two masses shown in fig. 2.32 is between the two masses and closer to the larger one.

Filename:tfigure8-syst-bods

Figure 2.32: Center of mass of a system consisting of two points.

Referring to equation 2.2,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\frac{\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{i}}m_{i}}{m_{tot}}$
	$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}% }m_{1}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}m_{2}}{m_{1% }+m_{2}}$
	$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}% }(m_{1}+m_{2})-{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}m_{% 2}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}m_{2}}{m_{1}+m_% {2}}$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}+% \underbrace{\left(\frac{m_{2}}{m_{1}+m_{2}}\right)}_{\text{the fraction of the distance that the cm is from ${\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{1}}$ to ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}$}}\underbrace{% \left({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}-{\accentset% {\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}\right)}_{\text{the vector % from ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}$ to ${\accentset% {\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}$.}}.$

The math agrees with common sense: the center-of-mass is on the line connecting the masses. If $m_{2}\gg m_{1}$ , then the center-of-mass is near $m_{2}$ . If $m_{1}\gg m_{2}$ , then the center-of-mass is near $m_{1}$ . If $m_{1}=m_{2}$ the center-of-mass is right in the middle at $({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}})/2$ .

Continuous systems

How do we evaluate integrals like $\displaystyle\int(\mbox{something})\,dm$ ? In center-of-mass calculations, $(\mbox{something})$ is position, but we will evaluate similar integrals where $(\mbox{something})$ is some other scalar or vector-valued function of position. Most often, we label the material by its spatial position, and evaluate $d m$ in terms of increments of position. For 3D solids $dm=\rho dV$ where $\rho$ is density (mass per unit volume). So $\int(\mbox{something})dm$ turns into a standard volume integral $\displaystyle\int_{V}(\mbox{something})\rho\,dV$

^†^†margin: Note: writing

\int_{m_{1}}^{m_{2}}(\mbox{something})\,dm

is nonsense because

m

is not a scalar parameter which labels points in a material (i.e., there is no point at

m=3\,\mathrm{kg}

Planar objects. For thin flat things like metal sheets we often take $\rho$ to mean mass per unit area $A$ so then $dm=\rho dA$ and $\int(\mbox{something})dm=\int_{A}(\mbox{something})\rho\,dA$ .

1D objects. For mass distributed along a line or curve we take $\rho$ to be the mass per unit length or arc length $s$ and so $dm=\rho ds$ and $\int(\mbox{something})dm=\int_{curve}(\mbox{something})\rho\,ds$ .

CoM of uniform rod. The center-of-mass of a uniform rod is naturally in the middle, as the calculations here show (see fig. 2.33a). Assume the rod has length $L=3\,\mathrm{m}$ and mass $m=7\,\mathrm{kg}$ .

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}=\frac{\int{% \accentset{\rightharpoonup}{\boldsymbol{r}}}\,dm}{m_{\mathrm{tot}}}=\frac{\int% _{0}^{L}x\hat{\boldsymbol{\imath}}\,\overbrace{\rho dx}^{dm}}{\int_{0}^{L}\rho% \,dx}=\frac{\rho(x^{2}/2)|^{L}_{0}}{\rho(1)|^{L}_{1}}\hat{\boldsymbol{\imath}}% =\frac{\rho(L^{2}/2)}{\rho L}\hat{\boldsymbol{\imath}}=(L/2)\hat{\boldsymbol{% \imath}}

Filename:sfig8-7-2

Figure 2.33: Where is the center-of-mass of a uniform rod? In the middle, as you can find calculating a few ways, or, more simply, by symmetry.

So ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}=(L/2)\hat{% \boldsymbol{\imath}}$ , or by dotting with $\hat{\boldsymbol{\imath}}$ (taking the x component) we get that the center-of-mass is on the rod a distance $d=L/2=1.5\,\mathrm{m}$ from the end.

The center-of-mass calculation is objective. It describes something about the object that does not depend on the coordinate system. In different coordinate systems the center-of-mass for the rod above will have different coordinates, but it will always be at the middle of the rod.

Example. Find the center-of-mass using the coordinate system with $s$ & $\hat{\boldsymbol{\lambda}}$ in fig. 2.33b:

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}=\frac{\int{% \accentset{\rightharpoonup}{\boldsymbol{r}}}\,dm}{m_{\mathrm{tot}}}=\frac{\int% _{0}^{L}s\hat{\boldsymbol{\lambda}}\,\rho ds}{\int_{0}^{L}\rho\,ds}\hat{% \boldsymbol{\lambda}}=\frac{\rho(s^{2}/2)|^{L}_{0}}{\rho(1)|^{L}_{0}}\hat{% \boldsymbol{\lambda}}=\frac{\rho(L^{2}/2)}{\rho L}\hat{\boldsymbol{\lambda}}=(% L/2)\hat{\boldsymbol{\lambda}},

again showing that the center-of-mass is in the middle.

Note, one can treat the center-of-mass vector calculations as separate scalar equations, one for each component. For example:

\hat{\boldsymbol{\imath}}\cdot\left\{{\accentset{\rightharpoonup}{\boldsymbol{% r}}}_{\!\mathrm{cm}}=\frac{\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}\,% dm}{m_{\mathrm{tot}}}\right\}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}r_{x\rm cm% }=x_{\rm cm}=\frac{\int x\,dm}{m_{\mathrm{tot}}}.

Finally, there is no law that says you have to use the best coordinate system. One is free to make trouble for oneself and use an inconvenient coordinate system.

CoM using awkward coordinates. Use the $x y$ coordinates of fig. 2.33c to find the center-of-mass of the rod.

	$\displaystyle x_{\rm cm}$	$\displaystyle=\frac{\int xdm}{m_{\mathrm{tot}}}=\frac{\int_{-\ell_{1}}^{\ell_{% 2}}\overbrace{s\cos\theta}^{x}\,\rho ds}{\int_{0}^{L}\rho\,ds}=\frac{\rho\cos% \theta\frac{s^{2}}{2}\|_{-\ell_{1}}^{\ell_{2}}}{\rho(1)\|_{-\ell_{1}}^{\ell_{2}}% }=\frac{\rho\cos\theta\frac{(\ell_{2}^{2}-\ell_{1}^{2})}{2}}{\rho(\ell_{1}+% \ell_{2})}$
		$\displaystyle=\frac{\cos\theta(\ell_{2}-\ell_{1})}{2}$

Similarly $y_{\rm cm}=\sin\theta(\ell_{2}-\ell_{1})/2$ so

\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}

\displaystyle=\frac{\ell_{2}-\ell_{1}}{2}(\cos\theta\hat{\boldsymbol{\imath}}+% \sin\theta\hat{\boldsymbol{\jmath}}),

which is at the middle of the rod.

The most commonly needed center-of-mass that can be found analytically but not so-simply from symmetry is that of a triangle (see box 2.2 on page 2.2). There are more examples of using integration to find the center-of-mass in your multi-variable text book.

Center of mass and centroid

For objects with uniform material density we have

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}=\frac{\int{% \accentset{\rightharpoonup}{\boldsymbol{r}}}dm}{m_{\mathrm{tot}}}=\frac{\int_{% V}{\accentset{\rightharpoonup}{\boldsymbol{r}}}\rho dV}{\int_{V}\rho dV}=\frac% {\rho\int_{V}{\accentset{\rightharpoonup}{\boldsymbol{r}}}dV}{\rho\int_{V}dV}=% \frac{\int_{V}{\accentset{\rightharpoonup}{\boldsymbol{r}}}dV}{V}.\\

The last expression is just the formula for geometric centroid. Analogous calculations hold for 2D and 1D geometric objects.

For objects with density that does not vary from point to point, the center-of-mass is at the same place as the geometric centroid.

Center of mass and symmetry

The center-of-mass respects any symmetry in the mass distribution of a system. If the word ‘middle’ has unambiguous meaning in English, then that is the location of the center-of-mass, as for the rod of fig. 2.33 and the other examples in fig. 2.34.

Filename:sfig8-7-2a

Figure 2.34: Both the center of mass and the geometric centroid share the symmetries of the object.

Example: Center of mass of a semicircle.

(see fig. 2.35)

Filename:sfig8-7-2again

Figure 2.35: The center of mass of a semicircular arc is about

0.6\,r

from the center of the circle.

The center of mass of a semicircular arc of mass $m$ and radius $r$ is on the $y$ axis at $x=0$ , by symmetry. The location of $y_{\textrm{G}}$ is found by

	$\displaystyle y_{\textrm{G}}$	$\displaystyle=$	$\displaystyle\frac{\int y\,dm}{m_{\mathrm{tot}}}=\frac{1}{m}\int_{0}^{\pi}% \overbrace{r\sin\theta}^{y}\overbrace{\rho\,ds}^{dm}\par=\frac{1}{m}\int_{0}^{% \pi}r\sin\theta\overbrace{\frac{m}{\pi r}}^{\rho}\,\overbrace{r\,d\theta}^{ds}$
		$\displaystyle=$	$\displaystyle\frac{r}{\pi}\int_{0}^{\pi}\sin\theta\,d\theta=\frac{2}{\pi}r% \approx 0.64\,r\par$

The center of mass of a semicircle is almost 2/3 of the way towards the perimeter from the center of the circle. You can see that G has to be above the halfway point by noticing how much more mass is near to $y=r$ (where the circular arc is nearly horizontal) than to $y=0$ (where the circular arc is running away from the $x$ axis).

Systems of systems and composite objects

Another way of interpreting the formula

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}=\frac{{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}m_{1}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}m_{2}+\cdots}{m_{1}+m_{2}+\cdots}

is that the $m$ ’s are the masses of subsystems, not just points, and that the ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i}}$ are the positions of the centers of mass of these systems. This subdivision, follows from the associated law of addition, and is justified in box 2.2 on page 2.2.

The center-of-mass of a single complex shaped object can be found by treating it as an assembly of simpler objects.

Filename:sfig8-7-2disks

Figure 2.36: Center of mass of two rods

Example: Two rods

The center-of-mass of two rods shown in fig. 2.36 can be found as

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}=\frac{{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}m_{1}+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}m_{2}}{m_{1}+m_{2}}

where ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}$ are the positions of the centers of mass of each rod and $m_{1}$ and $m_{2}$ are the masses.

Example: ‘L’ shaped plate

Consider the plate with uniform mass per unit area $\rho$ .

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G}}$	$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{I}% }m_{I}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{II}}m_{II}}{m_% {I}+m_{II}}$
	$\displaystyle=$	$\displaystyle\frac{(\frac{a}{2}\hat{\boldsymbol{\imath}}+a\hat{\boldsymbol{% \jmath}})(2\rho a^{2})+(\frac{3}{2}a\hat{\boldsymbol{\imath}}+\frac{a}{2}\hat{% \boldsymbol{\jmath}})(\rho a^{2})}{(2\rho a^{2})+(\rho a^{2})}$
	$\displaystyle=$	$\displaystyle\frac{5}{6}a(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}).$

Filename:sfig8-4-4

Figure 2.37: The center-of-mass of the ‘L’ shaped object can be found by thinking of it as a rectangle plus a square.

Composite objects using subtraction

It is sometimes useful to think of an object as composed of pieces, some of which have negative mass.

Example: ‘L’ shaped plate, again

Reconsider the plate from the previous example.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{G}}$	$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{I}% }m_{I}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{II}}m_{II}}{m_% {I}+m_{II}}$
	$\displaystyle=$	$\displaystyle\frac{(a\hat{\boldsymbol{\imath}}+a\hat{\boldsymbol{\jmath}})(% \rho(2a)^{2})+(\frac{3}{2}a\hat{\boldsymbol{\imath}}+\frac{3}{2}a\hat{% \boldsymbol{\jmath}})\overbrace{(-\rho a^{2})}^{m_{II}}}{(\rho(2a)^{2})+% \underbrace{(-\rho a^{2})}_{m_{II}}}$
	$\displaystyle=$	$\displaystyle\frac{5}{6}a(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}).$

Filename:sfig8-4-4a

Figure 2.38: Another way of looking at the ‘L’ shaped object is as a square minus a smaller square in its upper right-hand corner.

Box 2.5 Why can subsystems be treated like particles when finding the center-of-mass?

This is a theoretical aside, not needed for problem solving.

Filename:tfigure4-dottysubsys

Look at the collection of 47 particles above and then think of it as a set of three subsystems: I, II, and III with 2, 14, and 31 particles, respectively. Masses 1 and 2 as subsystem $I$ with center-of-mass ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{I}}$ and total mass $m_{I}$ . Similarly, we call subsystem $I I$ masses $m_{3}$ to $m_{16}$ , and subsystem $I I I$ , masses $m_{17}$ to $m_{47}$ . We can calculate the center-of-mass of the system by treating it as 47 particles, or we can re-arrange the sum as follows:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}\!\!$	$\displaystyle=$	$\displaystyle\!\!\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{1}}m_{1}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}m% _{2}+\cdots+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{46}}m_{46% }+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{47}}m_{47}}{m_{1}+m% _{2}+\cdots+m_{47}}$
	$\displaystyle=$	$\displaystyle\frac{\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{1}}m_{1}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}m% _{2}}{m_{1}+m_{2}}(m_{1}+m_{2})}{m_{1}+m_{2}+\cdots+m_{47}}$
		$\displaystyle+\frac{\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{3}}m_{3}+\cdots+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{16}}m_{16}}{m_{3}+\cdots+m_{16}}(m_{3}+\cdots+m_{16})}{m_{1}+m_{2}+% \cdots+m_{47}}$
		$\displaystyle+\frac{\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{17}}m_{17}+\cdots{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{47}}m_{47}}{m_{17}+\cdots+m_{47}}{(m_{17}+\cdots+m_{47})}}{{m_{1}+m_{2% }+\cdots+m_{47}}}$

		$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{I}% }m_{I}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{II}}m_{II}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{III}}m_{III}}{m_{I}+m_% {II}+m_{III}}\mbox{ , where}$
			$\displaystyle\textrm{where \ }{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% \!\mathrm{I}}=\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1% }}m_{1}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}m_{2}}{m_{% 1}+m_{2}},$
			$\displaystyle\phantom{\textrm{where \ }}m_{I}=m_{1}+m_{2}$
			$\displaystyle\phantom{\textrm{where \ }}{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{II}}=\textit{etc.}$

That is, the center of mass of the 47 particles is the same as the center of mass of three particles, where each of the three particles has the total mass of its subsystem located at its subsystem’s center of mass.

The reduction of subsystem of particles to one particle is easily generalized to the integral formulae as well like this.

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\frac{\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}\ dm}{\int dm}$
	$\displaystyle=$	$\displaystyle\frac{\int_{\mbox{\tiny region 1}}{\accentset{\rightharpoonup}{% \boldsymbol{r}}}\ dm+\int_{\mbox{\tiny region 2}}{\accentset{\rightharpoonup}{% \boldsymbol{r}}}\ dm+\int_{\mbox{\tiny region 3}}{\accentset{\rightharpoonup}{% \boldsymbol{r}}}\ dm+\cdots}{\int_{\mbox{\tiny region 1}}dm+\int_{\mbox{\tiny region% 2}}dm+\int_{\mbox{\tiny region 3}}dm+\cdots}$
	$\displaystyle=$	$\displaystyle\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{I}% }m_{I}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{II}}m_{II}+{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{III}}m_{III}+\cdots}{m% _{I}+m_{II}+m_{III}+\cdots}.$

The general idea of the calculations above is that center-of-mass calculations are basically big sums (addition), and addition is ‘associative.’

Center of gravity

The force of gravity on each little bit of an object is $gm_{i}$ where $g$ is the local gravitational ‘constant’ and $m_{i}$ is the mass of the bit. For objects that are small compared to the radius of the earth (a reasonable assumption for all but a few special engineering calculations) the gravity constant is indeed constant from one point on the object to another (see box 4.1 on page 4.1 for a discussion of the meaning and history of $g$ .)

Not only that, all the gravity forces point in the same direction, down. (For engineering purposes, the two intersecting lines that go from your two hands to the center of the earth are parallel.)

Let’s call this the $-\hat{\boldsymbol{k}}$ direction.

So the net force of gravity on an object is:

\begin{array}[]{ccccccccl}{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{net}}&=&\sum&{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!% \mathrm{i}}&=&\sum m_{i}g(-\hat{\boldsymbol{k}})&=&-mg\hat{\boldsymbol{k}}&% \mbox{(discrete systems)}\\ &=&\int&d{\accentset{\rightharpoonup}{\boldsymbol{F}}}&=&\int\underbrace{-g% \hat{\boldsymbol{k}}\,dm}_{d{\accentset{\rightharpoonup}{\boldsymbol{F}}}}&=&-% mg\hat{\boldsymbol{k}}&\mbox{(continuous systems)}\end{array}

That’s easy, the billions of gravity forces on an objects microscopic constituents add up to $m g$ pointed down. What about the net moment of the gravity forces? The answer turns out to be simple. The top line of the calculation below poses the question, the last line gives the lucky answer.^†^†margin: We do the calculation here using the $\int$ notation for sums. But it could be done just as well using $\sum$ .

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{C}}$	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}\times d{% \accentset{\rightharpoonup}{\boldsymbol{F}}}$	The net moment with respect to C.
	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{/C}}% \times\left(-g\hat{\boldsymbol{k}}\,dm\right)$	A force bit is gravity acting on a mass bit.
	$\displaystyle=$	$\displaystyle\left(\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{/C}}dm\right)\times\left(-g\hat{\boldsymbol{k}}\right)\,$	Distributive law ( $g$ & $\hat{\boldsymbol{k}}$ are constants).
	$\displaystyle=$	$\displaystyle({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm/C}}m% )\times\left(-g\hat{\boldsymbol{k}}\right)\,$	Definition of center-of-mass.
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm/C}}% \times\left(-mg\hat{\boldsymbol{k}}\right)\,$	Re-arranging terms.
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm/C}}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}\,$	Express in terms of net gravity force.

Thus the net moment is the same as for the total gravity force acting at the center-of-mass.

The near-earth gravity forces acting on a system are equivalent to a single force, $m g$ , acting at the system’s center-of-mass.

For the purposes of calculating the net force and moment from near-earth (constant g) gravity forces, a system can be replaced by a point mass at the center of gravity. The words ‘center-of-mass’ and ‘center of gravity’ both describe the same point in space.

Although the result we have just found seems plain enough, here are two things to ponder about gravity when viewed as an inverse square law (and thus not constant like we have assumed) that may make the result above seem less obvious.

•

The net gravity force on a sphere is indeed equivalent to the force of a point mass at the center of the sphere. It took the genius Isaac Newton 3 years to deduce this result and the reasoning involved is too advanced for this book.
•

The net gravity force on systems that are not spheres is generally not equivalent to a force acting at the center-of-mass (this is important for the understanding of tides as well as the orientational stability of satellites).

How to find the center-of-mass of a complex system

You find the center-of-mass of a complex system by knowing the masses and mass centers of its components. You find each of these centers of mass by

•

Treating it as a point mass, or
•

Treating it as a symmetric body and locating the center-of-mass in the middle, or
•

Using integration, or
•

Using the result of an experiment (which we will discuss in statics), or
•

Treating the component as a complex system itself and applying this very recipe.

The recipe is just an application of the basic definition of center-of-mass (eqn. 2.2) but with our accumulated wisdom that the locations and masses in that sum can be the centers of mass and total masses of complex subsystems.

One way to arrange one’s data is in a table or spreadsheet, like below.

Center of mass spreadsheet

Subsys#	1	2	3	4	5	6	7
Subsys 1	$x_{1}$	$y_{1}$	$z_{1}$	$m_{1}$	$m_{1}x_{1}$	$m_{1}y_{1}$	$m_{1}z_{1}$
Subsys 2	$x_{2}$	$y_{2}$	$z_{2}$	$m_{2}$	$m_{2}x_{2}$	$m_{2}y_{2}$	$m_{2}z_{2}$
$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$	$\vdots$
Subsys $N$	$x_{N}$	$y_{N}$	$z_{N}$	$m_{N}$	$m_{N}x_{N}$	$m_{N}y_{N}$	$m_{N}z_{N}$
Row $N+1$				$m_{\mathrm{tot}}=$
=sums				$\sum m_{i}$	$\sum m_{i}x_{i}$	$\sum m_{i}y_{i}$	$\sum m_{i}z_{i}$
divide row $N+1$ by $m_{\mathrm{tot}}$					$\displaystyle x_{\rm cm}=$	$\displaystyle y_{\rm cm}=$	$\displaystyle z_{\rm cm}=$
$\Rightarrow$ Result					$\displaystyle\frac{\sum^{\mathstrut}m_{i}x_{i}}{\mathstrut m_{\mathrm{tot}}}$	$\displaystyle\frac{\sum m_{i}y_{i}}{m_{\mathrm{tot}}}$	$\displaystyle\frac{\sum m_{i}z_{i}}{m_{\mathrm{tot}}}$

1.

The first four columns are the basic data. They are the $x, y$ , and $z$ coordinates of the subsystem center-of-mass locations (relative to some clear reference point), and the masses of the subsystems, one row for each of the $N$ subsystems.
2.

One next calculates three new columns (5,6, and 7) which come from each coordinate multiplied by its mass. For example the entry in the 6th row and 7th column is the $z$ component of the 6th subsystem’s center-of-mass multiplied by the mass of the 6th subsystem.
3.

Then one sums columns 4 through 7. The sum of column 4 is the total mass, the sums of columns 5 through 7 are the total mass-weighted positions.
4.

Finally the result, the system center of mass coordinates, are found by dividing columns 5-7 of row N+1 by column 4 of row N+1.

Of course, there are multiple ways of systematically representing the data. The spreadsheet-like calculation above is just one organization scheme.

Summary of center-of-mass

All discussions in mechanics make frequent reference to the concept of center of mass

$\displaystyle m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% \!\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{i}}m% _{i}$	for discrete systems or systems of systems
	$\displaystyle=$	$\displaystyle\int{\accentset{\rightharpoonup}{\boldsymbol{r}}}dm$	for continuous systems
where
$\displaystyle m_{\mathrm{tot}}$	$\displaystyle=$	$\displaystyle\sum\,\mathrm{m}_{i}$	for discrete systems or systems of systems
	$\displaystyle=$	$\displaystyle\int dm$	for continuous systems.

Who cares about the center of mass?

We showed that the gravity moment is calculated correctly by applying the net gravity force at the center-of-mass. These other facts about center-of-mass will be needed for dynamics.

For non-point-mass systems, the expressions for linear momentum, angular momentum, and kinetic energy are all simplified by using the center-of-mass.

Simple center-of-mass calculations also can serve as a check of a more complicated analysis. For example, after a computer simulation of a system with many moving parts is complete, one way of checking the calculation is to see if the whole system’s center of mass moves as would be expected by applying the net external force to the system.

Box 2.6 The COM of a uniform triangle is $h/3$ up from the base

The theory here is for the curious. One can just remember the result: The center-of-mass of a 2D uniform triangular region is the centroid of the area.

Filename:tfigure-triangle1

First, we consider a right triangle with perpendicular sides $b$ and $h$

Filename:tfigure-triangle2

and find the $x$ coordinate of the centroid as

$\displaystyle x_{\rm cm}A$	$\displaystyle=$	$\displaystyle\int x\,dA$
	$\displaystyle=$	$\displaystyle\int_{0}^{h}\left[\int_{0}^{{b\over h}x}x\,dy\right]\,dx=\left.% \int_{0}^{h}\left[xy\right]\right\|^{y={b\over h}x}_{y=0}\ dx$
$\displaystyle x_{\rm cm}\left({bh\over 2}\right)$	$\displaystyle=$	$\displaystyle\int_{0}^{h}x\left({b\over h}x\right)\,dx=\left.{b\over h}{x^{3}% \over 3}\right\|^{h}_{0}={bh^{2}\over 3}$

$\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\displaystyle x_{\rm cm}={2h% \over 3}$ , a third of the way to the left of the vertical base on the right. By similar reasoning, but in the $y$ direction, the centroid is a third of the way up from the base.

Filename:tfigure-triangle3

The center-of-mass of an arbitrary triangle can be found by treating it as the sum of two right triangles

Filename:tfigure-triangle4

so the centroid is a third of the way up from the base of any triangle. Finally, the result holds for all three bases. Summarizing, the centroid of a triangle is at the point one third up from each of the bases.

Filename:tfigure-triangle5

Non-calculus approach

Filename:tfigure-triangle6

M is the midpoint of the line segment BC. Divide triangle ABC into equal width strips that are parallel to AM. Group these strips into pairs, each a distance $s$ from AM. Because M is the midpoint of BC, by proportions each of these strips has the same length $\ell$ . What is the distance of the center-of-mass from the line AM? Because the strips are of equal area and equal distance from AM but on opposite sides, contributions to the sum come in canceling pairs. So the centroid is on AM. Likewise for all three sides. So the triangle’s centroid is at the intersection of the three side bisectors.

Why do the three side bisectors intersect a third of the way up each base? Look at the 6 triangles formed by the side bisectors.

Filename:tfigure-triangle7

The two triangles marked $a$ and $a$ have the same area (call it $a$ ) because they have the same height and bases of equal length (BM and CM). Likewise for the other side bisectors, so that the pairs marked $b$ have equal area as do the pairs $c$ . Triangle ABM has the same base and height and thus the same area as the triangle ACM. So $a+b+b=a+c+c$ . Thus $b=c$ and similarly $a=b$ : all six little triangles have equal area. Thus the area of ABC is 3 times the area of GBC. Because ABC and GBC share the base BC, ABC must have 3 times the height as GBC, and point G is thus a third of the way up from the base.

Where is the middle of a triangle?

We just showed that the centroid of a triangle is at the point that is at the intersection of: the three side bisectors; the three area bisectors (which are the side bisectors); and the three lines one third of the way up from the three bases.

And, if the triangle had three equal point masses on its vertices, and no mass anywhere else, the center of mass also lands on that same place. Thus the ‘middle’ of a triangle seems pretty well defined.

Yet, there is ambiguity. If the triangle were made of bars along each edge, each with equal cross sections, the center-of-mass would be in a different location (except for equilateral triangles). Also, the three angle bisectors of a triangle do not intersect at the centroid. Unless we define the middle to mean centroid, the “middle” of a triangle is not well defined.

SAMPLE 2.7

Filename:sfig8-4-4b

Figure 2.39:

Center of mass in 1-D: Three particles (point masses) of mass 2 $\,\mathrm{kg}$ , 3 $\,\mathrm{kg}$ , and 3 $\,\mathrm{kg}$ , are welded to a straight massless rod as shown in the figure. Find the location of the center-of-mass of the assembly.

Solution Let us select the first mass, $m_{1}=2\,\mathrm{kg}$ , to be at the origin of our co-ordinate system with the $x$ -axis along the rod. Since all the three masses lie on the $x$ -axis, the center-of-mass will also lie on this axis. Let the center-of-mass be located at $x_{cm}$ on the $x$ -axis. Then,

Filename:sfig8-4-4c

Figure 2.40:

$\displaystyle m_{\mathrm{tot}}x_{cm}$	$\displaystyle=$	$\displaystyle\sum_{i=1}^{3}m_{i}x_{i}=m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}$
	$\displaystyle=$	$\displaystyle m_{1}(0)+m_{2}(\ell)+m_{3}(2\ell)$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}x_{cm}$	$\displaystyle=$	$\displaystyle\frac{m_{2}\ell+m_{3}2\ell}{m_{1}+m_{2}+m_{3}}$
	$\displaystyle=$	$\displaystyle\frac{3\cancel{\,\mathrm{kg}}\cdot 0.2\,\mathrm{m}+3\cancel{\,% \mathrm{kg}}\cdot 0.4\,\mathrm{m}}{(2+3+3)\cancel{\,\mathrm{kg}}}$
	$\displaystyle=$	$\displaystyle\frac{1.8\,\mathrm{m}}{8}=0.225\,\mathrm{m}.$

Answer: $x_{cm}=0.225\,\mathrm{m}$

Filename:pfigure-s94h13p2

Figure 2.41:

Alternatively, we could find the center-of-mass by first replacing the two 3 $\,\mathrm{kg}$ masses with a single 6 $\,\mathrm{kg}$ mass located in the middle of the two masses (the center-of-mass of the two equal masses) and then calculate the value of $x_{cm}$ for a two particle system consisting of the 2 $\,\mathrm{kg}$ mass and the 6 $\,\mathrm{kg}$ mass (see Fig. 2.41):

x_{cm}=\frac{6\,\mathrm{kg}\cdot 0.3\,\mathrm{m}}{8\,\mathrm{kg}}=\frac{1.8\,% \mathrm{m}}{8}=0.225\,\mathrm{m}.

SAMPLE 2.8 Center of mass in 2-D: Two particles of mass $m_{1}=1\,\mathrm{kg}$ and $m_{2}=2\,\mathrm{kg}$ are located at coordinates (1m, 2m) and (-2m, 5m), respectively, in the $x y$ -plane. Find the location of their center-of-mass.

Solution Let ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}$ be the position vector of the center-of-mass. Then,

$\displaystyle m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% \!\mathrm{cm}}$	$\displaystyle=$	$\displaystyle m_{1}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}% }+m_{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\frac{m_{1}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{1}}+m_{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}}% {m_{\mathrm{tot}}}=\frac{m_{1}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% \!\mathrm{1}}+m_{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}% }}{m_{1}+m_{2}}$
	$\displaystyle=$	$\displaystyle\frac{1\cancel{\,\mathrm{kg}}(1\,\mathrm{m}\hat{\boldsymbol{% \imath}}+2\,\mathrm{m}\hat{\boldsymbol{\jmath}})+2\cancel{\,\mathrm{kg}}(-2\,% \mathrm{m}\hat{\boldsymbol{\imath}}+5\,\mathrm{m}\hat{\boldsymbol{\jmath}})}{3% \cancel{\,\mathrm{kg}}}$
	$\displaystyle=$	$\displaystyle\frac{(1\,\mathrm{m}-4\,\mathrm{m})\hat{\boldsymbol{\imath}}+(2\,% \mathrm{m}+10\,\mathrm{m})\hat{\boldsymbol{\jmath}}}{3}=-1\,\mathrm{m}\hat{% \boldsymbol{\imath}}+4\,\mathrm{m}\hat{\boldsymbol{\jmath}}.$

Filename:pfigure-f93f5

Figure 2.42:

Thus the center-of-mass is located at the coordinates(-1m, 4m).

Answer: $(x_{cm},y_{cm})=(-1\,\mathrm{m},4\,\mathrm{m})$

Geometrically, this is just a 1-D problem like the previous sample. The center-of-mass has to be located on the straight line joining the two masses. Since the center-of-mass is a point about which the distribution of mass is balanced, it is easy to see (see Fig. 2.42) that the center-of-mass must lie one-third way from $m_{2}$ on the line joining the two masses so that $2\,\mathrm{kg}\cdot(d/3)=1\,\mathrm{kg}\cdot(2d/3)$ .

SAMPLE 2.9

Filename:pfigure-s94h13p3

Figure 2.43:

Location of the center-of-mass. A structure is made up of three point masses, $m_{1}=1\,\mathrm{kg},\;m_{2}=2\,\mathrm{kg}$ and $m_{3}=3\,\mathrm{kg}$ , connected rigidly by massless rods. At the instant of interest, the coordinates of the three masses are (1.25 $\,\mathrm{m}$ , 3 $\,\mathrm{m}$ ), (2 $\,\mathrm{m}$ , 2 $\,\mathrm{m}$ ), and (0.75 $\,\mathrm{m}$ , 0.5 $\,\mathrm{m}$ ), respectively. At the same instant, the velocities of the three masses are $2\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\imath}},\;2\,\mathrm{m}/\mathrm{s}(% \hat{\boldsymbol{\imath}}-1.5\hat{\boldsymbol{\jmath}})$ and $1\,\mathrm{m}/\mathrm{s}\hat{\boldsymbol{\jmath}}$ , respectively. Find the coordinates of the center-of-mass of the structure.

Solution Just for fun, let us do this problem two ways — first using scalar equations for the coordinates of the center-of-mass, and second, using vector equations for the position of the center-of-mass.

1.

Scalar calculations: Let $(x_{cm},y_{cm})$ be the coordinates of the mass-center. Then from the definition of mass-center,

$\displaystyle x_{cm}$ $\displaystyle=$ $\displaystyle\frac{\sum m_{i}x_{i}}{\sum m_{i}}=\frac{m_{1}x_{1}+m_{2}x_{2}+m_% {3}x_{3}}{m_{1}+m_{2}+m_{3}}$

$\displaystyle=$ $\displaystyle\frac{1\,\mathrm{kg}\cdot 1.25\,\mathrm{m}+2\,\mathrm{kg}\cdot 2% \,\mathrm{m}+3\,\mathrm{kg}\cdot 0.75\,\mathrm{m}}{1\,\mathrm{kg}+2\,\mathrm{% kg}+3\,\mathrm{kg}}$

$\displaystyle=$ $\displaystyle\frac{7.5\cancel{\,\mathrm{kg}}\cdot\,\mathrm{m}}{6\cancel{\,% \mathrm{kg}}}=1.25\,\mathrm{m}.$

Similarly,

$\displaystyle y_{cm}$ $\displaystyle=$ $\displaystyle\frac{\sum m_{i}y_{i}}{\sum m_{i}}=\frac{m_{1}y_{1}+m_{2}y_{2}+m_% {3}y_{3}}{m_{1}+m_{2}+m_{3}}$

$\displaystyle=$ $\displaystyle\frac{1\,\mathrm{kg}\cdot 3\,\mathrm{m}+2\,\mathrm{kg}\cdot 2\,% \mathrm{m}+3\,\mathrm{kg}\cdot 0.5\,\mathrm{m}}{1\,\mathrm{kg}+2\,\mathrm{kg}+% 3\,\mathrm{kg}}$

$\displaystyle=$ $\displaystyle\frac{8.5\cancel{\,\mathrm{kg}}\cdot\,\mathrm{m}}{6\cancel{\,% \mathrm{kg}}}=1.42\,\mathrm{m}.$

Thus the center-of-mass is located at the coordinates (1.25 $\,\mathrm{m}$ , 1.42 $\,\mathrm{m}$ ).

Answer: $(1.25\,\mathrm{m},\;1.42\,\mathrm{m})$

Vector calculations: Let ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}$ be the position vector of the mass-center. Then,

	$\displaystyle m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% \!\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\sum_{i=1}^{3}m_{i}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_% {\!\mathrm{i}}=m_{1}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1% }}+m_{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}+m_{3}{% \accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{3}}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\frac{m_{1}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{1}}+m_{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}+% m_{3}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{3}}}{m_{1}+m_{2}% +m_{3}}$

Substituting the values of $m_{1},m_{2}$ , and $m_{3}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}=1.25\,\mathrm{m}% \hat{\boldsymbol{\imath}}+3\,\mathrm{m}\hat{\boldsymbol{\jmath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}~{}=~{}2\,\mathrm{% m}\hat{\boldsymbol{\imath}}+2\,\mathrm{m}\hat{\boldsymbol{\jmath}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{3}}=0.75\,\mathrm{m}% \hat{\boldsymbol{\imath}}+0.5\,\mathrm{m}\hat{\boldsymbol{\jmath}}$ , we get,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\frac{1\,\mathrm{kg}\cdot(1.25\hat{\boldsymbol{\imath}}+3\hat{% \boldsymbol{\jmath}})\,\mathrm{m}+2\,\mathrm{kg}\cdot(2\hat{\boldsymbol{\imath% }}+2\hat{\boldsymbol{\jmath}})\,\mathrm{m}+3\,\mathrm{kg}\cdot(0.75\hat{% \boldsymbol{\imath}}+0.5\hat{\boldsymbol{\jmath}})\,\mathrm{m}}{(1+2+3)\,% \mathrm{kg}}$
	$\displaystyle=$	$\displaystyle\frac{(7.5\hat{\boldsymbol{\imath}}+8.5\hat{\boldsymbol{\jmath}})% \cancel{\,\mathrm{kg}}\cdot\,\mathrm{m}}{6\cancel{\,\mathrm{kg}}}$
	$\displaystyle=$	$\displaystyle 1.25\,\mathrm{m}\hat{\boldsymbol{\imath}}+1.42\,\mathrm{m}\hat{% \boldsymbol{\jmath}}$

which, of course, gives the same location of the mass-center as above.

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}=1.25\,\mathrm{m}% \hat{\boldsymbol{\imath}}+1.42\,\mathrm{m}\hat{\boldsymbol{\jmath}}$

SAMPLE 2.10

Filename:p-s96-p3-3

Figure 2.44:

Center of mass of a bent bar: A uniform bar of mass 4 $\,\mathrm{kg}$ is bent in the shape of an asymmetric ’Z’ as shown in the figure. Locate the center-of-mass of the bar.

Solution

Filename:bikefork1-ang-mom

Figure 2.45:

Since the bar is uniform along its length, we can divide it into three straight segments and use their individual mass-centers (located at the geometric centers of each segment) to locate the center-of-mass of the entire bar. The mass of each segment is proportional to its length. Therefore, if we let $m_{2}=m_{3}=m$ , then $m_{1}=2m$ ; and $m_{1}+m_{2}+m_{3}=4m=4\,\mathrm{kg}$ which gives $m=1\,\mathrm{kg}$ . Now, from Fig. 2.45,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}$	$\displaystyle=$	$\displaystyle\ell\hat{\boldsymbol{\imath}}+\ell\hat{\boldsymbol{\jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}$	$\displaystyle=$	$\displaystyle 2\ell\hat{\boldsymbol{\imath}}+\frac{\ell}{2}\hat{\boldsymbol{% \jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{3}}$	$\displaystyle=$	$\displaystyle(2\ell+\frac{\ell}{2})\hat{\boldsymbol{\imath}}=\frac{5\ell}{2}% \hat{\boldsymbol{\imath}}$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\frac{m_{1}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{1}}+m_{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}+% m_{3}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{3}}}{m_{\mathrm{% tot}}}$
	$\displaystyle=$	$\displaystyle\frac{2m(\ell\hat{\boldsymbol{\imath}}+\ell\hat{\boldsymbol{% \jmath}})+m(2\ell\hat{\boldsymbol{\imath}}+\frac{\ell}{2}\hat{\boldsymbol{% \jmath}})+m(\frac{5\ell}{2}\hat{\boldsymbol{\imath}})}{4m}$
	$\displaystyle=$	$\displaystyle\frac{\cancel{m}\ell(2\hat{\boldsymbol{\imath}}+2\hat{\boldsymbol% {\jmath}}+2\hat{\boldsymbol{\imath}}+\frac{1}{2}\hat{\boldsymbol{\jmath}}+% \frac{5}{2}\hat{\boldsymbol{\imath}})}{4\cancel{m}}$
	$\displaystyle=$	$\displaystyle\frac{\ell}{8}(13\hat{\boldsymbol{\imath}}+5\hat{\boldsymbol{% \jmath}})$
	$\displaystyle=$	$\displaystyle\frac{0.5\,\mathrm{m}}{8}(13\hat{\boldsymbol{\imath}}+5\hat{% \boldsymbol{\jmath}})$
	$\displaystyle=$	$\displaystyle 0.812\,\mathrm{m}\hat{\boldsymbol{\imath}}+0.312\,\mathrm{m}\hat% {\boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}=0.812\,\mathrm{m% }\hat{\boldsymbol{\imath}}+0.312\,\mathrm{m}\hat{\boldsymbol{\jmath}}$

Geometrically, we could find the center-of-mass by considering two masses at a time, connecting them by a line and locating their mass-center on that line, and then repeating the process as shown in Fig. 2.46.

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Figure 2.46:

The center-of-mass of $m_{2}$ and $m_{3}$ (each of mass $m$ ) is at the mid-point of the line connecting the two masses. Now, we replace these two masses with a single mass $2m$ at their mass-center. Next, we connect this mass-center and $m_{1}$ with a line and find their combined mass-center at the mid-point of this line. The mass-center just found is the center-of-mass of the entire bar.

SAMPLE 2.11

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Figure 2.47:

Shift of mass-center due to cut-outs: A $2\,\mathrm{m}\times 2\,\mathrm{m}$ uniform square plate has mass $m=4\,\mathrm{kg}$ . A circular section of radius 250 mm is cut out from the plate as shown in the figure. Find the center-of-mass of the plate.

Solution

Let us use an $x y$ -coordinate system with its origin at the geometric center of the plate and the $x$ -axis passing through the center of the cut-out. Since the plate and the cut-out are symmetric about the $x$ -axis, the new center-of-mass must lie somewhere on the $x$ -axis. Thus, we only need to find $x_{cm}$ (since $y_{cm}=0$ ). Let $m_{1}$ be the mass of the plate with the hole, and $m_{2}$ be the mass of the circular cut-out. Clearly, $m_{1}+m_{2}=m=4\,\mathrm{kg}$ . The center-of-mass of the circular cut-out is at A, the center of the circle. The center-of-mass of the intact square plate (without the cut-out) must be at O, the middle of the square. Then,

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Figure 2.48:

	$\displaystyle m_{1}x_{cm}+m_{2}x_{A}$	$\displaystyle=$	$\displaystyle mx_{O}=0$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}x_{cm}$	$\displaystyle=$	$\displaystyle-\frac{m_{2}}{m_{1}}x_{A}.$

Now, since the plate is uniform, the masses $m_{1}$ and $m_{2}$ are proportional to the surface areas of the geometric objects they represent, i.e.,

$\displaystyle\frac{m_{2}}{m_{1}}$	$\displaystyle=$	$\displaystyle\frac{\pi r^{2}}{\ell^{2}-\pi r^{2}}=\frac{\pi}{\left(\frac{\ell}% {r}\right)^{2}-\pi}.$

$\displaystyle x_{cm}$	$\displaystyle=$	$\displaystyle-\frac{m_{2}}{m_{1}}d=-\frac{\pi}{\left(\frac{\ell}{r}\right)^{2}% -\pi}d$
	$\displaystyle=$	$\displaystyle-\frac{\pi}{\left(\frac{2\,\mathrm{m}}{.25\,\mathrm{m}}\right)^{2% }-\pi}\cdot 0.5\,\mathrm{m}$
	$\displaystyle=$	$\displaystyle-25.81\times{10}^{-3}\,\mathrm{m}=-25.81\,{\rm mm}$

Thus the center-of-mass shifts to the left by about 26 mm because of the circular cut-out of the given size.

Answer: $x_{cm}=-25.81\,{\rm mm}$

Comments: The advantage of finding the expression for $x_{cm}$ in terms of $r$ and $\ell$ as in eqn. (2.2) is that you can easily find the center-of-mass of any size circular cut-out located at any distance $d$ on the $x$ -axis. This is useful in design where you like to select the size or location of the cut-out to have the center-of-mass at a particular location.

SAMPLE 2.12

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Figure 2.49:

Center of mass of two objects: A square block of side $0.1\,\mathrm{m}$ and mass $2\,\mathrm{kg}$ sits on the side of a triangular wedge of mass $6\,\mathrm{kg}$ as shown in the figure. Locate the center-of-mass of the combined system.

Solution

The center-of-mass of the triangular wedge is located at $h/3$ above the base and $\ell/3$ to the right of the vertical side. Let $m_{1}$ be the mass of the wedge and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}$ be the position vector of its mass-center. Then, referring to Fig. 2.50,

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{1}}=\frac{\ell}{3}% \hat{\boldsymbol{\imath}}+\frac{h}{3}\hat{\boldsymbol{\jmath}}.

The center-of-mass of the square block is located at its geometric center $C_{2}$ . From geometry, we can see that the line AE that passes through $C_{2}$ is horizontal since $\angle OAB=45^{\circ}$ ( $h=\ell=0.3\,\mathrm{m}$ ) and $\angle DAE=45^{\circ}$ . Therefore, the coordinates of $C_{2}$ are ( $d/\sqrt{2},h$ ). Let $m_{2}$ and ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}$ be the mass and the position vector of the mass-center of the block, respectively. Then,

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}=\frac{d}{\sqrt{2}% }\hat{\boldsymbol{\imath}}+h\hat{\boldsymbol{\jmath}}.

Now, noting that $m_{1}=3m_{2}$ or $m_{1}=3m,\qquad\mbox{and}\qquad m_{2}=m$ where $m=2\,\mathrm{kg}$ , we find the center-of-mass of the combined system:

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Figure 2.50:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}$	$\displaystyle=$	$\displaystyle\frac{m_{1}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!% \mathrm{1}}+m_{2}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{2}}}% {(m_{1}+m_{2})}$
	$\displaystyle=$	$\displaystyle\frac{3m(\frac{\ell}{3}\hat{\boldsymbol{\imath}}+\frac{h}{3}\hat{% \boldsymbol{\jmath}})+m(\frac{d}{\sqrt{2}}\hat{\boldsymbol{\imath}}+h\hat{% \boldsymbol{\jmath}})}{3m+m}$
	$\displaystyle=$	$\displaystyle\frac{\cancel{m}[(\ell+\frac{d}{\sqrt{2}})\hat{\boldsymbol{\imath% }}+2h\hat{\boldsymbol{\jmath}}]}{4\cancel{m}}$
	$\displaystyle=$	$\displaystyle\frac{1}{4}(\frac{d}{\sqrt{2}}+\ell)\hat{\boldsymbol{\imath}}+% \frac{h}{2}\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle\frac{1}{4}(\frac{0.1\,\mathrm{m}}{\sqrt{2}}+0.3\,\mathrm{m})\hat% {\boldsymbol{\imath}}+\frac{0.3\,\mathrm{m}}{2}\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle 0.093\,\mathrm{m}\hat{\boldsymbol{\imath}}+0.150\,\mathrm{m}\hat% {\boldsymbol{\jmath}}.$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{cm}}=0.093\,\mathrm{m% }\hat{\boldsymbol{\imath}}+0.150\,\mathrm{m}\hat{\boldsymbol{\jmath}}$

Thus, the center-of-mass of the wedge and the block together is slightly closer to the side OA and higher up from the bottom OB than $C_{1}(0.1\,\mathrm{m},0.1\,\mathrm{m})$ . This is what we should expect from the placement of the square block.

Note that we could have, again, used a 1-D calculation by placing a point mass $3m$ at $C_{1}$ and $m$ at $C_{2}$ , connected the two points by a straight line, and located the center-of-mass $C$ on that line such that $CC_{2}=3CC_{1}$ . You can verify that the distance from $C_{1}(0.1\,\mathrm{m},0.1\,\mathrm{m})$ to $C(.093\,\mathrm{m},0.15\,\mathrm{m})$ is one third the distance from $C$ to $C_{2}(.071\,\mathrm{m},0.3\,\mathrm{m})$ .

Problems for 2.2 Center of mass and center of gravity

2.2.1 An otherwise massless structure is made of four point masses, $m$ , $2m$ , $3m$ and $4m$ , located at coordinates (0, 1 $\,\mathrm{m}$ ), (1 $\,\mathrm{m}$ , 1 $\,\mathrm{m}$ ), (1 $\,\mathrm{m}$ , $-1\,\mathrm{m}$ ), and (0, $-1\,\mathrm{m}$ ), respectively. Locate the center of mass of the structure. Answer: $(0.5\,\mathrm{m},\;-0.4\,\mathrm{m})$

2.2.2 3-D: The following data is given for a structural system modeled with five point masses in 3-D-space:

mass coordinates (in m)
$0.4\,\mathrm{kg}$	(1,0,0)
$0.4\,\mathrm{kg}$	(1,1,0)
$0.4\,\mathrm{kg}$	(2,1,0)
$0.4\,\mathrm{kg}$	(2,0,0)
$1.0\,\mathrm{kg}$	(1.5,1.5,3)

Locate the center of mass of the system.

2.2.3 Write a computer program to find the center of mass of a point-mass-system. The input to the program should be a table (or matrix) containing individual masses and their coordinates. (It is possible to write a single program for both 2-D and 3-D cases, write separate programs for the two cases if that is easier for you.) Check your program on Problems 2.2 and 2.2.

2.2.4 A cylinder of mass $m_{2}$ and radius $R$ rolls on a flat circular plate of mass $m_{1}$ and length $\ell$ . Let the position of the cylinder from the left edge of the plate be $x$ . Find the horizontal position of the center of mass of the system as a function of $x$ and a non-dimensional mass parameter $M=m_{1}/m_{2}$ .

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Figure 2.51

2.2.5 Two masses $m_{1}$ and $m_{2}$ are connected by a massless rod AB of length $\ell$ . In the position shown, the rod is inclined to the horizontal axis at an angle $\theta$ . Find the position of the center of mass of the system as a function of angle $\theta$ and the other given variables. Check if your answer makes sense by setting appropriate values for $m_{1}$ and $m_{2}$ .

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Figure 2.52

2.2.6 Find the center of mass of the following composite bars. Each composite shape is made of two or more uniform bars of length $0.2\,\mathrm{m}$ and mass $0.5\,\mathrm{kg}$ .

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Figure 2.53

2.2.7 A double pendulum consists of two uniform bars of length $\ell$ and mass $m$ each. The pendulum hangs in the vertical plane from a hinge at point O. Taking O as the origin of a $x y$ coordinate system, find the location of the center of gravity of the pendulum as a function of angles $\theta_{1}$ and $\theta_{2}$ .

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Figure 2.54

2.2.8 Find the center of mass of the following two objects [Hint: set up and evaluate the needed integrals.]

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Figure 2.55

2.2.9 A semicircular ring of radius $R=1\,\mathrm{m}$ and mass $m_{1}=0.1\,\mathrm{kg}$ rests in the vertical plane. A bead of mass $m_{2}=0.25\,\mathrm{kg}$ slides on the ring. Find the position of the center of mass of the ring-bead-system at an instant when $\theta=30^{{}^{\circ}}$ . How does the center of mass position change as $\theta$ changes?

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Figure 2.56

2.2.10 A uniform circular disk of mass $m$ and radius $R$ rolls on an inclined rectangular plate of mass $3m$ and dimensions $2R\times\ell$ . Point A is on the $y$ axis and point B is on the $x$ axis. Find the coordinates of the center of mass of the system for $m=1\,\mathrm{kg},\ell=1\,\mathrm{m},z=0.2\,\mathrm{m}$ , and $R=0.1\,\mathrm{m}$ .

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Figure 2.57

2.2.11 Find the center of mass of the following plates obtained from cutting out a section from a uniform circular plate of mass $1\,\mathrm{kg}$ (prior to removing the cutout) and radius $1/4\,\mathrm{m}$ .

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Figure 2.58

2.3 Free-body diagrams: interactions, showing forces, and partial FBDs

A free-body diagram is a sketch of the system of interest and the forces that act on the system.

A free-body diagram precisely defines the system to which you are applying mechanics equations and the forces to be considered. Any reader of your calculations needs to see your free-body diagrams. To put it directly, if you want to be right and be seen as right, then

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Figure 2.59: A sketch of a bicycle and a free-body diagram of the braked front wheel. A sketch of a person and a free-body diagram of the whole person.

^†^†margin: Free-body Perkins. In the 1950’s Cornell professor Harold C. Perkins was nick-named ‘Free-Body Perkins’. Perkins would stop students in the hall and say “You! Come in my office! Draw a free-body diagram of this!” Students drew free-body diagrams to please persnickety Perkins. And, so learned how to get mechanics problems right.

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Figure 2.60: The process of drawing a FBD is illustrated by the sequence shown.

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Figure 2.61: Pliers crushing a pencil. Some possible free-body diagrams (FBDs), neglecting gravity. With four major parts (upper jaw, lower jaw, pin & pencil) there are 15 possible subsystems: a) Whole system; b) Upper jaw; c) Pencil; d) Connecting pin; e) Lower jaw; f) Lower jaw with connecting pin; and nine others (e.g., pliers without pencil, upper jaw plus pin, both jaws plus pencil, etc.). For each system, the external forces are shown. g) shows a partial free-body diagram, showing the force on the pencil and upper jaw but not all of the forces on the pencil and upper jaw. See fig. 2.63 on page 2.63 for some bad FBDs of this system.

Draw a free-body diagram!

The concept of the free-body diagram is simple. In practice, however, drawing useful free-body diagrams takes some thought, even for those practiced at the art. Some basic tips are described below a few different ways.

How to draw a free-body diagram

We suggest the following procedure for drawing a free-body diagram, as shown schematically in fig. 2.60

1.

Define the system. In your own mind, define what system, or what collection of material, you would like to apply the laws of mechanics to. This ‘body’ may be just a part of your overall system of interest. Figure 2.61 on page 2.61 shows some possible systems when considering pliers.
2.

Sketch the system. Your sketch may include various cut marks to show how the ‘body’ is isolated from its environment. Imagine cutting the system free from its environment with a sharp scalpel, or with a chain-saw.
3.

Stare at each cut. Look at the picture at all of the places that the system interacts with material not shown in the picture, places where you made ‘cuts’.
4.

Fool the body. Use forces and torques to fool the system into thinking it has not been cut. For example, if, at a given contact point where you have cut the system free, the system is being pushed in a given direction, then show a force in that direction at that point. If a system is being prevented from rotating by a (cut) rod, then show a torque at that cut.
5.

Replace gravity with a force. To show that you have cut the system free from the earth’s gravity force, show the force of gravity on the system’s center-of-mass or on the centers of mass of its parts.

What shows on a free-body diagram? And what doesn’t?

Here are some more details about the elements of a good free-body diagram. Even though some of these are stylistic issues, we think they help with problem solving. So, unless you are confident that you know better, take the comments here as truth.

•

The system. A free-body diagram is a picture of the system for which you would like to apply the laws of mechanics (These are linear or angular momentum balance, force and moment balance being special cases, or power balance). The free-body diagram shows the system isolated (‘free’) from its environment. That is, the free-body diagram does not show things that are near to or touching the system of interest. The system must be free from its environment. See fig. 2.59.
•

The word ‘body’ means system. A free-body diagram may show one or more particles, rigid objects, deformable objects, or parts thereof such as a machine, a component of a machine, or a part of a component of a machine. You can draw a free-body diagram of any collection of material that you can identify. The word body connotes a standard object in some people’s minds. In the context of free-body diagrams, ‘body’ means system. The body in a free-body diagram may be a subsystem of the overall system of interest. For a system with $n$ parts there are $2^{n}-1$ sub-collections of parts. For the pliers of fig. 2.61 there are 4 parts and 15 possible FBDs (6 of which are shown).
•

Forces fool the system. The free-body diagram of a system shows the forces and moments that the surroundings impose on the system. That is, since the only method of mechanical interaction that Nature has invented is force (and moment), the free-body diagram shows what it would take to mechanically fool the system if it were, literally, cut free. That is,

The motion of the system would be totally unchanged if it were cut free and the forces shown on the free-body diagram were applied as a replacement for all of the actual external interactions^†^†margin: Technical caveat. What would be the same are the aspects of the motion determined by using the laws of mechanics on that free body diagram. To get all of the system’s distortions right, all of the possible free body diagrams, of all possible subsystems, need also to be drawn correctly. .
•

Each force has a source and a target. Every force shown on a FBD acts on the system (the body) and from another object according to some rule. For each force you should be able to name the target (the ‘free-body’), the source (e.g., a contacting body) and the rule (e.g., laws of gravity, a spring equation, the force sufficient to prevent interpenetration). Subscripts can help, such as $F_{ED}$ indicating the force is from E and on D (See fig. 2.61).
•

Place forces at cuts. The forces and moments are located on the free-body diagram at the points where they are applied. These are the places where you made ‘cuts’ to free the body.
•

Motion is caused by, or prevented by, forces. At places where the outside environment causes, or restricts, translation of the isolated system, a contact force is drawn on the free-body diagram.
•

Rotation is caused by, or prevented by, torques. At connections to the outside world that cause or restrict rotation of the system a contact torque (or couple or moment) is drawn. Draw this moment outside the system for viewing clarity. Look at fig. 2.62 and see how the moment on the block, due to the friction of the hinge, is best shown outside the block.
•

Draw contact forces outside the body. Draw the contact force outside the sketch of the system for viewing clarity. A block supported by a hinge with friction in fig. 2.62 illustrates how the reaction force on the block due to the hinge is best shown outside the block.

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Figure 2.62: A uniform block of mass $m$ supported by a hinge with friction in the presence of gravity. The free-body diagram on the right is correct, just less clear than the one on the left.
•

Draw body forces (e.g., gravity forces) inside the body. The free-body diagram shows the system, cut free from the source of any body forces that are applied to the system. Body forces are forces that act on the inside of a body from objects outside the body. Draw the body forces on the interior of the body at their effective (average) point of application; this is the center of mass for near-earth gravity forces. Figure 2.62 shows the cleanest way to represent the gravity force on the uniform block acting at the center-of-mass.

^†^†margin: Gravity and Center of Gravity. In this book, the only body force we consider is gravity. For near-earth gravity, gravity forces show on the free-body diagram as a single force at the center of gravity, or as a collection of forces each acting at the center of gravity of a system part. (For parts of electric motors and generators, not covered here in detail, electrostatic or electro-dynamic body forces also need to be considered.)
•

Internal forces are not drawn. The free-body diagram shows all external forces acting on the system but no internal forces — forces between objects within the body are not shown. See fig. 2.63 on page 2.63 for examples of what, despite temptation, not to do.
•

No velocity and no acceleration. The free-body diagram shows nothing about the motion

^†^†margin: Warning. A common error made by beginning dynamics students is to put velocity and/or acceleration arrows on the free-body diagram.

.

A free body diagram shows: no “centrifugal force”, no “acceleration force”, and no “inertial force” (Of course, for statics this is a non-issue because inertial terms are neglected for all purposes.), Repeating,

Velocities, accelerations and inertial forces do not show on a free-body diagram

^†^†margin: A white lie. The prescription that you not show inertial forces is a white lie. Actually, in the so-called d’Alembert approach to dynamics, a legitimate and intuitive approach for experts, one does show inertial forces on the free-body diagram. The d’Alembert approach is not followed in this book in any theory. Why not? Because of the frequent sign errors and mind-confusions it causes in beginners (translation for readers: “not allowed in homework or exams”). For those who are attracted to forbidden fruit, see the box titled “D’Alembert’s mechanics: beginners beware” in the Dynamics book.

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Figure 2.63: The pencil is totally crushed by the pliers. So it seems natural to show the crushing force on a sketch of the pliers crushing the pencil (a). And one might want to show the big force on the connecting pin between the pliers jaws (b). But these sketches are not Free-Body Diagrams. Free-body diagrams only show external forces on the system. For the systems shown in the sketches, (a) and (b) above, the forces on the pencil and on the pin and between the parts are internal forces and should not show on a free-body diagram. Similarly a partial free-body diagram (c) of part of the upper jaw and the pencil should not show the action and reaction pair (they are internal to the subsystem shown). See fig. 2.61 on page 2.61 for some good FBDs related to the pliers.

How to draw forces on free-body diagrams

How you draw a force on a free-body diagram depends on

•

How much you know about the force before your analysis. Do you know its direction? its magnitude? and
•

Your choice of notation (which may vary from vector to vector within one free-body diagram). See page 1.1 for a description of the ‘symbolic’ and ‘graphical’ vector notations.

Some of the possibilities are shown in fig. 2.64 when

(a)

Any ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ possible,
(b)

the direction of ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ is known, and
(c)

Everything about ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ is known.

In each case three different notations are shown.

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Figure 2.64: The various ways of notating a force on a free-body diagram. In column (a) nothing is known and everything is variable. In column (b) the direction is known and the magnitude isn’t. In column (c) everything is known. In one free-body diagram different notations can be used for different forces, as needed or convenient. Other unusual cases can be extrapolated, e.g., if the magnitude is known and the direction is unknown.

Using equivalent force systems to simplify

The concept of ‘fooling’ a system with forces is somewhat subtle. If the free-body diagram involves ‘cutting’ a rope what force should one show? A rope is made of many fibers so cutting the rope means cutting all of the rope fibers. Should one show hundreds of force vectors, one for each fiber that is cut? The answer is: yes and no. You would be correct to draw all of these hundreds of forces at the fiber cuts. But, since the equations that are used with any free-body diagram involve only the total force and total moment, you are also allowed to replace these forces with an equivalent force system (see section 2.1).

Any force system acting on a given free-body diagram can be replaced by an equivalent force and couple.

In the case of a rope, a single force directed nearly parallel to the rope and acting at about the center of the rope’s cross section is equivalent to the force system consisting of all the fiber forces. In the case of an ideal rope, the force is exactly parallel to the rope and acts exactly at its center.

Similarly the force of the net effect of the distributed ground forces on a shoe is often represented by a single force at “the center of pressure”.

Action and reaction

For some systems you will want to draw free-body diagrams of subsystems. For example, to study a machine, you may need to draw free-body diagrams of several of its parts; for a building, you may draw free-body diagrams of various structural components; and, for a biomechanics analysis, you may ‘cut up’ a human body (with your imagined scalpel). When separating a system into parts, you must take account of how the subsystems interact. Call the two touching parts of a machine $\cal A$ and $\cal B$ . Then,

If	$\cal A$	feels force	${\accentset{\rightharpoonup}{\boldsymbol{F}}}$	and couple	${\accentset{\rightharpoonup}{\boldsymbol{M}}}$	from $\cal B$ ,
then	$\cal B$	feels force	$-{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	and couple	$-{\accentset{\rightharpoonup}{\boldsymbol{M}}}$	from $\cal A$ .

To be precise, we must make clear that ${{\accentset{\rightharpoonup}{\boldsymbol{F}}}}$ and ${-{\accentset{\rightharpoonup}{\boldsymbol{F}}}}$ have the same line of action.

The principle of action and reaction doesn’t say anything about what force or moment acts on one object. It only says that the actor of a force and moment gets back the opposite force and moment.

It is easy to make mistakes when drawing free-body diagrams involving action and reaction. Box 2.3 on page 2.3 shows some correct and incorrect partial FBD’s of interacting bodies $\mathcal{A}$ and $\mathcal{B}$ . Use notation consistent with fig. 2.64 on page 2.64 for the action and reaction vectors.

Is the principle of action and reaction an independent law? That’s debatable^†^†margin: The principle of action and reaction can be derived from the momentum balance laws by drawing free-body diagrams of little slivers of material. Nonetheless, in practice you can think of the principle of action and reaction as a basic law of mechanics. Newton did. The principle of action and reaction is “Newton’s third law”. . But, no-one doubts the truth of it.

Which force is the action and which the reaction? Often one has a system in mind and the forces on the system are the actions and the forces on the surroundings by the system are the reactions

^†^†margin: Another, less-useful, interpretation of “the principle of Action and Reaction” is that some forces on a system are constraints, to be found, and some forces are more active (hence the word ‘action’). The ‘reactions’ are the constraint forces that hold the system in place. In this (not supported here) interpretation, the principle of Action and Reaction is really just a statement about equilibrium of the system.

For a complicated system, where one draws lots of free body diagrams of subsystems, just remember that forces come in pairs: A force on the system by the surroundings and a force on the surroundings by the system. It doesn’t matter which one you call “the reaction”.

Interactions

The way objects interact mechanically is by the transmission of a force or a set of forces. If you want to show the effect of body $\mathcal{B}$ on $\mathcal{A}$ , in the most general case you can expect a force and a moment. These two, together, are equivalent to the whole force set acting on $\mathcal{A}$ from $\mathcal{B}$ , however complex.

That is, the most general interaction of two bodies requires knowing

•

Three numbers in two dimensions (two force components and one moment), and
•

Six numbers in three dimensions (three force components and three moment components)

Most things don’t interact in this most general way so, usually, fewer numbers are required.

Some of the common ways in which mechanical things interact, at least ideally, are described in the following sections. As you read this, refer also to first three columns of the summary table on page Back tables. You should look frequently at this table, until you have absorbed it. You will use the forces and moments of these connections again and again.

Constrained motion and free motion

One general principle of interaction forces and moments concerns ‘geometric’ constraints.

Wherever a motion of $\mathcal{A}$ is either caused or prevented by $\mathcal{B}$ there is a corresponding force shown at the interaction point on the free-body diagram of $\mathcal{A}$ .

Similarly,

If $\mathcal{B}$ causes or prevents rotation there is a moment (or torque or couple) shown on the free-body diagram of $\mathcal{A}$ at the place of interaction.

The converse is also true. Many kinds of mechanical attachment gadgets are specifically designed to allow motion.

If an attachment allows free motion in some direction, a so called degree of freedom, then the free-body diagram shows no force in that direction. If the attachment allows free rotation about an axis then the free-body diagram shows no moment (couple or torque) about that axis.

You can think of each attachment point as having a variety of jobs to do. For every possible direction of translation and rotation, the attachment has to either allow free motion or restrict the motion. In every way that motion is restricted (or caused) by the connection, a force or moment is required. In every way that motion is free, there is no force or couple. Motion of body $\mathcal{A}$ is caused and restricted by forces and couples which act on $\mathcal{A}$ . Motion is freely allowed by the absence of such forces and couples.

Here, demonstrating the ideas above, are some of the common connections.

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Figure 2.65: A rigid connection: a cantilever structure on a building. At the point C where the cantilever structure is connected to the building all motions are restricted so every possible force needs to be shown on the free-body diagram cut at C.

Cuts at ‘rigid’ connections

Sometimes the body you draw in a free-body diagram is firmly attached to another. Figure 2.65 shows a cantilever structure on a building. The free-body diagram of the cantilever has to show all possible force and load components. Since we have used vector notation for the force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ and the moment ${\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt\hbox{$% \displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{C}$}}{\lower 1% .0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\scriptscriptstyle% \mathrm{C}$}}}$ we can be ambiguous about whether we are doing a two- or three-dimensional analysis.

Gravity is pointing down, so why do we show a horizontal reaction force at C? This is a reasonable question because a quick statics analysis shows that, for a stationary building and cantilever, that ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{C}$ must be vertical. There are two reasons to show the horizontal force anyway

1.

Mechanics includes both statics and dynamics. In dynamics the forces on a body do not add to zero. In fact, we forgot to tell you, the building shown in fig. 2.65 happens to be accelerating rapidly to the right due to the motions of a violent earthquake occurring at the instant pictured in the figure.
2.

Whether or not there is an earthquake, the attachment of the cantilever to the building at C in fig. 2.65 is surely intended to be rigid and prevent the cantilever from moving up or down (falling), and from moving sideways (and drifting into another building) and from rotating about point C. In most of the building’s life, the horizontal reaction at C is small. But, because the connection at C clearly prevents relative horizontal motion, it is probably best to draw a horizontal reaction force on the free-body diagram. Then, the same free-body diagram is good during earthquakes and during more boring times.

When you know a force is going to turn out to be zero, as for the sideways force in this example if treated as a statics problem, it is a matter of taste whether or not you show the sideways force on the free-body diagram (Box 2.3 on page 2.3 discusses just this issue).

Our general advice is

‘Better safe than sorry’. If you don’t know that a force or moment is going to turn out to be zero, leave it in the free-body diagram.

The situation with rigid connections, like the cantilever above, is shown more abstractly in both 3D and 2D in fig. 2.66.

Filename:sfig4-6-3

Figure 2.66: A rigid connection shown with partial free-body diagrams in two and three dimensions. One has a choice between showing the separate force components (top) or using the vector notation for forces and moments (bottom). On the moment vector, the double head is optional.

Cuts at hinges

A hinge, shown in fig. 2.68, allows rotation and prevents translation. Thus, the free-body diagram of an object cut at a hinge shows no torque about the hinge axis but does show the force or its components which prevent translation.

Filename:sfig4-6-3a

Figure 2.67: A door held by hinges. One must decide whether to model hinges as proper hinges or as ball-and-socket joints. The partial free-body diagram of the door at the lower right neglects the couples at the hinges, effectively idealizing the hinges as ball-and-socket joints. This idealization is generally quite accurate because the rotations that each hinge might resist are already resisted by there being two connection points.

Filename:pfigure4-4-rp12

Figure 2.68: A hinge with partial free-body diagrams in 2D and 3-D. A hinge joint is also called a pin joint because it is sometimes built by drilling a hole and inserting a pin.

There is some ambiguity about how to model pin joints (hinges) in three dimensions. The ambiguity is shown with reference to a hinged door (fig. 2.67) and discussed in detail below.

Clearly, one hinge, if the sole attachment, prevents rotation of the door about the $x$ and $y$ axes shown. So, it is natural to show a couple (torque or moment) in the $x$ direction, $M_{x}$ , and in the $y$ direction, $M_{y}$ . But, the hinge does not provide very stiff resistance to rotations in these directions compared to the resistance of the other hinge. That is, even if both hinges are modeled as ball-and-socket joints (see the next sub-section), offering no resistance to rotation, the door still cannot rotate about the $x$ and $y$ axes.

The stiffer constraint wins. If a connection between objects prevents relative translation or rotation that is already prevented by another stiffer connection, then the more compliant connection reaction is often neglected. Even without rotational constraints, the translational constraints at the hinges A and B restrict rotation of the door shown in fig. 2.67. Thus each of the two hinges are probably well-modeled — that is, they will lead to reasonably accurate calculations of forces and motions — by ball-and-socket joints at A and B.

In 2-D a ball-and-socket joint is equivalent to a hinge or pin joint (with the axis of the hinge orthogonal to the page).

Bearing alignment. If two connections both do the same job, for example the two door hinges above, they might not do it exactly the same way. And the incompatibility can be a structural problem. Thus, for example, door hinges need to be well aligned in order that the door opening is free and to prevent large forces and moments of the hinges fighting each other.

Ball-and-socket joint

Sometimes one wishes to attach two objects in a way that allows no relative translation but for which all rotation is free. The device that is used for this purpose is called a ‘ball-and-socket’ joint. It is constructed by rigidly attaching a sphere (the ball) to one of the objects and rigidly attaching a partial spherical cavity (the socket) to the other object.

Filename:pfigure-blue-38-2

Figure 2.69: A ball-and-socket joint allows all relative rotations and no relative translations so reaction forces, but not moments, are shown on the partial free-body diagrams. In two dimensions a ball-and-socket joint is just like a pin joint. The top partial free-body diagrams show the reaction in component form. The bottom illustrations show the reaction in vector form.

The human hip joint is a ball-and-socket joint (See fig. 2.70). At the upper end of the femur bone is the femoral head, a sphere to within a few thousandths of an inch. The hip bone has a spherical cup that accurately fits the femoral head.

Refer to caption — Figure 2.70: The ball part of an artificial hip joint. (Photo courtesy of Daniel Rutter, www.dansdata.com).

The human hip joint is not so different from engineered ball and socket joints

^†^†margin: True story. The Mann biomechanics lab at MIT put strain gauges in artificial hip joints, then surgically implanted these artificial joints in patients with bad bones to measure the hip forces (they measured contact pressure up to 18 Mpa

\approx

2600

\,{\rm lbf}/\in^{2}

). Dicky at the MIT boat house said he wanted a ball-and-socket joint for the base of the mast of the sailboat he was building. “Oh” said Crispin of the Mann lab, “we have a hip joint we don’t need”, and gave Dicky an uninstrumented metal hip to which Dicky welded this and that for use in one of his boat projects.

Car suspensions are constructed from a three-dimensional truss-like mechanism. Some of the parts need free relative rotation in three dimensions and thus use a joint called a ‘ball joint’ or ‘rod end’. A ‘rod-end is a ball-and-socket joint (Figure 2.71).

Since the ball-and-socket joint allows all rotations, no moment is shown at a cut ball-and-socket joint. Since a ball-and-socket joint prevents relative translation in all directions, the possibility of force in any direction is shown.

String, rope, wires, and light chain

One way to keep a radio tower from falling over is with wire, as shown in fig. 2.72.

If the weight of the wires seems small, and the wind resistance is negligible, it is common to assume they can only transmit forces along the line connecting their end points. Moments are not shown because ropes, strings, and wires are generally assumed to be so compliant in bending that the bending moments are negligible. For wires

tension is the force pulling away from a free-body diagram cut^†^†margin: Caution: Sometimes string-like things should not be treated as idealized strings. For example, Short wires can be stiff so bending moments may not be negligible. And the mass of chains can be significant so that the mass and weight may not be negligible. For a sagging chain, the direction of the tension force is not in the direction connecting the two chain endpoints. .

Filename:f92h7p1

Figure 2.72: A radio tower kept from falling with three wires. A partial free-body diagram of the tower is drawn two different ways. The upper figure shows three tensions that are parallel to the three wires. The lower partial free-body diagram is more explicit, showing the forces to be in the directions of the

\hat{\boldsymbol{\lambda}}

s, unit vectors parallel to the wires.

Box 2.7 How much mechanics reasoning should you use when you draw a free-body diagram?

Consider the simple symmetric truss with a load $W$ in the middle, a pin support at the left and a roller support at the right.

Filename:tfigure-FBDchoices-0

Here are various options for drawing a free-body-diagram of this truss.

Filename:tfigure-FBDchoices-a

(a) The simple prescription is to draw an unknown force every place a motion is (caused or) prevented and an unknown torque where rotation is (caused or) prevented, as shown above. In particular, there is an unknown force restricting both horizontal and vertical motion at B.

Someone thinking ahead and noting that $F_{\textrm{B}x}=0$ might say that the free-body diagram in (a) is wrong. It is not. In FBD (a) force $F_{\textrm{B}x}$ is not specified because it is not known from just looking at the FBD cut at the pin. That $F_{\textrm{B}x}=0$ turns out to be zero is consistent with FBD (a) because $F_{\textrm{B}x}=0$ is not specified and thus could have any value, including zero.

As a rule, we favor FBD (a).

Filename:tfigure-FBDchoices-b

(b) A person who knows some statics will quickly deduce that the horizontal force at B is zero and thus draw the free-body diagram in figure (b). This is also correct, although somewhat violating the philosophy of drawing FBDs and later using mechanics reasoning.

Filename:tfigure-FBDchoices-c

(c) Thinking ahead even more one could draw the free body diagram above. All three free-body diagrams above are correct. In particular diagram (a) is correct even though $F_{Bx}$ turns out to be zero and (b) is correct even though $F_{B}$ turns out to be equal to $F_{C}$ . FBD (c) has the most information in it, but also most violates the problem solving approach: FBD first, mechanics later.

Filename:tfigure-FBDchoices-d

(d) In contrast, the free-body diagram above explicitly and incorrectly assigns a non-zero value to $F_{Bx}$ , so it is wrong.

What to do? When in doubt we recommend following the naive rules yielding FBD (a). Then later use the force and momentum equations to find out more about the forces, e.g., FBD (c). You might then never explicitly draw FBD (c), as it will be implicit in your assignments of values to the forces (e.g., $F_{\textrm{B}x}=0$ ). If you are confident about the anticipated results, it might be a time saver to use diagrams analogous to (b) or (c) but

Beware of:

•

making assumptions that are not reasonable, rather than just being more naive and correct, and
•

wasting time trying to think ahead when the force and momentum balance equations will tell all in the end anyway.

A common error is to sloppily think through the mechanics laws and then incorrectly eliminate, or over-specify, forces on a FBD.

All this talk about force, what is force?

Force is the measure of mechanical interaction. It is a vector. It obeys the principle of action and reaction. Using forces on free-body diagrams, with

I.

constitutive laws, like $F=kx$ and $F=mg$ (see Pillar 1 on page 0.1) and
II.

mechanics laws, like $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{i}}={\accentset% {\rightharpoonup}{\boldsymbol{0}}}$ or ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ (see Pillar 3 on page 0.1)

we make accurate predictions. What is force? Its that quantity, that miraculously, has all these properties. What is force really? Beyond this constellation of relations, force is really …never mind, that’s just too deep.

Operationally, you can define force by how you can measure it. A force on a system can be measured by comparing its effect on the given system to

•

A weight suspended by a string which goes over a pulley and is attached to the system of interest instead of the force.
•

The effect of a calibrated spring on the system, or
•

The effect, and would be hard to arrange in practice, of an accelerating mass connected by pulleys and strings to the system. Of course if you have some way of moving the force around without changing its magnitude, you can apply it to a mass and measure the acceleration it causes.
•

Other contraptions that somehow show the effect of the questionable force on a suspended weight, a stretched spring or an accelerated mass.

Summary of free-body diagrams.

•

Draw one or more clear free-body diagrams!
•

Forces and moments on the free-body diagram show all mechanical interactions from outside the body.
•

Every point on the boundary of a body has a force in every direction that motion is either being caused or prevented. Similarly with torques.
•

If you do not know the direction of a force, use vector notation to show that the direction is yet to be determined.
•
Leave off the free-body diagram forces that you think are negligible such as, possibly:
- –
  
  The force of air on small slowly moving bodies;
- –
  
  Forces that prevent motion that is already prevented by a much stiffer means (as for the torques at each of a pair of hinges);
- –
  
  See the table on page Back tables to see the forces at various connections.

Box 2.8 Action and reaction on partial FBD’s

All students should master the notations here.

Objects $\cal A$ and $\cal B$ are interacting. You want to draw separate free-body diagrams (FBD’s) of each.

Filename:tfigure2-actionreaction-0

One force on the FBD of each shows the interaction force. The FBD of $\cal A$ shows the force of $\cal B$ on $\cal A$ . And, the FBD of $\cal B$ shows the force of $\cal A$ on $\cal B$ . There are various ways to show this, and many common errors for doing so. Below are several partial FBD’s of both $\cal A$ and $\cal B$ that show the principle of action and reaction. Figures (a - d) are correct and Figures (e - g) are not. See sample 1.1 on page 1.1 to resolve related subtleties about vector notation.

Correct partial FBD’s

(a) These are good partial FBD’s. The action and re-action vectors ( ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ and $-{\accentset{\rightharpoonup}{\boldsymbol{F}}}$ ) are equal in magnitude, opposite in sign, and applied on the same line of action. Because the symbolic notation takes precedence (see page 1.1), the direction and length of the drawn arrows, although drawn nicely here, are irrelevant.

(b) These partial FBD’s are also good: the opposite arrows multiplied by equal magnitude $F$ produce net vectors that are equal in magnitude and opposite in direction.

(c) The partial FBD’s may look wrong, and they are impractically misleading and not advised. But, technically, they are okay. Why? Because we take the vector notation to have precedence over the drawing inaccuracy.

(d) The partial FBD’s may look wrong, but because no vector notation is used, the forces should be interpreted as in the direction of the drawn arrows and multiplied by the shown scalars. Since the same arrow is multiplied by $F$ and $-F$ , the net vectors are actually equal (in magnitude) and opposite (in direction).

Wrong partial FBD’s

(e) These partial FBD’s are wrong because the vector notation ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ takes precedence over the drawn arrows. So, this shows the same force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ acting on both $\cal A$ and $\cal B$ , rather than the opposite force.

(f) Because the opposite arrow is multiplied by the negative scalars, the partial FBD’s here show the same force acting on both $\cal A$ and $\cal B$ . Treating a double-negative as a negative is a common beginner’s error.

(g) These partial FBD’s are obviously wrong. They again show the same force acting on $\cal A$ and $\cal B$ . These partial FBD’s would represent the principle of double action which applies to laundry detergents and not to mechanics.

SAMPLE 2.13

Filename:pfigure-blue-52-2

Figure 2.73:

A mass and a pulley. A block of mass $m$ is held up by applying a force $F$ through a massless pulley as shown in the figure. Assume the string to be massless. Draw free-body diagrams of the mass and the pulley separately and as one system.

Solution The free-body diagrams of the block and the pulley are shown in Fig. 2.74. Since the string is massless and we assume an ideal massless pulley, the tension in the string is the same on both sides of the pulley. Therefore, the force applied by the string on the block is simply $F$ . When the mass and the pulley are considered as one system, the force in the string on the left side of the pulley doesn’t show because it is internal to the system.

Filename:pfigure4-4-rp16

Figure 2.74: The free-body diagrams of the mass, the pulley, and the mass-pulley system. Note that for the purpose of drawing the free-body diagram we need not show that we know that

R=2F

. Similarly, we could have chosen to show two different rope tensions on the sides of the pulley and reasoned that they are equal as is done in the text.

SAMPLE 2.14

Filename:pfigure4-4-rp14

Figure 2.75:

Forces in strings. A block of mass $m$ is held in position by strings AB and AC as shown in Fig. 2.75. Draw a free-body diagram of the block and write the vector sum of all the forces shown on the diagram. Use a suitable coordinate system.

Solution To draw a free-body diagram of the block, we first free the block. We cut strings AB and AC very close to point A and show the forces applied by the cut strings on the block. We also isolate the block from the earth and show the force due to gravity. (See Fig. 2.76.)

Filename:pfigure4-4-rp15

Figure 2.76: Free-body diagram of the block and a diagram of the vector

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}

To write the vector sum of all the forces, we need to write the forces as vectors. To write these vectors, we first choose an $x y$ coordinate system with basis vectors $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ as shown in Fig. 2.76. Then, we express each force as a product of its magnitude and a unit vector in the direction of the force. So,

{\accentset{\rightharpoonup}{\boldsymbol{T}}}_{1}=T_{1}\hat{\boldsymbol{% \lambda}}_{\rm AB}=T_{1}\frac{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% \rm AB}}{|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AB}|},

where ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AB}$ is a vector from $A$ to $B$ and $|{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\rm AB}|$ is its magnitude. Neglecting the size of pulleys, we get, from the given geometry,

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{AB}}$	$\displaystyle=$	$\displaystyle-2\,\mathrm{m}\hat{\boldsymbol{\imath}}+2\,\mathrm{m}\hat{% \boldsymbol{\jmath}}$
	$\displaystyle\quad\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\hat{% \boldsymbol{\lambda}}_{\mathrm{AB}}$	$\displaystyle=$	$\displaystyle\frac{2\cancel{\,\mathrm{m}}(-\hat{\boldsymbol{\imath}}+\hat{% \boldsymbol{\jmath}})}{\sqrt{2^{2}+2^{2}}\cancel{\,\mathrm{m}}}=\frac{1}{\sqrt% {2}}(-\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}).$

Thus,

{\accentset{\rightharpoonup}{\boldsymbol{T}}}_{1}=T_{1}\frac{1}{\sqrt{2}}(-% \hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}).

Similarly,

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{T}}}_{2}$	$\displaystyle=$	$\displaystyle T_{2}\frac{1}{\sqrt{5}}(\hat{\boldsymbol{\imath}}+2\hat{% \boldsymbol{\jmath}})$
	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{g}}}$	$\displaystyle=$	$\displaystyle-mg\hat{\boldsymbol{\jmath}}.$

Now, we write the sum of all the forces:

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{T}}}_{1}+{\accentset{% \rightharpoonup}{\boldsymbol{T}}}_{2}+m{\accentset{\rightharpoonup}{% \boldsymbol{g}}}$
		$\displaystyle=$	$\displaystyle\left(-\frac{T_{1}}{\sqrt{2}}+\frac{T_{2}}{\sqrt{5}}\right)\hat{% \boldsymbol{\imath}}+\left(\frac{T_{1}}{\sqrt{2}}+\frac{2T_{2}}{\sqrt{5}}-mg% \right)\hat{\boldsymbol{\jmath}}.$

The $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ components of the net force depend on the values of the scalars (magnitudes) $T_{1}$ , $T_{2}$ and $m g$ .

Answer: $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=\left(-\frac{T_{1}}{\sqrt{2}% }+\frac{T_{2}}{\sqrt{5}}\right)\hat{\boldsymbol{\imath}}+\left(\frac{T_{1}}{% \sqrt{2}}+\frac{2T_{2}}{\sqrt{5}}-mg\right)\hat{\boldsymbol{\jmath}}$

SAMPLE 2.15

Filename:tfigure4-spherical-rotaxis

Figure 2.77: Two carts connected by a massless spring

Two bodies connected by a massless spring. Two carts $A$ and $B$ are connected by a massless spring. The carts are pulled to the left with a force $F$ and to the right with a force $T$ as shown in Fig. 2.77. Assume the wheels of the carts to be massless and frictionless. Draw free-body diagrams of

•

cart $A$ ,
•

cart $B$ , and
•

carts $A$ and $B$ together.

Solution The three free-body diagrams are shown in Fig. 2.78 (a) and (b). In Fig. 2.78 (a) the force $F_{s}$ is applied by the spring on the two carts. Why is this force the same on both carts? In Fig. 2.78(b) the spring is a part of the system. Therefore, the forces applied by the spring on the carts and the forces applied by the carts on the spring are internal to the system. Therefore these forces do not show on the free body diagram.

Note that the normal reaction of the ground can be shown either as separate forces on the two wheels of each cart or as a resultant reaction.

Filename:tfigure5-7

Figure 2.78: Free-body diagrams of (a) cart A and cart B separately and (b) cart A and B together

SAMPLE 2.16

Filename:tfigure5-gen-rigid-body

Figure 2.79: Two carts connected by massless pulleys.

Two carts connected by pulleys. The two masses shown in Fig. 2.79 have frictionless bases and round frictionless pulleys. The inextensible massless cord connecting them is always taut. Mass A is pulled to the left by force $F$ and mass B is pulled to the right by force $P$ as shown in the figure. Draw free-body diagrams of each mass.

Solution Let the tension in the cord be $T$ . Since the pulleys and the cord are massless, the tension is the same in each section of the cord. This equality is clearly shown in the Free-body diagrams of the two masses below.

Filename:tfigure5-term1-a

Figure 2.80: Free-body diagrams of the two masses.

Comments: We have shown unequal normal reactions on the wheels of mass B. In fact, the two reactions would be equal only if the forces applied by the cord on mass B satisfy a particular condition. Can you see what condition must be satisfied for, say, $N_{A_{1}}=N_{A_{2}}$ .

[Hint: think about the moment balance about the center-of-mass A.]

SAMPLE 2.17

Filename:tfigure5-term1-b

Figure 2.81:

Structures with pin connections. A horizontal force $T$ is applied on the structure shown in the figure. The structure has pin connections at A and B and a roller support at C. Bars AB and BC are rigid. Draw free-body diagrams of each bar and the structure including the spring.

Solution The free-body diagrams are shown in figure 2.82. Note that there are both vertical and horizontal forces at the pin connections because pins restrict translation in any direction. At the roller support at point C there is only vertical force from the support ( $T$ is an externally applied force).

Filename:sfig1-2-12

Figure 2.82: Free-body diagrams of (a) the individual bars and (b) the structure as a whole.

SAMPLE 2.18

Filename:sfig4-6-5

Figure 2.83: A four bar linkage. Force

F

is the only load. Pin B has non-negligible friction, the other pins have negligible friction.

The four-bar linkage. The four-bar linkage^†^†margin: Although there are only three bars here, it is called a four-bar linkage because the rigid ground connection AD is also counted as another ”bar” in the linkage. shown in the figure is pushed to the right with a force $F$ . Pins A, C & D have negligible friction but joint B is rusty and resists rotation (has non-negligible friction). Draw free-body diagrams of each of the bars separately and of the whole structure. Use consistent notation for the interaction forces and moments. Clearly mark the action-reaction pairs. Neglect gravity.

Solution An ideal pin resists any relative translation of the pinned parts by exerting forces on them. We could draw three separate free-body diagrams of the pin, the first part, and the second part. Usually, however, we let the pin be a part of one of the objects and just draw two free-body diagrams.

Because rusty joint B resists relative rotation we also show a moment at point B acting on rods AB and BC.

Filename:sfig4-6-5a

Figure 2.84: Style 1, components: Free-body diagrams of the structure and the individual bars. The forces shown in (a) and (b) are the same.

Figure 2.84 shows the forces in terms of their $x$ - and $y$ -components. The directions of the force components are shown by the arrows and the scalar multipliers are labeled as $A_{x}$ , $A_{y}$ , etc. Here the word scalar is sloppily called ‘magnitude’, even though it can be positive or negative. Therefore, a force, shown as an arrow in the positive x-direction with magnitude $A_{x}$ , is the same as that shown as an arrow in the negative x-direction with ‘magnitude’ $-A_{x}$ . Thus, the free-body diagrams in Fig. 2.84(a) show exactly the same forces as in Fig. 2.84(b).

In Fig. 2.85, we show the forces by an arrow in an arbitrary direction. The corresponding labels represent their magnitudes. The angles represent the unknown directions of the forces.

Filename:sfig4-6-5b

Figure 2.85: Style 2, magnitude times drawn unit vector: The forces in (a) have arcs indicating angles which would show the directions of the forces. The FBDs in (b) are more informal and not recommended. Why not? There is no visible notation showing the directions of the forces.

In Fig. 2.86, we show yet another way of drawing and labeling the free-body diagrams, where the forces are labeled as vectors.

Filename:sfig4-6-5c

Figure 2.86: Style 3, vector notation: The labels of forces as vectors indicate both their magnitude and direction. The arrows are arbitrary and merely indicate that a force or a moment acts at those locations.

Note: Bar CD is a two-force member. We could have used that to simplify all of the free-body diagrams above by showing equal and opposite forces at C and D that were parallel to the line CD.

Problems for 2.3 Interactions, Partial FBDs

Preparatory Problems

2.3.1 How does one know what forces and moments to use in

(a)

the statics force balance and moment-balance equations? Answer: The forces and moments that show on a free-body diagram, the external forces and moments.
(b)

the dynamics linear momentum balance and angular momentum balance equations? Answer: The forces and moments that show on a free-body diagram, the external forces and moments. No “inertial” or “acceleration” forces show.

2.3.2 In a free-body diagram of a whole man standing with his right hand extended how do you show the force of his right arm on his body? Answer: You don’t.

2.3.3 Reproduce the first column of the table in fig. 2.64 on page 2.64 for the force acting on your right foot from the ground as you step on a stair.

2.3.4 Reproduce the second column of the table in fig. 2.64 on page 2.64 for a force in the direction of $3\hat{\boldsymbol{\imath}}+4\hat{\boldsymbol{\jmath}}$ but with unknown magnitude.

2.3.5 Reproduce the third column of the table in fig. 2.64 on page 2.64 for a $50\,{\mbox{N}}$ force in the direction of the vector $3\hat{\boldsymbol{\imath}}+4\hat{\boldsymbol{\jmath}}$ .

More-Involved Problems

2.3.6 Simple massless pulleys. Draw free-body diagrams of

(a)

mass A with a little bit of rope,
(b)

mass B with a little bit of rope,
(c)

Pulley C with three bits of rope,
(d)

Pulley D with three bits of rope, and
(e)

The system consisting of everything below the ceiling

Filename:sfig4-6-5d

Figure 2.87:

2.3.7 For the block and pulley arrangement shown in the figure, assume negligible friction at the wheels. Draw the free-body diagrams of

(a)

the upper mass A,
(b)

the lower mass B, and
(c)

the pulley C.

Filename:sfig4-5-6

Figure 2.88:

2.3.8 Multiple pulleys. A goods container of mass $m$ is pulled to the right using a force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ and the pulley arrangement shown in the figure. Draw the free-body diagram of

(a)

the massless block B, and
(b)

the container A along with the two pulleys attached to it.

Filename:sfig4-5-6a

Figure 2.89:

2.3.9 Pulleys on inclined planes. Draw the free-body diagram of mass $m_{2}$ and write the expression for each force vector acting on the mass.

Filename:sfig4-6-8

Figure 2.90:

2.3.10 Nested pulleys. In the nested arrangement of pulleys shown in the figure, assume that all the pulleys are massless. Draw the free-body diagram of

(a)

mass A,
(b)

mass E, and
(c)

pulley D.

Write the expression for the net force on pulley D.

Filename:sfig4-6-8a

Figure 2.91:

2.3.11 A point mass $m$ at G is attached to a piston by two inextensible cables. There is gravity. Draw a free-body diagram of the mass with a little bit of the cables and write the vector expression for the net force acting on G.

Filename:sfig4-6-8b

Figure 2.92:

2.3.12 A uniform rod of mass $m$ rests in the back of a flatbed truck as shown in the figure. Draw a free-body diagram of the rod.

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Figure 2.93:

2.3.13 A spring mounted pinned rod. The uniform rigid rod shown in the figure hangs in the vertical plane with the support of the spring shown. In this position the spring is stretched by $\Delta s$ from its rest length.

(a)

Draw a free body diagram of the spring.
(b)

Draw a free-body diagram of the rod.
(c)

Write an expression for the net force on the rod.
(d)

Write the expression for the net moment on the rod about its center of mass.

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Figure 2.94:

2.3.14 Two stacked blocks sliding without friction. Two frictionless blocks sit stacked on a frictionless surface. A force $F$ is applied to the top block.

(a)

Draw free-body diagrams of each block separately.
(b)

Draw a free-body diagram of the two blocks together.

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Figure 2.95:

2.3.15 A disk in a frictionless groove. A disc of mass $m$ sits in a wedge shaped groove. There is gravity but no friction.

(a)

Draw a free-body diagram of the disk.
(b)

Write vector expressions for the reaction forces from the two walls.
(c)

Write the expression for the net force on the disk.
(d)

What is the net moment on the disk about its mass center?

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Figure 2.96:

2.3.16 FBD of an arm throwing a ball. An arm throws a ball up. A crude model of an arm is that it is made of four rigid bodies (shoulder, upper arm, forearm and a hand) that are connected with hinges. At each hinge there are muscles that apply torques between the links. Draw a FBD of

(a)

the system consisting of the whole arm (three parts, but not the shoulder) and the ball.
(b)

the ball,
(c)

the hand, and
(d)

the fore-arm,
(e)

the upper arm,

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Figure 2.97:

2.3.17 A uniform rectangular board of mass $m$ sits on a cart supported by a rod on one corner and a pin at the diagonally opposite corner as shown in the figure. Draw a free-body diagram of the board and write an appropriate vector expression for the force exerted by the rod on the board.

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Figure 2.98:

2.3.18 Cantilevered truss A cantilever truss, shown in the figure, is made up of identical horizontal and vertical bars of length $d$ . A vertical force $F$ is applied at point A. The truss is pinned to the wall at joints S and R.

(a)

Draw a free-body diagram of the entire truss.
(b)

Cut the truss to the right of bar GE by cutting rods GD, EC and ED, and draw a free-body diagram of that portion of the truss to the right of bar GE.
(c)

Draw a free-body diagram of bar IE.
(d)

Draw a free-body diagram of the joint at I with a small length of the bars protruding from I.

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Figure 2.99:

2.3.19 An X structure. Two rods are pinned together in the middle to form a structure in the shape of ‘X’ as shown in the figure. A freebody diagram of the joint J with a little bit of the bars near J is shown. Draw free-body diagrams of each bar and of the whole structure. Answer: Note, no couples show on any of the free-body diagrams asked for.

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Figure 2.100:

2.3.20 The strings connected to winches at B, C, and D hold up the mass $m=3\,\mathrm{kg}$ at A. The relevant dimensions are shown in the figure. There is gravity. Draw a free-body diagram of the mass and express the string forces as appropriate vectors.

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Figure 2.101:

2.3.21 Mass on inclined plane. A block of mass $m$ rests on a frictionless inclined plane. It is supported by two stretched springs. The mass is pulled down the plane by an amount $\delta$ and released. Draw a FBD of the mass just after it is released.

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Figure 2.102:

2.3.22 Hanging a shelf. A shelf with negligible mass supports a $0.5$ $\,\mathrm{kg}$ mass at its center. The shelf is supported at one corner with a ball and socket joint and the other three corners with strings. Draw a FBD of the shelf.

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Figure 2.103:

2.3.23 Billboard in the back of a truck. Draw a free-body diagram of the billboard sitting in the back of the truck. There are ball and socket joints at G and O. IH is a rod. Write all forces shown on your diagram as appropriate vectors in terms of unit vectors $\hat{\boldsymbol{\imath}}$ , $\hat{\boldsymbol{\jmath}}$ and $\hat{\boldsymbol{k}}$ .

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Figure 2.104:

2.4 Contact forces: Sliding, rolling, and collisions

The primary mechanical interaction between intermediate-sized objects, say much smaller than the earth and much larger than an atom, is through contact ^†^†margin: In common engineering, the most-often encountered non-contact force is gravity, especially the gravity force from the earth on terrestrial objects like cars, buildings, and rain drops. .

Contact between two objects restricts their possible motions and causes forces on the objects. Objects cause contact forces on each other when and where they touch. While some contact situations are modeled in standard ways that we have discussed, including contact at a pin-joint, ball-and-socket, hinge, weld, and tied string, here, we consider objects that press against each other in ways not necessarily well-idealized with one of those standard mechanical connections.

Generally, people categorize contact as being one of the three major types: friction, rolling, or collision

^†^†margin: In practice it is not always clear how to make the distinctions between sliding and rolling or between sliding and collision. But, at least for a first pass it is a useful conceptual distinction to think of sliding, rolling, and collisions as three different kinds of contact.

Discussion of collisional free-body diagrams is postponed until the Dynamics book.

To solve problems, we need rules that will help us find the forces during sliding and rolling contact. There are many candidate descriptions for friction and rolling. They vary in their conceptual simplicity, their ease of use in analytical or numerical calculations, and their accuracy and applicability. Here, we will present the simplest useful rules, describe some of their short-comings, and then give some guidance towards more sophisticated contact rules.

A rule for finding the forces of interaction in terms of the objects’ positions and velocities is called, as mentioned in Chapter 1, a constitutive law or constitutive relation (see particularly page 0.1). This section is about constitutive laws for contact.

Contact laws are all rough approximations

We must emphasize at the outset:

Constitutive laws for contact are rough approximations. For at least some of the quantities of interest, theory and practice typically differ by at least 10%, and up to a factor of two or more.

That is, equations for contact are of a lower class than the momentum equations. For most engineering purposes the momentum balance equations (the basic mechanics laws) are extremely accurate, with error of well less than a part per billion. Newton’s law of gravitational attraction is a similarly accurate law. And the laws of Euclidean (non-Riemannian) geometry and calculus (the kinds you studied) are also extremely accurate (See chapter 0.2 page 0.2).

The laws for springs and dashpots are less accurate. But still, accuracies of one part per thousand are possible for measuring spring stiffness, and perhaps parts per hundred for some dashpot constants.

But the laws for the contact interactions of solids are much less accurate. Not only can’t you know the coefficient of friction between any two pieces of steel with any certainty, you also can’t even trust the concept of a coefficient of friction to be very accurate. Because you will see contact-force equations in books, and be asked to learn them, it is easy to forget their inaccuracy. Once we see an equation in print, we are too-easily tempted into believing it is ‘true.’ So, a common mistake by beginning engineers is to use contact constitutive equations with confidence, as if accurate. Rather, all contact simple equations are only rough approximations, at best.

Friction

When two objects are in contact, and one is sliding with respect to the other, we call the force which resists this sliding friction.

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Figure 2.105: Two bodies in contact. The forces between them satisfy the law of action and re-action. It is often convenient to decompose the force of interaction into a part

F

tangent to the surface of interaction and a part

N

perpendicular to the surface of interaction.

Frictional contact is usually assumed to be either ‘lubricated’ or ‘dry.’ When bodies are in lubricated contact they are not in real contact at all, a thin layer of liquid or gas separates them. For example, most of the nominally metal-to-metal contact in a car engine is so lubricated. The contact of the car tires with the road is ‘dry’ unless the car is ‘hydroplaning’ on worn-smooth tires on a very wet road. The friction forces in lubricated contact are very small compared to forces of unlubricated contact. There is no quick way to estimate these small lubricated slip forces. The accurate estimation of lubricated friction forces requires use of lubrication theory, a part of fluid mechanics. For many purposes lubricated friction forces are so small that they are negligible. Because lubricated sliding forces are so small and often negligible, and because estimating them is a more advanced topic, we have no more to say about them here.

Dry friction forces are usually not small and thus cannot be sensibly neglected in mechanics problems involving sliding contact. Thus, we need, even if not accurate, friction laws. The simplest model for friction forces (simplest friction law) is called Coulomb’s law of friction or just Coulomb friction. Despite that Coulomb friction is often a high-school physics topic, even this simplest of friction laws, has subtleties.

‘Smooth’ and ‘Rough’ are common misnomers for low-friction and high-friction

As a simplification, when we think friction is not important, we sometimes neglect it by setting $\mu=\phi=0$ . In many books this neglect is named “perfectly smooth”.

Smooth surfaces separated by a little fluid (say water between your feet and the bathroom tile, or oil between pieces of a bearing) do slide easily by each other. And even without a lubricant, sometimes slipping forces can be reduced by smoothing a surface. But, making a surface progressively smoother does not diminish the friction to zero. Rather, extremely smooth surfaces sometimes have anomalously high friction (extremely clean flat surfaces can even bond to each other). In general, there is no reliable correlation between smoothness and low friction.

Similarly many books use the phrase “perfectly rough” to mean perfectly high friction ( $\mu\to\infty$ and $\phi\to 90^{{}^{\circ}}$ ) and hence that no slip is allowed. This is misleading twice over. First, as just stated, rougher surfaces do not reliably have more friction than smooth ones. Second, even when $\mu\to\infty$ slip can proceed in some situations (see, for example, box 5.3 on page 5.3).

We use the phrase frictionless or negligible friction to mean that there is no tangential force component. We use the phrase no slip to mean that no tangential motion is allowed and that there is some unknown tangential force. So

Because they do not correspond to reality, we do not use the words smooth and rough in this book to indicate low and high friction.

Coulomb friction

‘Coulomb’s’ law of friction is also sometimes attributed to Amonton and sometimes to da Vinci. It summarized by the deceptively simple equation:

F=\mu N.

(2.6)

This equation, like many other simple equations, needs some descriptive words to be useful. What is the direction of ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ ? When does this equation apply, or not?

The direction of the force $F$ on body $\cal A$ is in the opposite direction of the slip velocity of $\cal A$ relative to $\cal B$ . By the principle of action and reaction we deduce that the force on body $\cal B$ is in the opposite direction. This force is also opposite to the relative slip velocity of $\cal B$ relative to $\cal A$ . That is, $F$ resists relative motion between $\cal A$ and $\cal B$ .

The friction force $F$ is proportional to the normal force $N$ with the proportionality constant $\mu$ . The constant $\mu$ is assumed to be independent of the area of contact between bodies $\cal A$ and $\cal B$ . In the simplest renditions of Coulomb’s law, $\mu$ is assumed to be independent of slip distance, slip velocity, time of contact, etc. The friction coefficient is assumed to be a property of the two contacting materials. Thus, one can look up $\mu$ for the contact between various pairs of materials.

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Figure 2.106: Coulomb friction. The relation between friction force

F

and relative slip rate

\dot{\delta}

is described by the dark line. Since there is a jump from

-\mu N

\mu N

in the friction force when the slip rate goes from negative to positive the relation is not a proper mathematical function between

F

and

\dot{\delta}

. Instead the relation is a curve in the

F,\dot{\delta}

plane.

When contacting bodies are not sliding the meaning of the friction equation 2.6 changes. Friction still resists slip, it is the presence of the friction force that prevents slip. But when there is no slip eqn. (2.6) doesn’t describe the force, but rather an upper limit on the possible size of the force. That is, the friction force must be less than or equal to $\mu N$ in magnitude during non-sliding contact.

|F|\leq\mu N

(2.7)

All of the discussion above can be summarized with the following equations for the friction force

	$\displaystyle\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\mbox{% on $\cal A$ from $\cal B$}}}^{\text{The friction force}}$	$\displaystyle={-\mu\overbrace{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \cal A/B}}^{\text{Relative slip velocity.}}\over{\|{\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{\cal A/B}\|}}N$	during slip
	$\displaystyle\underbrace{\|{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\mbox% {on $\cal A$ from $\cal B$}}\|}_{\text{The magnitude of the friction force}}$	$\displaystyle\leq\underbrace{\mu N}_{\text{An upper bound on the friction force}}$	during stationary contact

For two-dimensional problems where slip can only be in one direction (or the opposite) this pair of functions describes the dark line in the friction graph of fig. 2.106 in which $\dot{\delta}$ is the speed of relative slip.

The simplest friction law, the one we use in this book, uses a single constant coefficient of friction $\mu$ . Almost always $.05\leq\mu\leq 1.2$ and more commonly $.2\leq\mu\leq 1$ . In this book, we do not distinguish the static coefficient $\mu_{s}$ from the dynamic coefficient $\mu_{d}$ or $\mu_{k}$ .

2 in ^†^†margin: How dynamic friction differs from static friction, and how that matters in mechanics, is interesting. If you are wildly curious, here is a random paper on the topic: Slip instability and state variable friction laws, A Ruina - JGR, 1983. That is $\mu=\mu_{s}=\mu_{k}=\mu_{d}$ for our purposes. We use only this simplest law for a few reasons.

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Figure 2.107: Object

\cal A

does not slide relative to the plane

\cal B

5 in

•

All friction laws used are quite approximate, no matter how complex. Unless the distinction between static and dynamic coefficients of friction is essential to the engineering calculation, using $\mu_{s}\neq\mu_{k}$ doesn’t add to the calculation’s usefulness.
•

The concept of a static coefficient of friction that is larger than a dynamic coefficient is, it turns out, not well defined if the contacting objects have more than one point of contact. And, contacting objects most-often do have more than one point of contact (See page LABEL:box:staticfriction.)
•

Students learning mechanics are often confused about friction. Because the more complex friction laws are of questionable accuracy and usefulness anyway, it seems time is better spent understanding the simplest friction laws.

See page LABEL:box:Coulombcritique for more discussion of the pros and cons of the Coulomb-friction approximation.

In summary, the simple model of friction we use is:

Friction resists relative slipping motion. During slip, the friction force opposes relative motion and has magnitude $F=\mu N$ . When there is no slip the magnitude of the friction force $F$ cannot be determined from the friction law, except that it cannot exceed $\mu N$ , that is $F\leq\mu N$ .

Friction angle

Sometimes people describe the friction coefficient with a friction angle $\phi$ rather than the coefficient of friction (see fig. 2.108).

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Figure 2.108: Two ways of characterizing friction: the friction coefficient

\mu

and friction angle

\phi

The friction angle is the angle between the net interaction force (normal force plus friction force) and the normal to the sliding surface when slip is occurring. The relation between the friction coefficient $\mu$ and the friction angle $\phi$ is

\tan\phi\equiv\mu.

The use of $\phi$ or $\mu$ to describe friction is equivalent. Which you use is a matter of taste and convenience. Sometimes analytic formulas in problems come out simpler looking with one or the other of $\mu$ and $\phi$ used to describe the friction.

Rolling contact

An idealization for the non-skidding contact of balls, wheels, and the like is pure rolling.

Objects $\cal A$ and $\cal B$ are in pure rolling contact when their (relatively convex) contacting points have equal velocity. They are not slipping, separating, or interpenetrating.

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Figure 2.109: Rolling contact: Points of contact on adjoining bodies have the same velocity,

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{B}

Most often, we are interested in cases where the contacting bodies have some non-zero relative angular velocity — a ball sitting still on level ground may be technically in rolling contact, but not interestingly so.

The simplest common example is the rolling of a round wheel on a flat surface in two dimensions. See fig. 2.110.

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Figure 2.110: Pure rolling of a round wheel on a flat slope in two dimensions.

In practice, there is often confusion about the direction and magnitude of the force $F$ shown in the free-body diagram in fig. 2.110. Here is a recipe:

1.)

Draw F as shown in any direction which is tangent to the surface.
2.)

Solve the statics or dynamics problem and find the scalar $F$ . (It may turn out to be a negative, which is fine.)
3.)

Check that rolling is really possible; that is, that slip would not occur. If the force is greater than the frictional strength, $|F|>\mu N$ , the assumption of rolling contact is not appropriate. In this case, you must assume that $F=\mu N$ or $F=-\mu N$ and that slip occurs; then, re-solve the problem.

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Figure 2.111: Rolling ball in 3-D. The force

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

and moment

{\accentset{\rightharpoonup}{\boldsymbol{M}}}

are applied loads from, say, wind, gravity, and any attachments.

N

is the normal reaction and

F_{1}

and

F_{2}

are the in plane components of the frictional reaction. One must check the no-slip condition,

\mu^{2}N^{2}\geq F_{1}^{2}+F_{2}^{2}

In three-dimensional rolling contact, we have a free-body diagram that again looks like a free-body diagram for non-slipping frictional contact. Consider, for example, the ball shown in fig. 2.111. For the friction force to be less than the friction coefficient times the normal force, we have the no slip condition

\sqrt{F_{1}^{2}+F_{2}^{2}}\ \ \leq\ \ \mu N\quad\mbox{or}\quad F_{1}^{2}+F_{2}% ^{2}\ \ \leq\ \ \mu^{2}N^{2}

Rolling is just a special case of frictional contact. It is the case where bodies contact at a single point (or on a line, as with cylinders) and have relative rotation yet have no relative velocity at their contacting points.

Rolling resistance

Non-ideal rolling contact includes provision for rolling resistance. This resistance is simply represented by either moving the location of the point of contact force or by a contact couple. Rolling resistance leads to subtle questions which we skip here

^†^†margin: Note that the tangential forces in fig. 2.112 and fig. 2.4 are not rolling resistance.

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Figure 2.112: Partial free-body diagrams of wheel in a braking or accelerating car that is pointed and moving to the right. The force of the ground on the tire is shown. But, for simplicity, the forces of the axle, gravity, and brakes on the wheel are not shown (that’s why it’s a partial FBD). An ideal point-contact wheel is assumed. There is no ‘rolling resistance’ here.

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Figure 2.113: An ideal wheel is round, massless, rigid, undriven, and rolls on flat rigid ground with no rolling resistance. Free body diagrams of ideal undriven wheels are shown in two and three dimensions. The force

F

shown in the three-dimensional picture is perpendicular to the path of the wheel. The lateral moment

M_{L}

keeps the wheel from falling over sideways. (b) 2D free-body diagram of a wheel with mass, possibly driven or braked. If the wheel has mass but is not driven or braked the figure is unchanged but for the moment

M

being zero.

Ideal wheels

An ideal wheel is an approximation of a real wheel.

It is a sensible approximation if the mass of the wheel is negligible, bearing friction is negligible, and rolling resistance is negligible. Free-body diagrams of undriven ideal wheels in two and three dimensions are shown in fig. 2.113. This idealization is rationalized in chapter 4 in box 5.3 on page 5.3. Note that if the wheel is not massless, the 2-D free-body diagram looks more like the one in fig. 2.113b with $F_{\mbox{friction}}\leq\mu N$ .

Extended contact

When things touch each other over an extended region, like the block on the plane of fig. 2.114a, it is not clear what forces to put where on the free-body diagram. On the one hand one imagines reality to be somewhat reflected by millions of small forces as in fig. 2.114b which may or may not be divided into normal ( $n_{i}$ ) and frictional ( $f_{i}$ ) components. But one generally is not interested in such detail, and even if interested one cannot find it easily.

A simple approach is to replace the detailed force distribution with a single equivalent force, as shown in fig. 2.114c broken into components. The location of this force is not relevant for some problems.

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Figure 2.114: The contact forces of a block on a plane can be sensibly modeled in various ways.

^†^†margin: In 3D, contact force distributions cannot always be replaced with an equivalent force at an appropriate location (see section 2.1). A couple may be required. Nonetheless, many people often make the approximation that a contact force distribution can be replaced by a force at an appropriate location. For example, this is the “center of pressure” approach used to describe the location of an imagined-equivalent ground force on a robot’s foot. This approximation neglects any frictional resistance to twisting about the normal to the contact plane.

If one wants to make clear that the contact forces serve to keep the block from rotating, one may replace the contact force distribution with a pair of contacts at the corners as in fig. 2.114d.

Collisional free-body diagrams

As noted earlier, there are special conventions for drawing free-body diagrams of objects that are in the process of colliding. These we treat in the relevant dynamics portions of the book.

SAMPLE 2.19

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Figure 2.115: Two blocks held in place on an inclined surface

Stacked blocks at rest on an inclined plane. Blocks $A$ and $B$ with masses $m$ and $M$ , respectively, rest on a frictionless inclined surface with the help of force $T$ as shown in Fig. 2.115. There is friction between the two blocks. Draw free-body diagrams of each of the two blocks separately and a free-body diagram of the two blocks as one system.

Solution The three free-body diagrams are shown in Fig. 2.116 (a) and (b). Note the action and reaction pairs between the two blocks; the normal force $N_{\rm A}$ and the friction force $F_{\rm f}$ between the two bodies A and B. If we consider the two blocks together as a system, then the forces $N_{\rm A}$ and $F_{\rm f}$ do not show on the free-body diagram of the system (See Fig. 2.116(b)), because now they are internal to the system.

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Figure 2.116: Free-body diagrams of (a) block

A

and block

B

separately and (b) blocks

A

and

B

together.

SAMPLE 2.20

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Figure 2.117: Two blocks slide down a frictional inclined plane. The blocks are connected by a light rigid rod.

Two blocks slide down a frictional inclined plane. Two blocks of identical mass but different material properties are connected by a massless rigid rod. The system slides down an inclined plane which provides different friction to the two blocks. Draw free-body diagrams of the two blocks separately and of the system (two blocks with the rod).

Solution The Free-body diagrams are shown in Fig. 2.118. Note that the friction forces on the two blocks are different because the coefficients of friction are different for the two blocks. The normal reaction of the plane, however, is the same for each block (why?).

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Figure 2.118: Free-body diagrams of (a) the two blocks and the rod as a system and (b) the two blocks separately.

SAMPLE 2.21

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Figure 2.119: A cart with pulleys

Massless pulleys. A force $F$ is applied to the pulley arrangement connected to the cart of mass $m$ shown in Fig. 2.119. All the pulleys are massless and frictionless. The wheels of the cart are also massless but there is friction between the wheels and the horizontal surface. Draw a free-body diagram of the cart, its wheels, and the two pulleys attached to the cart, all as one system.

Solution The free-body diagram of the cart system is shown in Fig. 2.120. The force in each part of the string is the same because it is the same string that passes over all the pulleys.

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Figure 2.120: Free-body diagram of the cart.

SAMPLE 2.22

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Figure 2.121: The unicyclist

A unicyclist in action. A unicyclist weighing 160 lbs exerts a force on the front pedal with a vertical component of $30\,{\rm lbf}$ at the instant shown in figure 2.121. The rear pedal barely touches the other foot. Assume the wheel and the frame are massless. Draw free-body diagrams of the cyclist and the cycle. Make other reasonable assumptions if required.

Solution Let us assume, there is friction between the seat and the cyclist and between the pedal and the cyclist’s foot. Let’s also assume a 2-D analysis. The free-body diagrams of the cyclist and the cycle are shown in Fig. 2.122. We assume no couple interaction at the seat.

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Figure 2.122: Free-body diagram of the cyclist and the cycle.

Problems for 2.4 Contact and friction

2.4.1 A block on an inclined plane. A block of mass $m$ sits on an inclined plane. The coefficient of friction between the block and the plane is $\mu$ . Draw the free-body diagram of the block and write the expression for the force(s) applied by the incline on the block in terms of incline angle $\phi$ .

Filename:sfig4-7-1

Figure 2.123

2.4.2 A block of mass $m$ sits on a surface supported at points $A$ and $B$ . A horizontal force $P$ acts at point E. There is gravity. The block is sliding to the right. The coefficient of friction between the block and the ground is $\mu$ . Draw a free-body diagram of the block.

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Figure 2.124

2.4.3 A block sliding on level ground. A block of mass $10\,\mathrm{kg}$ is pulled by an inextensible cable over the pulley.

(a)

Assuming the block remains on the floor, draw a free-body diagram of the block. (There are various correct answers depending how you model the interaction of the bottom of the block with the ground. See fig. 2.114 on page 2.114)
(b)

Draw a free-body diagram of the pulley with a little bit of the cable extending to both sides.

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Figure 2.125

2.4.4 A ladder standing still. A ladder of mass $m$ rests against a frictionless wall and a floor with more than enough friction to prevent slip. There is gravity.

(a)

Draw a free-body diagram of the ladder.
(b)

What is force on the ladder at point B? Find the direction of this force at B assuming the coefficient of friction to be $\mu$ and the ladder to be in a state of impending slip. Does the direction of the net force at B depend on the relative positions of A and B (again, assuming impending slip)?

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Figure 2.126

2.4.5 For the system shown in the figure draw free-body diagrams of each mass separately and of the system of two blocks.

(a)

Assume there is friction with coefficient $\mu$ . At the time of interest block $\cal B$ is sliding to the right and block $\cal A$ is sliding to the left relative to $\cal B$ .
(b)

Assume there is so much friction that neither block slides.

Filename:sfig4-7-2a

Figure 2.127

2.4.6 Two blocks A and B, with mass $m_{A}$ and $m_{B}$ respectively, are held on an inclined plane as shown in the figure. Draw a free-body diagram of block A and find the net force acting on the block.

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Figure 2.128

2.4.7 A spool rolling up an inclined plane. Draw a free-body diagram of the spool shown, including a bit of the rope. Assume the spool does not slip on the ramp.

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Figure 2.129

2.4.8 A bead riding on a rotating wire. A bead of mass $m$ is free to slide on a wire bent in the shape of a parabola. The coefficient of friction between the wire and the bead is $\mu$ . The wire rotates about the $y$ -axis. A snapshot during the motion captures the wire in the $x y$ -plane. At this instant, assume the position of the bead to be $(x,y)$ .

(a)

Draw a free-body diagram of the bead.
(b)

Write the vector expression for the force on the bead from the wire. [Hint: you need to find the normal vector to the wire at the position of the bead.]

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Figure 2.130

2.4.9 A collar sliding up a rotating rod. A uniform rod OA of negligible mass rotates in the plane about point O. A collar B of mass $m$ slides on the rod but faces friction with coefficient $\mu=0.1$ . At the instant shown, draw the free-body diagram of the rod and the collar separately evaluating the force of their interaction as explicitly as possible. Ignore gravity.

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Figure 2.131

2.4.10 A rack and a pinion. In the rack and pinion arrangement shown in the figure, the pinion C is welded to the disk D. The rack is pulled up with a force $F$ . As a result the pinion rotates clockwise along with disk D. Assume that the rack and the pinion mesh perfectly together and that there is no slip between them. You can include or ignore gravity.

(a)

Draw a free-body diagram of the rack.
(b)

Draw a free-body diagram of the pinion along with the disk D.

Filename:ch4-7-c

Figure 2.132

2.4.11 Two racks with one pinion A pinion of mass 5 kg and radius 15 cm meshes with two massless racks. The left rack is pushed up with force $F$ making the pinion rotate clockwise. Assuming no slipping between the meshing teeth, draw the free-body diagrams of each rack and the pinion separately. Ignore gravity.

Filename:pfigure-blue-87d-1

Figure 2.133

2.4.12 A bicycle with unequal wheels. For the bicycle shown in the figure, assume the mass of the bicycle (and possibly the rider) to be a point mass located at C. A vertical downward force $F$ is applied on the front pedal.

(a)

Draw a free-body diagram of the front wheel.
(b)

Draw a free-body diagram of the back wheel.
(c)

Draw a free-body diagram of the entire bicycle.
(d)

What assumptions have you made in modeling the interaction force of the ground with the wheels?

Filename:Mikef91p3

Figure 2.134

$\displaystyle x_{cm}$	$\displaystyle=$	$\displaystyle\frac{\sum m_{i}x_{i}}{\sum m_{i}}=\frac{m_{1}x_{1}+m_{2}x_{2}+m_% {3}x_{3}}{m_{1}+m_{2}+m_{3}}$
	$\displaystyle=$	$\displaystyle\frac{1\,\mathrm{kg}\cdot 1.25\,\mathrm{m}+2\,\mathrm{kg}\cdot 2% \,\mathrm{m}+3\,\mathrm{kg}\cdot 0.75\,\mathrm{m}}{1\,\mathrm{kg}+2\,\mathrm{% kg}+3\,\mathrm{kg}}$
	$\displaystyle=$	$\displaystyle\frac{7.5\cancel{\,\mathrm{kg}}\cdot\,\mathrm{m}}{6\cancel{\,% \mathrm{kg}}}=1.25\,\mathrm{m}.$

$\displaystyle y_{cm}$	$\displaystyle=$	$\displaystyle\frac{\sum m_{i}y_{i}}{\sum m_{i}}=\frac{m_{1}y_{1}+m_{2}y_{2}+m_% {3}y_{3}}{m_{1}+m_{2}+m_{3}}$
	$\displaystyle=$	$\displaystyle\frac{1\,\mathrm{kg}\cdot 3\,\mathrm{m}+2\,\mathrm{kg}\cdot 2\,% \mathrm{m}+3\,\mathrm{kg}\cdot 0.5\,\mathrm{m}}{1\,\mathrm{kg}+2\,\mathrm{kg}+% 3\,\mathrm{kg}}$
	$\displaystyle=$	$\displaystyle\frac{8.5\cancel{\,\mathrm{kg}}\cdot\,\mathrm{m}}{6\cancel{\,% \mathrm{kg}}}=1.42\,\mathrm{m}.$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{net}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{D}}% \mathrel{\mathop{=}\limits^{\surd}}-20\,{\mbox{N}}\hat{\boldsymbol{\jmath}}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{B}}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\!\mathrm{D/B}}% \times{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{D}}$
	$\displaystyle=$	$\displaystyle 0.5\,\mathrm{m}\hat{\boldsymbol{\imath}}\times(-20\,{\mbox{N}}% \hat{\boldsymbol{\jmath}})=-10\,{\rm N}\mathord{\cdot}{\rm m}\hat{\boldsymbol{% k}}\mathrel{\mathop{=}\limits^{\surd}}{\accentset{\rightharpoonup}{\boldsymbol% {M}}}_{\mathrm{B_{net}}}.$