Introduction to Mechanics for Engineersby Andy Ruina & Rudra Pratap13 Many particles in space 15 Circular motion of a particle

Chapter 14 Constrained straight-line motion

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14.0Introductionstudy PDF
14.11D motion and pulleysstudy PDF + solutions
14.21D motion w/ 2D & 3D forcesstudy PDF + solutions

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Here is an introduction to kinematic constraint in its simplest context, systems that are constrained to move without rotation in a straight line. In one dimension, pulley problems provide the main example. Two- and three-dimensional problems are covered, such as finding structural support forces in accelerating vehicles and the slowing or incipient capsize of a braking car or bicycle. Angular momentum balance is introduced as a needed tool but without the complexities of rotational kinematics.

Refer to caption — Figure 14.1: A truck or car running on straight level road is in straight-line motion, neglecting, of course, the wheel rotation, the bouncing, the moving engine parts, and the wandering eyes of the passengers.

In the previous chapters you learned to write the equations of motion for a particle or collection of a few particles, assuming you have a model for the forces on the particles in terms of their positions, velocities, and time.

Some caveats to using that approach for engineering systems that don’t seem to behave like isolated particles, but rather are composed of many particles were listed at the start of chapter 13 on page 13.

One way to finesse these problems is to make kinematic assumptions about how particles and collections of particles move.

Why?

Sometimes, often actually, the simplest model of mechanical interaction is not a law for force as a function of position, velocity and time, but just a geometric description of the relative positions or velocities of points. The reasons for this geometric, instead of force-based, approach are two-fold:

•

The minute details of the motion are often not of interest and therefore not worth tracking. For example, the vibrations of a solid, or relative motions of atoms in a solid might be of a smaller scale than the overall motion of interest, and
•

Often one does not know an accurate force law. For example, at the microscopic level one does not know the details of atomic interactions; or, at the machine level, one may not know exactly the relations between the small motions of one part relative to another with which it makes contact. For example, even though one knows that the axle being in a hole restricts the relative motion of the axle with the train one may not know in detail how the contact forces depend on the exact position of the axle in its hole.

Kinematic constraints

Much mechanical modeling involves the replacement of force-interaction rules with assumptions about the geometry of the motions. Idealizing an interaction force as causing a definite geometric restriction on motion is called imposing a kinematic constraint.

A kinematic constraint is an equation that describes a restriction on allowed positions, velocities or accelerations of parts in a system. Kinematic constraints are always accompanied by one or more a priori unknown ‘constraint’ forces that maintain the geometric constraint relations.

The basic laws of forces and mechanics apply to all systems, no matter how they are or are not constrained. But, if objects are treated as kinematically constrained the methods in mechanics have a slightly different flavor. To get the idea we start with simple systems that have simple constraints and that move in simple ways. In this short chapter, we will discuss the mechanics of things where every point in each object has the same velocity and acceleration as every other point (so called parallel motion) and with the further restriction that every point moves in a straight line.

Example: Train on Straight Level Tracks

Consider a train on straight level tracks. If we focus on the body of the train, we can approximate the motion as parallel straight-line motion. All parts move the same amount, with the same velocities and accelerations in the same fixed direction.

We start with 1-D mechanics and constraint with string and pulleys, and then move on to 2-D and 3-D mechanics (of systems in 1D motion).

14.1 1-D constrained motion and pulleys

In this section masses are connected together with bars or ropes. These connections are idealized as being inextensible.

Consider a car towing another with a strong light chain. We may not want to consider the elasticity of the chain but instead idealize the chain as having a fixed length. Although the idealization of zero deformation is a simplification, it is a simplification that requires special treatment. It is the simplest example of a kinematic constraint.

Filename:tfigure8-ang-vel-ex

Figure 14.2: A schematic of one car pulling another, or of a boat pulling a barge. Also shown are FBDs of the bodies separately. Because our analysis is only in one spatial dimension, forces with no component in

\hat{\boldsymbol{\imath}}

direction are not shown.

Figure 14.2 shows a schematic of one car pulling another. One-dimensional free-body diagrams are also shown. The force $F$ is the force transmitted from the road to the front car through the tires. The tension $T$ is the tension in the connecting chain. From linear momentum balance for each of the objects (modeled as particles):

T=m_{1}\ddot{x}_{1}\hbox to36.135pt{\hfil and\hfil}F-T=m_{2}\ddot{x}_{2}.

(14.1)

These equations are exactly the same as for cars connected by a spring, a dashpot, or any idealized-as-massless connector. And all these systems have the same free-body diagrams but different motions. If the connection were with a spring or dashpot the equations above would be supplemented with

T=k(x_{2}-x_{1}-\ell_{0})\quad\mbox{or}\quad T=c(\dot{x}_{2}-\dot{x}_{1})

In this case we need our equations to somehow indicate that the two particles are not allowed to move independently. We need a constraint equation to replace these constitutive laws.

Kinematic constraint: two approaches

There are two basic ways of dealing with kinematic constraints:

1.

Use separate free-body diagrams and equations of motion for each particle and then add extra kinematic constraint equations, or
2.

do something clever to avoid having to find the constraint forces.

Method 1: Finding the accelerations and the constraint forces together

The geometric (or kinematic) constraint that two masses move together is

x_{1}=x_{2}+{\rm Constant}.

We can differentiate the kinematic constraint twice to get

\ddot{x}_{1}=\ddot{x}_{2}.

(14.2)

If we take $F$ and the two masses as given, equations 14.1 and 14.2 are three equations for the unknowns $\ddot{x}_{1},\ddot{x}_{2}$ , and $T$ . In matrix form, we have:

\left[\begin{array}[]{ccc}m_{1}&0&-1\\ 0&m_{2}&1\\ -1&1&0\end{array}\right]\left[\begin{array}[]{c}\ddot{x}_{1}\\ \ddot{x}_{2}\\ T\end{array}\right]=\left[\begin{array}[]{c}0\\ F\\ 0\end{array}\right].

We can solve these equations to find $\ddot{x}_{1},\ddot{x}_{2}$ , and $T$ in terms of $F$ .

Method 2: Finesse having to find the constraint force

On the other hand, if all we are interested in are the accelerations of the cars it would be nice to avoid even having to think about the constraint force. One way to avoid dealing with the constraint force is to draw a free-body diagram of the entire system as in

Filename:tfigure8-ang-vel

Figure 14.3: A free-body diagram of the whole system. Note that the unknown tension (constraint) force does not show. As usual for 1D mechanics, vertical forces are left off for simplicity (although it would be more correct to include them).

fig. 14.3. If we just call the acceleration of the system $\ddot{x}$ we get, from linear momentum balance, one equation in one unknown:

F=(m_{1}+m_{2})\ddot{x}.

Kinematic constraints

A generalization of the 1D inextensible-cable constraint example above is the rigid-object constraint where not just two, but many particles are assumed to keep constant distance from one another, and in one, two or three dimensions. Another important constraint is an ideal hinge connection between two objects. Much of the theory of mechanics after Newton has been motivated by a desire to deal easily with these and other kinematic constraints. In fact, one way of characterizing the primary difficulty of dynamics as a subject is the difficulty of dealing with kinematic constraints.

Pulleys

Pulleys are used to redirect force to amplify or attenuate force and to amplify or attenuate motion. Like a lever, a pulley system is an example of a mechanical transmission. Objects connected by inextensible ropes around ideal pulleys are also examples of kinematic constraint.

Constant length and constant tension

Problems with pulleys are solved by using two facts about idealized strings. First, an ideal string is inextensible so the sum of the string lengths, over the different inter-pulley sections, adds to a constant (not varying in time):

\ell_{1}+\ell_{2}+\ell_{3}+\ell_{4}+...=\mbox{constant}.

(14.3)

Second, for round pulleys of negligible mass and no bearing friction, tension is constant along the length of the string

^†^†margin: See fig. 5.50 on page 5.50 and the related text which shows why

T_{1}=T_{2}

for one round pulley idealized as frictionless and massless.

.The tension on one side of a pulley is the same as the tension on the other side:

T_{1}=T_{2}=T_{3}\ldots.

(14.4)

Example: Length of string calculation

Filename:tfigure8-ang-accel

Figure 14.4: One mass, one pulley, and one string

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Figure 14.5: The four classic cases: (a) no pulley, (b) a pulley system with no mechanical advantage, (c) a pulley system that multiplies force and attenuates motion, and (d) a pulley system that attenuates force and amplifies motion.

We use the trivial pulley example in fig. 14.4.

Starting from point $A$ , we add up the lengths of string

\ell_{tot}=x_{A}+\pi r+x_{B}\equiv constant.

(14.5)

One portion of the string touches half of the pulley circumference, $\pi r$ , even if $x_{A}$ and $x_{B}$ change in time and different portions of string wrap around the pulley at different times.

We now formally deduce the relations between the velocities and accelerations of points $A$ and $B$ . Differentiating equation 14.5 with respect to time once and then again, we get

$\displaystyle\dot{\ell}_{tot}=0$	$\displaystyle=$	$\displaystyle\dot{x}_{A}+0+\dot{x}_{B}$
$\displaystyle\Rightarrow\dot{x}_{A}$	$\displaystyle=$	$\displaystyle-\dot{x}_{B}$
$\displaystyle\Rightarrow\ddot{x}_{A}$	$\displaystyle=$	$\displaystyle-\ddot{x}_{B}$	(14.6)

When point $A$ is displaced to the right by an amount $\Delta x_{A}$ , point $B$ is displaced exactly the same amount but to the left; that is, $\Delta x_{A}=-\Delta x_{B}$ . Note, to find the kinematic relations 14.6 for the pulley system, we never need to know the total length of the string, only that it is constant in time. The constant-in-time quantities (the pulley half-circumference and the string length) get ‘killed’ in the process of differentiation.

Commonly we think of pulleys as small and thus never account for the pulley-contacting string length. Luckily this approximation generally leads to no error because we most often are interested in displacements, velocities, and accelerations. And in these cases, as in the example above, the pulley contact length drops out of the equations anyway.

The classic simple uses of pulleys

First imagine trying to move a load with no pulley as in fig. 14.5a. The force you apply goes right to the mass. This is like direct drive with no transmission.

Now you would like to use pulleys to help you move the mass. In the cases we consider here the mass is on a frictionless support and we are trying to accelerate it. But the concepts are the same if there are also resisting forces on the mass. What can we do with one pulley? Three possibilities are shown in fig. 14.5b-d which might, at a blinking glance, look roughly the same. But they are quite different. Here we discuss each design qualitatively.

In fig. 14.5b we pull one direction and the mass accelerates the other way. This illustrates

the simplest use of a pulley, to redirect an applied force.

The force on the mass has magnitude $|{\accentset{\rightharpoonup}{\boldsymbol{F}}}|$ and there is no mechanical advantage.

Figure 14.5c shows

the classic use of a pulley, to multiply a force.

Here’s a detailed solution to this problem.

Example: Pulley in figure 14.5c

The methods here will solve every pulley problem

^†^†margin: How to solve pulley problems Follow these rules, and you can solve all pulley problems (see the example from fig. 14.5c in the text). 1. Draw a free-body diagram of each pulley and each mass, taking account that the tension in any given rope is constant along its length. 2. Write linear momentum balance for each pulley and mass (this might just be force balance if you neglect the mass of, say, a pulley). 3. Do length accounting for each rope, taking account that its length is constant. To avoid sign errors, use a reference point that is off to the side of the whole mechanism, even if there is no such point in the original problem (the wall at 0 in the solution to fig. 14.5c above). 4. Differentiate the string length equations (from the step above), twice. 5. Solve the equations from parts 2 and 4 (above) for desired unknowns.

First draw the FBDs and rope geometry sketch (fig. 14.6).

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Figure 14.6: FBDs and kinematics for the mechanism in fig. 14.5c

First linear momentum balance (and force balance for the negligible-mass pulley):

\textrm{LMB}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}m\ddot{x}_{A}=T_{A}\qquad% \mbox{and}\qquad T_{A}=2T_{B}.

(14.7)

Next the rope length kinematics:

\ell_{1}=x_{C}-x_{A}\qquad\mbox{and}\qquad\ell_{2}=(x_{B}-x_{C})+(x_{D}-x_{C})% +\pi r

Differentiate the rope-length equations twice, remembering that $\ell_{1},\ell_{2},r$ and $x_{D}$ are all constants, to get

\ddot{x}_{C}=\ddot{x}_{A}\qquad\mbox{and}\qquad\ddot{x}_{B}=2\ddot{x}_{C}.

(14.8)

The second equation above ( $\ddot{x}_{B}=2\ddot{x}_{C}$ ) is the only one that you couldn’t write down at a glance. Solve these equations in terms of $F=T_{B}$ to get:

T_{A}=2F\qquad\mbox{and}\qquad\ddot{x}_{A}=2F/m\qquad\mbox{and}\qquad\ddot{x}_% {B}=4F/m.

Figure 14.5d shows

a less common use of a pulley, to multiply motion.

In this case an analysis similar to that above (that you are asked to do in problem 14.25 on page 14.25) shows that:

T_{A}=F/2\qquad\mbox{and}\qquad\ddot{x}_{A}=F/(2m)\qquad\mbox{and}\qquad\ddot{% x}_{B}=F/(4m)

Despite the superficial similarity, this setup is the opposite to the use in fig. 14.5c.

Force amplification is motion attenuation

In fig. 14.5c a force $F$ is applied at B. The resulting force on the mass at A is $2F$ and point A moves with half the acceleration of point B. However, in fig. 14.5d the resulting force on the mass at A is $F/2$ , and point A moves with twice the acceleration of point B.

These illustrate a general duality rule

If force is amplified then motion is equally attenuated. If motion is amplified the force is equally attenuated ^†^†margin: Caveats: Negligible dissipation and negligible mass of the transmission parts. .

High gear and low gear. This rule applies to all frictionless passive transmissions (e.g., levers, gear trains, hydraulic systems) with negligible inertia: In a low gear in a car (or bicycle) the force at the wheel is large for a given force of the engine (leg muscle) but the wheel doesn’t turn much for a given displacement of the engine (foot). In a high gear the force at the wheel is small but the wheel turns a lot for a given amount of engine rotation (or foot displacement). These are, of course, special cases of the much more general rule: you can’t win!

Power balance

In every special case we can derive the duality rule (above) using momentum balance and kinematics. But the general result is best understood with energy balance:

the power of the applied force is the power applied to the mass.

Example: Power balance and the pulley in figure 14.5c

There are various ways of setting up the energy (or power) balance equations. Here is one:

\textrm{Power into the transmission}=\textrm{Power out of the transmission}

The power in is $T_{B}\cdot\dot{x}_{B}$ . The power out, delivered to A, is $T_{A}\cdot\dot{x}_{A}$ . So,

T_{B}\cdot\dot{x}_{B}=T_{A}\cdot\dot{x}_{A}\hbox to36.135pt{\hfil$\Rightarrow$% \hfil}\frac{T_{A}}{T_{B}}=\frac{\dot{x}_{B}}{\dot{x}_{A}}.

But because $\dot{x}_{a}/\dot{x}_{B}$ is a constant, differentiating we get $\ddot{x}_{A}/\ddot{x}_{B}=\dot{x}_{A}/\dot{x}_{B}$ . So we also have

\frac{T_{A}}{T_{B}}=\frac{\ddot{x}_{B}}{\ddot{x}_{A}}.

The ‘effective mass’ of a point of force application

The feel of the machine is of concern for machines that people handle. One aspect of feel is the effective mass (sometimes called ‘reflected inertia’) is defined by the response of a point to an applied force.

m_{\rm eff}={|{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\!\!\mathrm{B}}|% \over|{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\mathrm{B}}|}.

For the case of fig. 14.5a and fig. 14.5b the effective mass of point B is just $m$ . For the case of fig. 14.5c the block at A has $2|{\accentset{\rightharpoonup}{\boldsymbol{F}}}|$ acting on it and point B has twice the acceleration of point A. So the acceleration of point B is $4F/m=F/(m/4)$ and the effective mass of point B is $m/4$ . For the case of fig. 14.5d, the mass only has $|{\accentset{\rightharpoonup}{\boldsymbol{F}}}|/2$ acting on it and point B only has half the acceleration of point A, so the effective mass is $4m$ .

These special cases exemplify the general rule:

The effective mass of one end of a transmission is the mass of the other end multiplied by the square of the motion amplification ratio.

In terms of the effective mass, the systems shown in fig. 14.5c and fig. 14.5d which look so similar to a novice, actually differ by a factor of $2^{2}\cdot 2^{2}=16$ . With a given $F$ and $m$ point B in fig. 14.5c has 16 times the acceleration of point B in fig. 14.5d.

SAMPLE 14.1

Filename:sfig8-2-3

Figure 14.7:

Find the motion of two cars. One car is towing another of equal mass on level ground. The thrust of the wheels of the first car is $F$ . The second car rolls frictionlessly. Find the acceleration of the system two ways:

1.

using separate free-body diagrams,
2.

using a system free-body diagram.

Solution

1.

The free-body diagram of each car is shown below, in fig. 14.8.

Filename:sfig8-2-3a

Figure 14.8: Partial free-body diagrams of the two cars (the vertical ground reactions are not shown as they are of no interest to us for the horizontal motion.

From the linear momentum balance of each car, we get

$\displaystyle m\ddot{x}_{1}$ $\displaystyle=$ $\displaystyle T$ (14.9)

$\displaystyle F-T$ $\displaystyle=$ $\displaystyle m\ddot{x}_{2}$ (14.10)

The kinematic constraint of towing (the cars move together, i.e., no relative displacement between the cars) gives

$\ddot{x}_{1}-\ddot{x}_{2}=0$ (14.11)

Solving eqns. (14.9), (14.10), and (14.11) simultaneously, we get

$\ddot{x}_{1}=\ddot{x}_{2}=\frac{F}{2m}\quad(T=\frac{F}{2})$
2.

The free-body diagram of the two cars together is shown below, in fig. 14.9.

Filename:sfig8-2-3b

Figure 14.9:

From the linear momentum balance of the two cars as one system, we get

$\displaystyle m\ddot{x}+m\ddot{x}$ $\displaystyle=$ $\displaystyle F$

$\displaystyle\ddot{x}$ $\displaystyle=$ $\displaystyle F/2m$

Answer: $\ddot{x}=\ddot{x}_{1}=\ddot{x}_{2}=F/2m$

SAMPLE 14.2

Filename:sfig8-3-1

Figure 14.10:

Filename:sfig8-3-1a

Figure 14.11: Free-body diagram of the hammer and pile system.

F_{r}

is the total resistance of the ground.

Driving a pile into the ground. A cylindrical wooden pile of mass 10 kg and cross-sectional diameter 20 cm is driven into the ground with the blows of a hammer. The hammer is a block of steel with mass 50 kg which is dropped from a height of 2 m to deliver the blow. At the $n$ th blow the pile is driven into the ground by an additional 5 cm. Assuming the impact between the hammer and the pile to be totally inelastic (i.e., the two stick together), find the average resistance of the soil to penetration of the pile.

Solution Let $F_{r}$ be the average (constant over the period of driving the pile by 5 cm) resistance of the soil. From the free-body diagram of the pile and hammer system, we have

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=-mg\hat{\boldsymbol{\jmath}}% -Mg\hat{\boldsymbol{\jmath}}+N\hat{\boldsymbol{\jmath}}+F_{r}\hat{\boldsymbol{% \jmath}}.

But $N$ is the normal reaction of the ground, which from static equilibrium, must be equal to $mg+Mg$ . Thus,

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=F_{r}\hat{\boldsymbol{\jmath% }}.

Therefore, from linear momentum balance ( $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}$ ),

{\accentset{\rightharpoonup}{\boldsymbol{a}}}=\frac{F_{r}}{M+m}\hat{% \boldsymbol{\jmath}}.

Now we need to find the acceleration from given conditions. Let $v$ be the speed of the hammer just before impact and $V$ be the combined speed of the hammer and the pile immediately after impact. Then, treating the hammer and the pile as one system, we can ignore all other forces during the impact (none of the external forces: gravity, soil resistance, ground reaction, is comparable to the impulsive impact force, see page 17.5). The impact force is internal to the system. Therefore, during impact, $\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}$ which implies that linear momentum is conserved. Thus

	$\displaystyle-Mv\hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle-(m+M)V\hat{\boldsymbol{\jmath}}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\quad V$	$\displaystyle=$	$\displaystyle\left(\frac{M}{m+M}\right)v=\frac{50\,\mathrm{kg}}{60\,\mathrm{kg% }}v=\frac{5}{6}v.$

The hammer speed $v$ can be easily calculated, since it is the free fall speed from a height of 2 m:

v=\sqrt{2gh}=\sqrt{2\cdot(9.81\,\mathrm{m}/\mathrm{s}^{2})\cdot(2\,\mathrm{m})% }\quad=6.26\,\mathrm{m}/\mathrm{s}\hbox to36.135pt{\hfil$\Rightarrow$\hfil}V=% \frac{5}{6}v=5.22\,\mathrm{m}/\mathrm{s}.

The pile and the hammer travel a distance of $s=5\,\mathrm{cm}$ under the deceleration $a$ . The initial speed $V=5.22\,\mathrm{m}/\mathrm{s}$ and the final speed = 0. Plugging these quantities into the one-dimensional kinematic formula

$\displaystyle v^{2}$	$\displaystyle=$	$\displaystyle v_{0}^{2}+2as,$
we get,
$\displaystyle 0$	$\displaystyle=$	$\displaystyle V^{2}-2as\quad\mbox{ (Note that $a$ is negative)}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}a$	$\displaystyle=$	$\displaystyle\frac{V^{2}}{2s}=\frac{(5.22\,\mathrm{m}/\mathrm{s})^{2}}{2\times 0% .05\,\mathrm{m}}=272.48\,\mathrm{m}/\mathrm{s}^{2}.$

Thus ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=272.48\,\mathrm{m}/\mathrm{s}^{2% }\hat{\boldsymbol{\jmath}}$ . Therefore,

F_{r}=(m+M)a=(60\,\mathrm{kg})\mathord{\cdot}(272.48\,\mathrm{m}/\mathrm{s}^{2% })=1.635\times 10^{4}\,{\mbox{N}}

Answer: $F_{r}\approx 16.35\,{\rm kN}$

SAMPLE 14.3

Filename:sfig8-5-wiper

Figure 14.12:

Pulley kinematics. For the masses and ideal-massless pulleys shown in figure 14.12, find the acceleration of mass A in terms of the acceleration of mass B. Pulley C is fixed to the ceiling and pulley D is free to move vertically. All strings are inextensible.

Filename:sfig8-5-wiper-a

Figure 14.13:

Solution Let us measure the position of the two masses from a fixed point, say the center of pulley C. (Since C is fixed, its center is fixed too.) Let $y_{A}$ and $y_{B}$ be the vertical distances of masses A and B, respectively, from the chosen reference (C). Then the position vectors of A and B are:

{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{A}=y_{A}\hat{\boldsymbol{\jmath% }}\quad\mbox{and}\quad{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B}=y_{B}% \hat{\boldsymbol{\jmath}}.

Therefore, the velocities and accelerations of the two masses are

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}$	$\displaystyle=$	$\displaystyle\dot{y}_{A}\hat{\boldsymbol{\jmath}},\quad\quad{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{B}=\dot{y}_{B}\hat{\boldsymbol{\jmath}},$
	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}$	$\displaystyle=$	$\displaystyle\ddot{y}_{A}\hat{\boldsymbol{\jmath}},\quad\quad{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{B}=\ddot{y}_{B}\hat{\boldsymbol{\jmath}}.$

Since all quantities are in the same direction ( $\hat{\boldsymbol{\jmath}}$ ), we can drop $\hat{\boldsymbol{\jmath}}$ from our calculations and just do scalar calculations. We are asked to relate $\ddot{y}_{A}$ to $\ddot{y}_{B}$ .

In all pulley problems, the trick in doing kinematic calculations is to relate the variable positions to the fixed length of the string. Here, the length of the string $\ell_{tot}$ is:

^†^†margin: We have done an elaborate calculation of

\ell_{tot}

here. Usually, the constant lengths over the pulleys and some constant segments such as aa^′ are ignored in calculating

\ell_{tot}

. These constant length segments can be ignored because they drop out of the equation when we take time derivatives to relate velocities and accelerations of different points, such as B and D here.

$\displaystyle\ell_{tot}$	$\displaystyle=$	$\displaystyle\mbox{ab + bc + cd + de + ef}={\rm{constant}}$
where ab	$\displaystyle=$	$\displaystyle\underbrace{\mbox{aa${}^{\prime}$}}_{\rm{constant}}+\underbrace{% \mbox{a${}^{\prime}$b}}_{(=\mbox{cd}=y_{D})}$
bc	$\displaystyle=$	string over the pulley D = constant
de	$\displaystyle=$	string over the pulley C = constant
ef	$\displaystyle=$	$\displaystyle y_{B}$
$\displaystyle\mbox{thus}\quad\ell_{tot}$	$\displaystyle=$	$\displaystyle 2y_{D}+y_{B}+\overbrace{\mbox{constant}}^{\rm{(aa^{\prime}+bc+de% )}}.$

Taking the time derivative on both sides, we get

	$\displaystyle\overbrace{\frac{d}{dt}(\ell_{tot})}^{\text{$=0$, because $\ell_{% tot}$ does not change with time}}=2\dot{y}_{D}+\dot{y}_{B}$	$\Rightarrow$	$\displaystyle\dot{y}_{D}=-\frac{1}{2}\dot{y}_{B}$		(14.12)
		$\Rightarrow$	$\displaystyle\ddot{y}_{D}=-\frac{1}{2}\ddot{y}_{B}.$		(14.13)

	$\displaystyle\mbox{But}\quad\quad y_{D}=y_{A}-\mbox{AD and AD = constant}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad\dot{y}_{D}=\dot{y}% _{A}\quad\mbox{and}\quad\ddot{y}_{D}=\ddot{y}_{A}.$

Thus, substituting $\dot{y}_{A}$ and $\ddot{y}_{A}$ for $\dot{y}_{D}$ and $\ddot{y}_{D}$ in (14.12) and (14.13) we get

\dot{y}_{A}=-\frac{1}{2}\dot{y}_{B}\quad\mbox{and}\quad\ddot{y}_{A}=-\frac{1}{% 2}\ddot{y}_{B}

Answer: $\ddot{y}_{A}=-\frac{1}{2}\ddot{y}_{B}$

SAMPLE 14.4

Filename:sfig8-5-2disks

Figure 14.14: A two-mass pulley system.

A two-mass pulley system. The two masses shown in Fig. 14.14 have frictionless bases and round frictionless pulleys. The inextensible cord connecting them is always taut. Given that $F=130\,{\mbox{N}},\;m_{A}=m_{B}=m=40\,\mathrm{kg}$ , find the acceleration of the two blocks using:

1.

linear momentum balance and
2.

energy balance.

Solution

1.

Using Linear Momentum Balance:

The free-body diagrams of the two masses A and B are shown in Fig. 14.15.

Filename:sfig8-4-1

Figure 14.15: Free-body diagrams of the two masses.

Linear momentum balance for mass A gives (assuming ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=a_{A}\hat{\boldsymbol{\imath}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=a_{B}\hat{\boldsymbol{\imath}}$ ):

$\displaystyle(2T-F)\hat{\boldsymbol{\imath}}+(2N_{A}-mg)\hat{\boldsymbol{% \jmath}}$ $\displaystyle=$ $\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=ma_{A}\hat{% \boldsymbol{\imath}}$

$\displaystyle\mbox{(dotting with $\hat{\boldsymbol{\jmath}}$)}\hbox to36.135pt% {\hfil$\Rightarrow$\hfil}\qquad 2N_{A}$ $\displaystyle=$ $\displaystyle mg$

$\displaystyle\mbox{(dotting with $\hat{\boldsymbol{\imath}}$)}\hbox to36.135pt% {\hfil$\Rightarrow$\hfil}2T-F$ $\displaystyle=$ $\displaystyle ma_{A}$ (14.14)

Similarly, linear momentum balance for mass B gives:

$\displaystyle-3T\hat{\boldsymbol{\imath}}+(2N_{B}-mg)\hat{\boldsymbol{\jmath}}$ $\displaystyle=$ $\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=ma_{B}\hat{% \boldsymbol{\imath}}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\qquad 2N_{B}$ $\displaystyle=$ $\displaystyle mg$

$\displaystyle\hbox to36.135pt{and\hfil}-3T$ $\displaystyle=$ $\displaystyle ma_{B}.$ (14.15)

From (14.14) and (14.15) we have three unknowns: $T,\,a_{A},\,a_{B}$ , but only 2 equations!. We need an extra equation to solve for the three unknowns.

^†^†margin: You may be tempted to use angular momentum balance (AMB) to get an extra equation. In this case AMB could help determine the vertical reactions, but offers no help in finding the rope tension or the accelerations.

We can get the extra equation from kinematics. Since A and B are connected by a string of fixed length, their accelerations must be related. For simplicity, and since these terms drop out anyway, we neglect the radius of the pulleys and the lengths of the little connecting cords. We use the left wall as the reference position to get

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Figure 14.16: Pulley kinematics. The left wall is added as a reference point, off to the side of all pulleys and masses, to help prevent sign errors.

$\displaystyle\ell_{tot}$ $\displaystyle\equiv$ length of the string connecting A and B

$\displaystyle=$ $\displaystyle 3(x_{B}-X_{C})+2(X_{C}-x_{A})$

$\displaystyle=$ $\displaystyle 3x_{B}-X_{C}-2x_{A}$

Now, differentiating the expression for $\ell_{tot}$ with time and noting that the total length of the string remains constant and that $X_{C}$ is a fixed location in space, we get

$\displaystyle\overbrace{\frac{d}{dt}(\ell_{tot})}^{0}\;$ $\displaystyle=$ $\displaystyle 3{\dot{x}}_{B}-0-2{\dot{x}}_{A}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad{\ddot{x}}_{B}$ $\displaystyle=$ $\displaystyle\frac{2}{3}{\ddot{x}}_{A}$ (14.16)

Since

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}$ $\displaystyle=$ $\displaystyle v_{A}\hat{\boldsymbol{\imath}}={\dot{x}}_{A}\hat{\boldsymbol{% \imath}},$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}$ $\displaystyle=$ $\displaystyle a_{A}\hat{\boldsymbol{\imath}}={\ddot{x}}_{A}\hat{\boldsymbol{% \imath}},$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$ $\displaystyle=$ $\displaystyle v_{B}\hat{\boldsymbol{\imath}}=\ {\dot{x}}_{B}\hat{\boldsymbol{% \imath}},\mbox{ and}$

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$ $\displaystyle=$ $\displaystyle a_{B}\hat{\boldsymbol{\imath}}=\ {\ddot{x}}_{B}\hat{\boldsymbol{% \imath}},$

we get

$a_{B}=\frac{2}{3}a_{A}.$ (14.18)

Substituting (14.18) into (14.15), we get

$9T=-2ma_{A}.$ (14.19)

Now solving (14.14) and (14.19) for $T$ , we get

$T=\frac{2F}{13}=\frac{2\cdot 130\,{\mbox{N}}}{13}=20\,{\mbox{N}}.$

Therefore,

$\displaystyle a_{A}$ $\displaystyle=$ $\displaystyle-\frac{9T}{2m}=-\frac{9\cdot 20\,{\mbox{N}}}{2\cdot 40\,\mathrm{% kg}}=-2.25\,\mathrm{m}/\mathrm{s}^{2}$

$\displaystyle a_{B}$ $\displaystyle=$ $\displaystyle\frac{2}{3}a_{A}=-1.5\,\mathrm{m}/\mathrm{s}^{2}$

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=-2.25\,\mathrm{m}/\mathrm{s}% ^{2}\hat{\boldsymbol{\imath}},\;{\accentset{\rightharpoonup}{\boldsymbol{a}}}_% {B}=-1.5\,\mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{\imath}}$ .
2.

Using Power Balance (III): We have,

$P=\dot{{E_{\mathrm{K}}}}.$

The power balance equation becomes

$\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot{\accentset{% \rightharpoonup}{\boldsymbol{v}}}=m\,a_{A}\,v_{A}\,+m_{B}\,a_{B}\,v_{B}.$

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Figure 14.17: 1D free-body diagram of the whole system. Note that except $F$ , no other forces do any work.

Because the force at A is the only force that does work on the system, ^†^†margin: It may not be obvious why $T$ shown in the FBD in fig. 14.17 does no work. $T$ acts on a material point q’ somewhere on the inextensible string from the fixed point ‘c’. Even though mass B moves, point ‘q’ on the string has no displacement. Because the material point $q$ on which it acts does not move, the work done by $T$ is zero. when we apply power balance to the whole system (see the FBD in fig. 14.17), we get,

$\displaystyle-Fv_{A}-T\overbrace{v_{q}}^{=v_{c}=0}$ $\displaystyle=$ $\displaystyle m_{A}v_{A}a_{A}+m_{B}v_{B}a_{B}$

$\displaystyle\hbox to36.135pt{or\hfil}F$ $\displaystyle=$ $\displaystyle-ma_{A}-m\frac{v_{B}}{v_{A}}a_{B}$

$\displaystyle=$ $\displaystyle-a_{A}(m+m\frac{v_{B}}{v_{A}}\frac{a_{B}}{a_{A}}).$

Substituting $a_{B}=2/3a_{A}$ and $v_{B}=2/3v_{A}$ from eqn. (14.18),

$\displaystyle a_{A}$ $\displaystyle=$ $\displaystyle\frac{-F}{m+\frac{4}{9}m}$

$\displaystyle=$ $\displaystyle\frac{-130\,{\mbox{N}}}{40\,\mathrm{kg}(1+\frac{4}{9})}$

$\displaystyle=$ $\displaystyle-2.25\,\mathrm{m}/\mathrm{s}^{2},$

and since $a_{B}=2/3a_{A}$ ,

$a_{B}=-1.5\,\mathrm{m}/\mathrm{s}^{2},$

which are the same accelerations as found before.

Answer: $a_{A}=-2.25\,\mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{\imath}},\;a_{B}=-1.5\,% \mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{\imath}}$

SAMPLE 14.5

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Figure 14.18:

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Figure 14.19: Free-body diagram of the two masses as one system in static equilibrium (this special case could be skipped as it follows from the free-body diagram below).

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Figure 14.20: Free-body diagrams of the individual masses.

In static equilibrium the spring in fig. 14.18 is compressed by $y_{s}$ from its unstretched length $\ell_{0}$ . Now, the spring is compressed by an additional amount $y_{0}$ and released with no initial velocity.

1.

Find the force on the top mass $m$ exerted by the lower mass $M$ .
2.

When does this force become minimum? Can this force become zero?
3.

Can the force on $m$ due to $M$ ever be negative?

Solution

1.

The free-body diagram of the two masses is shown in Figure 14.19 when the system is in static equilibrium. From linear momentum balance we have

$\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}={\accentset{\rightharpoonup}% {\boldsymbol{0}}}\quad\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad ky_{s}=(m% +M)g.$ (14.20)

The free-body diagrams of the two masses at an arbitrary position $y$ during motion are shown in Figure 14.20. Since the two masses oscillate together, they have the same acceleration. From linear momentum balance for mass $m$ we get (note that we have chosen $y$ to be positive downwards),

$mg-N=m\ddot{y}.$ (14.21)

We are interested in finding the normal force $N$ . Clearly, we need to find $\ddot{y}$ to calculate $N$ . Now, from linear momentum balance for mass $M$ we get

$Mg+N-k(y+y_{s})=M\ddot{y}.$ (14.22)

Adding eqn. (14.21) with eqn. (14.22) we get

$(m+M)g-ky-ky_{s}=(m+M)\ddot{y}.$

But $ky_{s}=(m+M)g$ from eqn. (14.20). Therefore, the equation of motion of the system is

$\displaystyle-ky=(m+M)\ddot{y}$

$\displaystyle\mbox{or}\quad\ddot{y}+\frac{k}{(m+M)}y=0.$ (14.23)

As you recall from your study of the harmonic oscillator, the general solution of this differential equation is

$\displaystyle y(t)$ $\displaystyle=$ $\displaystyle A\sin\lambda t+B\cos\lambda t$ (14.24)

$\displaystyle\mbox{where }\quad\lambda$ $\displaystyle=$ $\displaystyle\sqrt{\frac{k}{m+M}}.$ (14.25)

The constants $A$ and $B$ are to be determined from the initial conditions. From eqn. (14.24) we obtain

$\dot{y}(t)=A\lambda\cos\lambda t-B\lambda\sin\lambda t.$ (14.26)

Substituting the given initial conditions $y(0)=y_{0}$ and $\dot{y}(0)=0$ in eqns. (14.24) and (14.26), respectively, we get

$\displaystyle\overbrace{y(0)}^{y_{0}}$ $\displaystyle=A\sin(\lambda\cdot 0)+B\cos(\lambda\cdot 0)\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}$ $\displaystyle B=y_{0}$

$\displaystyle\underbrace{\dot{y}(0)}_{0}$ $\displaystyle=A\lambda\cos(\lambda\cdot 0)-B\lambda\sin(\lambda\cdot 0)\hbox to% 36.135pt{\hfil$\Rightarrow$\hfil}$ $\displaystyle A=0.$

Thus,

$y(t)=y_{0}\cos\lambda t.$ (14.27)

Now we can find the acceleration by differentiating eqn. (14.27) twice :

$\ddot{y}=-y_{0}\lambda^{2}\cos\lambda t.$

Substituting this expression in eqn. (14.21) we get the force applied by mass $M$ on the smaller mass $m$ :

$\displaystyle mg-N$ $\displaystyle=$ $\displaystyle m\overbrace{(-y_{0}\lambda^{2}\cos\lambda t)}^{\ddot{y}}$

$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad N$ $\displaystyle=$ $\displaystyle mg+my_{0}\lambda^{2}\cos\lambda t$ (14.28)

$\displaystyle=$ $\displaystyle mg\left(1+\frac{y_{0}\lambda^{2}}{g}\cos\lambda t\right)$

Answer: $N=mg\left(1+\frac{y_{0}\lambda^{2}}{g}\cos\lambda t\right)$
2.

Since $\cos\lambda t$ varies between $\pm 1$ , the value of the force $N$ varies between $mg\pm y_{0}\lambda^{2}$ . Clearly, $N$ attains its minimum value when $\cos\lambda t=-1,$ i.e., when $\lambda t=\pi$ . This condition is met when the spring is fully stretched and the mass is at its highest vertical position. At this point,

$N\equiv N_{\rm min}=mg\left(1-\frac{y_{0}\lambda^{2}}{g}\right).$

If $y_{0}$ , the initial displacement from the static equilibrium position, is chosen such that $y_{0}\lambda^{2}=g$ (that is, the amplitude of the harmonically varying acceleration equals $g$ ), then $N=0$ when $\cos\lambda t=-1$ , i.e., at the topmost point in the vertical motion. This condition, $N=0$ , means that the two masses momentarily lose contact with each other; and it happens precisely when they are about to begin their downward motion.
3.

From eqn. (14.28) we can get a negative value of $N$ when $\cos\lambda t=-1$ and $y_{0}\lambda^{2}>g$ . However, a negative value for $N$ is nonsense unless the blocks are glued. Without glue the bigger mass $M$ cannot apply a negative force (or a compression) on $m$ , i.e., it cannot “suck” $m$ . When $y_{0}\lambda^{2}>g$ then $N$ becomes zero before $\cos\lambda t$ decreases to $-1$ . That is, assuming no bonding, the two masses lose contact on their way to the highest vertical position but before reaching the highest point. Beyond that point, the equations of motion derived above are no longer valid for unglued blocks because the equations assume contact between $m$ and $M$ . Equation (14.28) is inapplicable when $N\leq 0$ .

Problems for 14.1 1D constrained motion and pulleys

For all problems, unless stated otherwise, treat all strings as inextensible, flexible and massless. Treat all pulleys and wheels as round, frictionless and massless. Assume all massive objects are prevented from rotating (e.g., wheels stay on the ground, etc.). When numbers are called for use $g=10\,\mathrm{m}/\mathrm{s}^{2}$ or $g=32\,{\mathrm{ft}}/{\mathrm{s}}^{2}$ .

Preparatory Problems

14.1.1 A motor at $B$ allows the block of mass $m=3\,\mathrm{kg}$ shown in the figure to accelerate downwards at $2\,\mathrm{m}/\mathrm{s}^{2}$ . There is gravity. What is the tension in the string AB?

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Figure 14.21:

14.1.2 Two masses connected by an inextensible string hang from an ideal pulley.

(a)

Find the downward acceleration of mass $B$ . Answer in terms of any or all of $m_{A}$ , $m_{B}$ , $g$ , and the present velocities of the blocks. As a check, your answer should give $a_{B}=g$ when $m_{A}=0$ and $a_{B}=0$ when $m_{A}=m_{B}$ . Answer: $a_{B}=\left(\frac{m_{B}-m_{A}}{m_{A}+m_{B}}\right)g$ .
(b)

Find the tension in the string. As a check, your answer should give $T=m_{B}g=m_{A}g$ when $m_{A}=m_{B}$ and $T=0$ when $m_{A}=0$ . Answer: $T=2\frac{m_{A}m_{B}}{m_{A}+m_{B}}g$ .

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Figure 14.22:

14.1.3 The blocks shown are released from rest.

(a)

What is the acceleration of block A at $t=0^{+}$ (just after release)?
(b)

What is the speed of block B after it has fallen 2 meters?

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Figure 14.23:

14.1.4 What is the acceleration of block A? Use $g=10\,\mathrm{m}/{\rm s}^{2}$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm A}=-2\,\mathrm{m}/\mathrm{s% }^{2}\hat{\boldsymbol{\jmath}}$

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Figure 14.24:

14.1.5 For the system shown in problem 14.21, find the acceleration of mass $B$ using energy balance $(P={\dot{E}_{\mathrm{K}}})$ .

14.1.6 For the various situations pictured, find the acceleration of mass A and point B. Clearly define any variables, coordinates or sign conventions that you use.

Answer: (a) ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{B}=\frac{F}{m}\hat{\boldsymbol{\imath}}$ , where $\hat{\boldsymbol{\imath}}$ is parallel to the ground and pointing to the right., (b) ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=\frac{2F}{m}\hat{\boldsymbol% {\imath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=\frac{4F}{m}\hat{\boldsymbol% {\imath}}$ , (c) ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=\frac{F}{2m}\hat{\boldsymbol% {\imath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=\frac{F}{4m}\hat{\boldsymbol% {\imath}}$ , (d) ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=\frac{F}{m}\hat{\boldsymbol{% \imath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=-\frac{F}{m}\hat{\boldsymbol% {\imath}}$ .

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Figure 14.25:

Four different ways to pull a mass.

14.1.7 For each of the situations in problem 14.1 find the acceleration of the mass using energy balance $(P={\dot{E}_{\mathrm{K}}})$ . Define any variables, coordinates, or sign conventions that you need to do your calculations and to define your solution.

14.1.8 What is the ratio of the acceleration of point A to that of point B in each configuration? $m=m$ and $F=F$ . Answer: $\frac{a_{A}}{a_{B}}=81$ .

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Figure 14.26:

14.1.9 Find the acceleration of points A and B in terms of $F$ and $m$ . Answer: $a_{A}=2F/(5m)$ , $a_{B}=F/(5m)$ to the right.

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Figure 14.27:

14.1.10 For the situation pictured in problem 14.26 find the accelerations of the two masses using energy balance $(P={\dot{E}_{\mathrm{K}}})$ . Define any variables, coordinates, or sign conventions that you need to do your calculations and to define your solution.

14.1.11 The point of application $A$ of the force moves twice as fast as the mass. At some instant in time $t$ , the speed of the mass is $\dot{x}$ to the left. Find the input power to the system at time $t$ . Answer: $P=2F\dot{x}$

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Figure 14.28:

More-Involved Problems

14.1.12 A train engine of mass $m$ pulls and accelerates $N$ cars, each of mass $m$ . The power of the engine is $P_{t}$ and its speed is $v_{t}$ . Find the tension $T_{n}$ between car $n$ and car $n$ +1. Assume there is no resistance and the ground is level. Assume the cars are connected with rigid links. Answer: $T_{n}=\frac{P_{t}}{v_{t}}\frac{n}{N}$ .

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Figure 14.29:

14.1.13 A cart of mass $M$ , initially at rest, can move horizontally along a frictionless track. When $t=0$ , a force $F$ is applied as shown to the cart. During the acceleration of $M$ by the force $F$ , a small box of mass $m$ slides along the cart from the front to the rear. The coefficient of friction between the cart and the box is $\mu$ , and it is assumed that the acceleration of the cart is sufficient to cause sliding.

(a)

Draw free-body diagrams of the cart, the box, and the cart and box together.
(b)

Write the equation of linear momentum balance for the cart, the box, and the system of cart and box.
(c)

Show that the equations of motion for the cart and box can be combined to give the equation of motion of the mass center of the system of two bodies.
(d)

Find the displacement of the cart at the time when the box has moved a distance $\ell$ along the cart. Answer: displacement of cart $=\frac{F-\mu mg}{F-\mu(m+M)g}\ell$

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Figure 14.30:

14.1.14 For the situations pictured, find the accelerations of mass A and of point B. Clearly define any variables, coordinates or sign conventions that you use.

(a)

A single mass and four pulleys. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=\frac{5F}{m}\hat{\boldsymbol% {\imath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=\frac{25F}{m}\hat{% \boldsymbol{\imath}}$ , where $\hat{\boldsymbol{\imath}}$ is parallel to the ground and points to the right.
(b)

Two masses and two pulleys. Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=\frac{g}{(4m_{1}+m_{2})}(2m_% {2}-\sqrt{3}m_{2})\hat{\boldsymbol{\lambda}}_{1}$ , ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=-\frac{g}{2(4m_{1}+m_{2})}(2% m_{1}-\sqrt{3}m_{2})\hat{\boldsymbol{\lambda}}_{2}$ , where $\hat{\boldsymbol{\lambda}}_{1}$ is parallel to the slope that mass $m_{1}$ travels along, pointing down and to the left, and $\hat{\boldsymbol{\lambda}}_{2}$ is parallel to the slope that mass $m_{2}$ travels along, pointing down and to the right.
(c)

A single mass and four pulleys. Answer: $a_{A}=\frac{5F}{4m}$ and $a_{B}=\frac{25F}{16m}$ in the direction of F.

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Figure 14.31:

Various pulley arrangements.

14.1.15 For the situations pictured in problem 14.1, find the acceleration of the mass using energy balance $(P={\dot{E}_{\mathrm{K}}})$ .

14.1.16 A person of mass $m$ , modeled as a rigid object, is sitting on a cart of mass $M>m$ and pulling the string towards herself. The coefficient of friction between her seat and the cart is $\mu$ . Point B is attached to the cart and point A is attached to the rope.

(a)

If you are given that she is pulling rope in with acceleration $a_{0}$ relative to herself (that is, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A/B}\equiv{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{A}-{\accentset{\rightharpoonup}{\boldsymbol% {a}}}_{B}=-a_{0}\hat{\boldsymbol{\imath}}$ ) and that she is not slipping relative to the cart, find ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}$ . (Answer in terms of some or all of $m,M,g,\mu,\hat{\boldsymbol{\imath}}$ and $a_{0}$ .) Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=-\frac{a_{0}}{2}\hat{% \boldsymbol{\imath}}$
(b)

Find the largest possible value of $a_{0}$ without the person slipping off the cart? (Answer in terms of some or all of $m, M, g$ and $\mu$ . You may assume her legs get out of the way if she slips backwards.)
(c)

If instead, $M<m$ , what is the largest possible value of $a_{0}$ without the person slipping off the cart? (Answer in terms of some or all of $m, M, g$ and $\mu$ . You may assume her legs get out of the way if she slips backwards.)

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Figure 14.32:

Pulley.

14.1.17 Two blocks and a pulley. Two identical blocks are stacked and tied together by the pulley as shown. Find

(a)

the acceleration of point A, and
(b)

the tension in the line.

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Figure 14.33:

14.1.18 What is the natural frequency of vibration of this system? Include gravity. $x$ measures the vertical position of the lower mass from equilibrium. $y$ measures the vertical position of the upper mass from equilibrium. Answer: angular frequency of vibration $\equiv\lambda=\sqrt{\frac{64k}{65m}}$ .

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Figure 14.34:

14.1.19 For the situation pictured, find the acceleration of mass A and points B and C shown. [Hint: the situation with point C is subtle.] Answer: The assembly at C has, with the idealizations given, no mass and no net force acting on it. The equation ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup}{% \boldsymbol{a}}}$ says $0=0$ and, within the assumptions made, the acceleration is indeterminate. If you built such a machine, the acceleration of C would be determined by the effects which are neglected here, such as the mass of the central assembly and the friction in the pulley bearings. If you built such an assembly you would see that only small additional forces are needed to move point C most any which way.

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Figure 14.35:

14.1.20 For the situation pictured in problem 14.34, find the acceleration of point A using energy balance $(P={\dot{E}_{\mathrm{K}}})$ . Define any variables, coordinates, or sign conventions that you need to do your calculations and to define your solution.

14.1.21 Design a pulley system. You are to design a pulley system to move a mass. There is no gravity. Point A has a force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=F\hat{\boldsymbol{\imath}}$ pulling it to the right. Mass B has mass $m_{B}$ . You can connect point A to the mass with any number of ideal strings and ideal pulleys. You can make use of rigid walls or supports anywhere you like (say, to the right or left of the mass). You must design the system so that mass B accelerates to the left with ${F\over 2m_{B}}$ (i.e., ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=-{F\over 2m_{B}}\hat{% \boldsymbol{\imath}}$ ).

(a)

Draw the system clearly. Justify your answer with enough words or equations so that a reasonable person, say a grader, can tell that you understand your solution.
(b)

Find the acceleration of point A.

14.1.22 Design a pulley system. You are to design a pulley system to move a mass. There is no gravity. Point A has a force ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=F\hat{\boldsymbol{\imath}}$ pulling it to the right. Mass B has mass $m_{B}$ . You can connect the point A to the mass with any number of ideal strings and ideal pulleys. You can make use of rigid walls or supports anywhere you like (say, to the right or left of the mass). Draw the system clearly. Justify your answer with enough words or equations to convince a skeptical person that your solution is correct. You must design the system so that the mass B accelerates .

(a)

to the left with ${F\over m_{B}}$ (i.e., ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=-{F\over m_{B}}\hat{% \boldsymbol{\imath}}$ )
(b)

to the left with ${2F\over m_{B}}$
(c)

to the left with ${F\over 2m_{B}}$
(d)

to the right with ${2F\over m_{B}}$
(e)

to the right with ${F\over 2m_{B}}$
(f)

to the left with ${8F\over m_{B}}$
(g)

to the right with ${F\over 5m_{B}}$

14.1.23 Pulley and spring. For the hanging mass find the period of oscillation. Only vertical motion is of interest. There is gravity.

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Figure 14.36:

14.1.24 The spring-mass system shown (m = 10 slugs ( $\equiv\,{\rm lb}\cdot sec^{2}/ft$ ), $k=10\,{\rm lb}/ft$ ) is excited by moving the free end of the cable vertically according to $\delta(t)=4\sin(\omega t)\in$ , as shown in the figure.

(a)

Derive the equation of motion for the block in terms of the displacement $x$ from the static equilibrium position, as shown in the figure.
(b)

If $\omega=0.9\,\mathrm{rad}/\mathrm{s}$ , check to see if the pulley is always in contact with the cable (ignore the transient solution).

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Figure 14.37:

14.1.25 The block of mass $m$ hanging on the spring with constant $k$ and a string shown in the figure is forced by $F=A\sin(\omega t)$ . Do not neglect gravity. The pulley has negligible mass.

(a)

What is the differential equation governing the motion of the block? You may assume that the only motion is vertical motion. Answer: $m\ddot{x}+4kx=A\sin\omega t+mg$ , where $x$ is the distance measured from the mass position when the spring is unstretched.
(b)

Given $A$ , $m$ and $k$ , for what values of $\omega$ would the string go slack at some point in the cyclical motion?
(The common assumption in such problems, which you can use, is to neglect the homogeneous solution to the differential equation. It is assumed that the damping, small enough to be neglected in the governing equations is large enough so that the particular solution will have damped out at the time of observation.) Answer: The string will go slack if $\omega>\sqrt{\frac{4k}{m}\left(1-\frac{A}{mg}\right)}$ .

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Figure 14.38:

14.1.26 Block A, with mass $m_{A}$ , is pulled to the right a distance $d$ from the position it would have if the spring were relaxed. It is then released from rest. Assume ideal string, pulleys and wheels. The spring has constant $k$ .

(a)

What is the acceleration of block A just after it is released (in terms of $k$ , $m_{A}$ , and $d$ )? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=-\frac{9kd}{m_{A}}\hat{% \boldsymbol{\imath}}$ .
(b)

What is the speed of the mass when the mass passes through the position where the spring is relaxed? Answer: $v=3d\sqrt{\frac{k}{m_{A}}}$ .

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Figure 14.39:

14.1.27 What is the static displacement of the mass from the position where the spring is just relaxed?

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Figure 14.40:

14.1.28 For the two situations pictured, find the acceleration of point A shown using balance of linear momentum $(\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{% \rightharpoonup}{\boldsymbol{a}}})$ . Assuming both masses are deflected an equal distance from the position where the spring is just relaxed, how much smaller or bigger is the acceleration of block (b) than that of block (a). Define any variables, coordinate system origins, coordinates or sign conventions that you need to do your calculations and to define your solution.

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Figure 14.41:

14.1.29 For each of the situations pictured in problem 14.40, find the acceleration of the mass using energy balance $(P={\dot{E}_{\mathrm{K}}})$ . Define any variables, coordinates, or sign conventions that you need to do your calculations and to define your solution.

14.2 1D motion with 2D and 3D forces

Even if all the motion is in a single direction, an engineer may still have to consider two- or three-dimensional forces.

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Figure 14.42: (a) shows a piston in a cylinder. (b) shows a free-body diagram of the piston. To draw this FBD, we have assumed: (1) a coefficient of friction

\mu

between the piston and cylinder wall, and (2) negligible mass for the connecting rod, and (3) ignored the spatial extent of the cylinder.

Example: Piston in a cylinder.

A piston slides vertically in a cylinder with coefficient of friction $\mu$ between the piston and the cylinder wall. Assume the connecting rod has negligible mass so it can be treated as a two-force member as discussed in section 5.2b. The free-body diagram of the piston (with a bit of the connecting rod) is shown in fig. 14.42.

The piston is moving up, so the friction force resists the motion and points down. Linear momentum balance for this system is:

	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{i}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$
	$\displaystyle-N\hat{\boldsymbol{\imath}}-\mu N\hat{\boldsymbol{\jmath}}+T\hat{% \boldsymbol{\lambda}}_{rod}$	$\displaystyle=$	$\displaystyle m_{piston}\,a\hat{\boldsymbol{\jmath}}.$

If we assume that the acceleration $a\hat{\boldsymbol{\jmath}}$ of the piston is known, as is its mass $m_{piston}$ , the coefficient of friction $\mu$ , and the orientation of the connecting rod $\hat{\boldsymbol{\lambda}}_{rod}$ , then we can solve for the rod tension $T$ and the normal reaction $N$ .

Note: even though the piston moves in one direction, the momentum balance equation is a two-dimensional vector equation.

Unlike the 1D mechanics of the previous chapter, in this section on 1D motion, the momentum balance equations are 2D and 3D vector equations. Compared to more general 2D and 3D motion, the 1-D motions we assume in this chapter allow an easy introduction to 2D and 3D dynamics calculations.

Highly constrained bodies

This chapter is about rigid objects that move in straight lines. Most objects will not agree to be the topic of such discussion without being forced into doing so. Without being held in place they would rotate and move in a curvy way. To keep an object that is subject to various forces from rotating or curving takes some constraint by wires, rods, rails, hinges, welds, etc.. Of course the presence of constraint is not always associated with the disallowance of rotation — constraints could even cause rotation. But in this chapter, constraints keep a rigid object in straight-line motion.

Constraint forces are of interest

Of common interest is making sure that static and dynamic loads do not cause failure of parts that enforce constraints. For example, suppose a truck hauls a very heavy load that is held down by chains or straps. When the truck accelerates, what is the tension in the chains, and will it exceed their strength limits?

1D mechanics with 1D forces, moving on

This all is in contrast with the situation in 1D ”unconstrained” dynamics of the previous chapter. For one-dimensional mechanics we assumed that everything of interest mechanically happened in, say, the $\hat{\boldsymbol{\imath}}$ ( $x$ ) direction. That is, we ignored all torques and angular momenta, and only considered the $\hat{\boldsymbol{\imath}}$ components of the forces (i.e., ${\accentset{\rightharpoonup}{\boldsymbol{F}}}\cdot\hat{\boldsymbol{\imath}}$ ) and linear momentum ( ${\accentset{\rightharpoonup}{\boldsymbol{L}}}\cdot\hat{\boldsymbol{\imath}}$ ), namely $F_{x}$ and $L_{x}$ .

Kinematics of parallel motion and straight line motion

Let’s consider a set of points in the system of interest. Let’s call them $A$ to $G$ , or generically, $P$ . For convenience we distinguish a reference point $O^{\prime}$ . $O^{\prime}$ may be the center-of-mass, the origin of a local coordinate system, or a fleck of dirt that serves as a marker. By parallel motion, we mean that the system happens to move in such a way that ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{O^{\prime}}$ , and ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{P}={\accentset{\rightharpoonup}% {\boldsymbol{v}}}_{O^{\prime}}$ (fig. 14.43).

That is,

{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{B}={\accentset{\rightharpoonup}{\boldsymbol{a}}}_{C}={% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{D}={\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{E}={\accentset{\rightharpoonup}{\boldsymbol{a}}}_{F}={% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}={\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{P}={\accentset{\rightharpoonup}{\boldsymbol{a}}}_{O^{\prime}}

at every instant in time. We also assume that ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}=\ldots={\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{P}={\accentset{\rightharpoonup}{\boldsymbol% {v}}}_{O^{\prime}}$ .

A special case of parallel motion is straight-line motion.

a system moves with straight-line motion if it moves like a non-rotating rigid body, in a straight line.

For straight-line motion, the velocity of the body is in a fixed unchanging direction. If we call a unit vector in that direction $\hat{\boldsymbol{\lambda}}$ , then we have

{\accentset{\rightharpoonup}{\boldsymbol{v}}}(t)=v(t)\hat{\boldsymbol{\lambda}% },\qquad{\accentset{\rightharpoonup}{\boldsymbol{a}}}(t)=a(t)\hat{\boldsymbol{% \lambda}}\qquad\mbox{and}\qquad{\accentset{\rightharpoonup}{\boldsymbol{r}}}(t% )={\accentset{\rightharpoonup}{\boldsymbol{r}}}_{0}+s(t)\hat{\boldsymbol{% \lambda}}

for every point in the system. ${\accentset{\rightharpoonup}{\boldsymbol{r}}}_{0}$ is the position of a point at time 0 and $s$ is the distance the point moves in the $\hat{\boldsymbol{\lambda}}$ direction. Every point in the system has the same $s$ , $v$ , $a$ , and $\hat{\boldsymbol{\lambda}}$ as the other points. There are a variety of problems of practical interest that can be idealized as fitting into this class, notably, the motions of things constrained to move on belts, roads, and rails, like the train on a straight track.

Filename:pfigure-s94h13p3

Figure 14.43: Parallel motion: all points on the body have the same acceleration

{\accentset{\rightharpoonup}{\boldsymbol{a}}}=a\hat{\boldsymbol{\lambda}}

. For straight-line motion:

\hat{\boldsymbol{\lambda}}(t)

=constant in time and

{\accentset{\rightharpoonup}{\boldsymbol{v}}}=v\hat{\boldsymbol{\lambda}}

Example: Parallel swing is not straight-line motion

Filename:p-s96-p3-3

Figure 14.44: A swing showing instantaneous parallel motion which is curvilinear. At every instant, each point has the same velocity as the others, but the motion is not in a straight line.

The swing shown does not rotate — all points on the swing have the same velocity. The velocities of all particles are parallel but, since paths are curved, this motion is not straight-line motion. Such curvilinear parallel motion will be discussed later in the book.

Velocity of a point

The velocity of any point P on a non-rotating rigid body (such as for straight-line motion) is the same as that of any reference point on the body (see Fig. 14.45).

Filename:bikefork1-ang-mom

Figure 14.45: A non-rotating body

\cal B

with points

O^{\prime}

and

P

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm P}={\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\rm O^{\prime}}

A more general case, which you will learn in later chapters, is shown as 5b in Table II at the back of the book. This formula concerns rotational rate which we will measure with the vector ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}$ . For now all you need to know is that ${\accentset{\rightharpoonup}{\boldsymbol{\omega}}}={\accentset{\rightharpoonup% }{\boldsymbol{0}}}$ when something is not rotating. In 5b in Table II, if you set $\,{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\mathcal{B}}=0\,$ and $\,{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{{\rm P}/{\cal B}}={\accentset% {\rightharpoonup}{\boldsymbol{0}}}\,$ it says that $\,{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm P}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{\rm O^{\prime}/O}\,$ or in shorthand, $\,{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\rm P}={\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\rm O^{\prime}}\,$ , as we have written above.

Acceleration of a point

Similarly, the acceleration of every point on a non-rotating rigid body is the same as every other point. The more general case, not needed in this chapter, is shown as entry 5c in Table II at the back of the book.

Box 14.1 Angular momentum for straight-line motion

For straight-line motion, and parallel motion in general, we can derive the simplification in the calculation of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/C% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/C}$}}}$ as follows:

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$	$\displaystyle\equiv$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i\mathrm{/C}}% \times\,\mathrm{m}_{i}{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{i}\mbox{(% definition)}$
	$\displaystyle=$	$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i\mathrm{/C}}% \times\,\mathrm{m}_{i}{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{% cm}}\quad\mbox{(since, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{i}={% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}$)}$
	$\displaystyle=$	$\displaystyle\left(\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i\mathrm% {/C}}\,\mathrm{m}_{i}\right)\times{\accentset{\rightharpoonup}{\boldsymbol{v}}% }_{\mathrm{cm}},$
	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{\mathrm{cm/C}}% \times(m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{% cm}}),$
		( since, $\sum{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i\mathrm{/C}}\,\mathrm{m}_{% i}\equiv m_{\mathrm{tot}}{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{% \mathrm{cm/C}}$ ).

The derivation that $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/C}$}}}={\accentset{\rightharpoonup}{\boldsymbol{r% }}}_{\mathrm{cm/C}}\times(m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{% \mathrm{cm}})$ follows from $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/C}$}}}\equiv\sum{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{i\mathrm{/C}}\times\,\mathrm{m}_{i}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{i}$ by the same reasoning.

General results

Before we proceed with discussion of the details of the mechanics of straight-line motion we present some ideas that are also more generally applicable. That is, the concept of the center-of-mass allows some useful simplifications of the general expressions for ${\accentset{\rightharpoonup}{\boldsymbol{L}}}$ , $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/C% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/C}$}}}$ , $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/C}$}}}$ and ${E_{\mathrm{K}}}$ .

Linear momentum ${\accentset{\rightharpoonup}{\boldsymbol{L}}}$ and its rate of change $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$

Although we are dealing with zillions of atoms in a given object, the linear momentum and angular momentum are simple to evaluate:

{\accentset{\rightharpoonup}{\boldsymbol{L}}}=m_{\mathrm{tot}}{\accentset{% \rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}\qquad\mbox{and}\qquad\dot{% \accentset{\rightharpoonup}{\boldsymbol{L}}}=m_{\mathrm{tot}}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{\mathrm{cm}}.

Actually, as the front inside cover states, these formulas are good for any motion of any system. The nice simplification for the straight-line motion of this chapter is that all points on a given object have the same velocity and acceleration. So we don’t need to find or track the center of mass, but can track the motion of any point on the object.

Angular momentum ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/C% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/C}$}}}$ and its rate of change, $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{C}$ for straight-line motion

For the motions in this chapter, where ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{i}={\accentset{\rightharpoonup}% {\boldsymbol{a}}}_{cm}$ and thus ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{i/cm}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ , angular momentum considerations are simplified, as explained in Box 14.2 on page 14.2

^†^†margin: Calculating rate of change of angular momentum will get more difficult as the book progresses. For a rigid body

\cal B

in more general motion, the calculation of rate of change of angular momentum involves the angular velocity

{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}

, its rate of change

\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}_{\cal B}

, and the moment of inertia matrix

[\boldsymbol{I}^{\mathrm{cm}}]

. If you look in the back of the book at Table I, entries 6c and 6d, you will see formulas that reduce to the formulas below if you assume no rotation and thus use

\,{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}\,

and

\,\dot{\accentset{\rightharpoonup}{\boldsymbol{\omega}}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}

. On the other hand, rate of change of linear momentum is simple, at least in concept, in this chapter, as well as in the rest of this book. For all motions of all systems we have

\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}=m_{tot}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{cm}.

{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/C% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/C}$}}}={\accentset{\rightharpoonup}{\boldsymbol{r}}% }_{\!\mathrm{cm/C}}\times{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{% \mathrm{cm}}m_{\mathrm{tot}}\qquad\mbox{and}\qquad\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\mathrm{C}}={\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{\!\mathrm{cm/C}}\times{\accentset{\rightharpoonup}{% \boldsymbol{a}}}_{\mathrm{cm}}m_{\mathrm{tot}}

But for straight-line motion (and, slightly more generally, for any parallel motion), the calculations turn out to be the same as we would get if we put a single point mass at the center-of-mass^†^†margin: Caution: The special motions in this chapter are almost the only cases where the angular momentum and its rate of change are so easy to calculate. :

	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$	$\displaystyle\equiv\sum({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i% \mathrm{/C}}\times m_{i}{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{i})$	$\displaystyle=\ {{\accentset{\rightharpoonup}{\boldsymbol{r}}}}_{cm/C}\times(m% _{\mbox{total}}{{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{\mathrm{cm}}}),$
	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt% \mathchoice{\lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \textstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1% .0pt\hbox{$\scriptscriptstyle\mathrm{/C}$}}}$	$\displaystyle\equiv\sum({\accentset{\rightharpoonup}{\boldsymbol{r}}}_{i% \mathrm{/C}}\times m_{i}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{i})$	$\displaystyle=\ {{\accentset{\rightharpoonup}{\boldsymbol{r}}}}_{cm/C}\times(m% _{\mbox{total}}{{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm}}).$

Note, there is some subtlety in the definition of ${\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{\lower 1% .0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{/C% }$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{/C}$}}}$ , as explained in section LABEL:sec:KonigSchmonig.

Kinetic energy

Generally things will not be so simple, but for straight-line motion, or any parallel motion where all points on an object have the same velocity and acceleration, kinetic energy and its rate of change are also easy to calculate:

{E_{\mathrm{K}}}=m_{\mathrm{tot}}v_{\mathrm{cm}}^{2}/2\qquad\mbox{and}\qquad{% \dot{E}_{\mathrm{K}}}=m_{\mathrm{tot}}v_{\mathrm{cm}}a_{\mathrm{cm}}.

The kinetic energy works the same as if all the mass was concentrated at the center of mass. This result does not generalize to more complex motions.

Approach

To study systems in straight-line motion (as always) we:

•

draw a free-body diagram, showing the appropriate forces and couples at places where connections are ‘cut’,
•

state reasonable kinematic assumptions based on the motions that the constraints allow,
•

write linear and/or angular momentum balance equations and/or energy balance, and
•

solve for quantities of interest.

Angular momentum balance about a judiciously chosen axis is a particularly useful tool for reducing the number of equations that need to be solved.

Example: Plate on a cart

Filename:bikefork-ang-mom

Figure 14.46: Uniform plate supported by a hinge and a rod on an accelerating cart.

A uniform rectangular plate $A B C D$ of mass $m$ is supported by a light rigid rod $D E$ and a hinge joint at point $B$ . The dimensions are as shown. The cart has acceleration $a_{x}\hat{\boldsymbol{\imath}}$ due to a force $F\hat{\boldsymbol{\imath}}$ and the constraints of the wheels. Referring to the free-body diagram in fig. 14.46 and writing angular momentum balance for the plate about point $B$ , we can get an equation for the tension in the rod $T_{DE}$ in terms of $m$ and $a_{x}$ :

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{/B}$	$\displaystyle=$	$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{/B}$
$\displaystyle\left\{{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{D/B}\times% \par(T_{DE}\hat{\boldsymbol{\lambda}}_{DE})+{\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{G/B}\times(-mg\hat{\boldsymbol{\jmath}})\right.$	$\displaystyle=$	$\displaystyle\left.{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{G/B}\times(% ma_{x}\hat{\boldsymbol{\imath}})\right\}$
$\displaystyle\{\,\}\cdot\hat{\boldsymbol{k}}\Rightarrow\qquad T_{DE}$	$\displaystyle=$	$\displaystyle\frac{\sqrt{5}}{7}m(a_{x}-\frac{3}{2}g).$

Summarizing note:

angular momentum balance is important even when there is no rotation.

Sliding and pseudo-sliding objects

A car coming to a stop can be roughly modeled as a rigid body that translates and does not rotate. That is, at least for a first approximation, the rotation of the car due to the suspension and tire deformation, can be neglected. The free-body diagram will show various forces with lines of action that do not all act through a single point so that angular momentum balance must be used to analyze the system.

Similarly, a bicycle which is braking or a box that is skidding (if not tipping) may be analyzed by assuming straight-line motion.

Example: Car skidding

Consider the accelerating four-wheel drive car in fig. 14.47.

Filename:summer95f-5-a

Figure 14.47: A four-wheel drive car accelerating but not tipping. See fig. 2.112 on page 2.112 for more about FBDs involving wheel contact.

The motion quantities for the car are $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}=m_{car}{\accentset{% \rightharpoonup}{\boldsymbol{a}}}_{car}$ and $\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\kern-2.0pt\mathchoice{% \lower 1.0pt\hbox{$\displaystyle\mathrm{/C}$}}{\lower 1.0pt\hbox{$\textstyle% \mathrm{/C}$}}{\lower 1.0pt\hbox{$\scriptstyle\mathrm{/C}$}}{\lower 1.0pt\hbox% {$\scriptscriptstyle\mathrm{/C}$}}}={\accentset{\rightharpoonup}{\boldsymbol{r% }}}_{cm/C}\times{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{car}m_{car}$ . We could calculate angular momentum balance relative to the car’s center of mass in which case $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{cm}}=\dot{% \accentset{\rightharpoonup}{\boldsymbol{H}}}_{\mathrm{cm}}={\accentset{% \rightharpoonup}{\boldsymbol{0}}}$ (because the position of the center-of-mass relative to the center-of-mass is ${\accentset{\rightharpoonup}{\boldsymbol{0}}}$ ).

As mentioned, it is often useful to calculate angular momentum balance of sliding objects about points of contact (such as where tires contact the road) or about points that lie on lines of action of applied forces when writing angular momentum balance to solve for forces or accelerations. To do so usually eliminates some unknown reactions from the equations to be solved. For example, the angular momentum balance equation about the rear-wheel contact of a car does not contain the rear-wheel contact forces.

Wheels

The function of wheels is to allow easy sliding-like (pseudo-sliding) motion between objects, at least in the direction they are pointed. On the other hand, wheels do sometimes slip due to:

•

being overpowered (as in a screeching accelerating car),
•

being braked hard, or
•

having very bad bearings (like a rusty toy car).

How wheels are treated when analyzing cars, bikes, and the like depends on both the application and on the level of detail one requires. In this chapter, we will always assume that wheels have negligible mass. Thus, when we treat the special case of un-driven and un-braked wheels our free-body diagrams will be as in fig. 2.113a on page 2.113 and not like the one in fig. 2.113b. With the ideal wheel approximation, all of the various cases for a car traveling to the right are shown with partial free-body diagrams of a wheel in fig. 2.112.

For the purposes of actually solving problems, we have accepted Coulomb’s law of friction as a model for contacting interaction (see page 2.4 in sec. 2.4).

3-D forces in straight-line motion

The ideas we have discussed apply as well in three dimensions as in two. As you learned from doing statics problems, working out the details in 3D, where vector methods must be used carefully, is more involved than in 2D. As for statics, three-dimensional problems often yield simple results and simple intuitions by considering angular momentum balance about an axis.

Angular momentum balance about an axis

The simplest way to think of angular momentum balance about an axis is to look at angular momentum balance about a point and then take a dot product with a unit vector along an axis:

\hat{\boldsymbol{\lambda}}\cdot\left\{\sum{\accentset{\rightharpoonup}{% \boldsymbol{M}}}_{\mathrm{/C}}=\dot{\accentset{\rightharpoonup}{\boldsymbol{H}% }}_{\rm/C}\right\}.

Note that the axis need not correspond to any mechanical device in any way resembling an axle. The equation above applies for any point C and any vector $\hat{\boldsymbol{\lambda}}$ . If you choose C and $\hat{\boldsymbol{\lambda}}$ judiciously many terms in your equations may drop out.

SAMPLE 14.6 Force in braking. A front-wheel-drive car of mass $m=1200\,\mathrm{kg}$ is cruising at $v=60\,\mathrm{mph}$ on a straight road when the driver slams on the brake. The car slows down to $20\,\mathrm{mph}$ in $4\,\mathrm{s}$ while maintaining its straight path.

1.

What is the average force (average in time) applied on the car during braking?
2.

What is the average power of braking?

Solution

Filename:pfigure4-2-rp10

Figure 14.48: Free-body diagram of a front-wheel-drive car during braking. Note that we have (arbitrarily) pointed

F_{x}

to the right. The algebra in this problem will tell us that

F_{x}<0

Let us assume that we have an $x y$ coordinate system in which the car is traveling along the $x$ -axis during the entire time under consideration. Then, the velocity of the car before braking, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{1}$ , and after braking, ${\accentset{\rightharpoonup}{\boldsymbol{v}}}_{2}$ , are

{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{1}=v_{1}\hat{\boldsymbol{\imath% }}=60\,\mathrm{mph}\ \hat{\boldsymbol{\imath}}\hbox to36.135pt{\hfil and\hfil}% {\accentset{\rightharpoonup}{\boldsymbol{v}}}_{2}=v_{2}\hat{\boldsymbol{\imath% }}=20\,\mathrm{mph}\ \hat{\boldsymbol{\imath}}.

The linear impulse during braking is ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{\rm ave}\Delta t$ where ${\accentset{\rightharpoonup}{\boldsymbol{F}}}\equiv F_{x}\hat{\boldsymbol{% \imath}}$ (see free-body diagram of the car). Now, from the impulse-momentum relationship,

{\accentset{\rightharpoonup}{\boldsymbol{F}}}\Delta t={\accentset{% \rightharpoonup}{\boldsymbol{L}}}_{2}-{\accentset{\rightharpoonup}{\boldsymbol% {L}}}_{1},

where ${\accentset{\rightharpoonup}{\boldsymbol{L}}}_{1}$ and ${\accentset{\rightharpoonup}{\boldsymbol{L}}}_{2}$ are linear momenta of the car before and after braking, respectively, and ${\accentset{\rightharpoonup}{\boldsymbol{F}}}$ is the average applied force. Therefore,

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{F}}}$	$\displaystyle=$	$\displaystyle\frac{1}{\Delta t}({\accentset{\rightharpoonup}{\boldsymbol{L}}}_% {2}-{\accentset{\rightharpoonup}{\boldsymbol{L}}}_{1})=\frac{m}{\Delta t}({% \accentset{\rightharpoonup}{\boldsymbol{v}}}_{2}-{\accentset{\rightharpoonup}{% \boldsymbol{v}}}_{1})$
	$\displaystyle=$	$\displaystyle\frac{1200\,\mathrm{kg}}{4\,\mathrm{s}}(20-60)\,\mathrm{mph}\hat{% \boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle-12000\frac{\,\mathrm{kg}}{\,\mathrm{s}}\cdot\frac{\cancel{\,% \mathrm{mi}}}{\cancel{\,{\mathrm{hr}}}}\cdot\frac{1600\,\,\mathrm{m}}{1\,% \cancel{\,\mathrm{mi}}}\cdot\frac{1\,\cancel{\,{\mathrm{hr}}}}{3600\,\mathrm{s% }}\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle-{16,000\over 3}\,\mathrm{kg}\cdot\,\mathrm{m}/\mathrm{s}^{2}\hat% {\boldsymbol{\imath}}=-5.33\,{\rm kN}\hat{\boldsymbol{\imath}}.$
Thus
$\displaystyle F_{x}\hat{\boldsymbol{\imath}}$	$\displaystyle=$	$\displaystyle-5.33\,{\rm kN}\hat{\boldsymbol{\imath}}\hbox to36.135pt{\hfil$% \Rightarrow$\hfil}F_{x}=-5.33\,{\rm kN}.$

Answer: $F_{x}=-5.33\,{\rm kN}$

2.

Let the average power during braking be $P_{\rm ave}$ . Then the work done during braking is $W=\int P_{\rm ave}dt$ . From work-energy principle, we have

$\displaystyle W$ $\displaystyle=$ $\displaystyle\Delta{E_{\mathrm{K}}}$

$\displaystyle\int_{t_{1}}^{t_{2}}P_{\rm ave}dt$ $\displaystyle=$ $\displaystyle\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})$

$\displaystyle P_{\rm ave}(t_{2}-t_{1})$ $\displaystyle=$ $\displaystyle\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})$

$\displaystyle P_{\rm ave}$ $\displaystyle=$ $\displaystyle\frac{m}{2\Delta t}(v_{2}^{2}-v_{1}^{2})$

Substituting $m=1200\,\mathrm{kg}$ , $\Delta t=4\,\mathrm{s}$ , $v_{1}=60\,\mathrm{mph}=26.67\,\mathrm{m}/\mathrm{s}$ and $v_{2}=20\,\mathrm{mph}=8.89\,\mathrm{m}/\mathrm{s}$ , we get

$P_{\rm ave}=-94815\,{\rm N}\mathord{\cdot}{\rm m}/\,\mathrm{s}=-94.815\,{\rm kW}.$

It is easy to check that if we take the average force $F_{\rm ave}$ calculated above and the average speed $v_{\rm ave}=(v_{1}+v_{2})/2=40\,\mathrm{mph}=17.77\,\mathrm{m}/\mathrm{s}$ , then

$P_{\rm ave}=F_{\rm ave}v_{\rm ave}=-5.33\,{\rm kN}\cdot 17.77\,\mathrm{m}/% \mathrm{s}=-94.815\,{\rm kW},$

as obtained above.

Answer: $P_{\rm ave}=-94.815\,{\rm kW}$

SAMPLE 14.7

Filename:pfigure-blue-125-2

Figure 14.49: A suitcase in motion.

A suitcase skidding on frictional ground. A suitcase of mass $m$ is pushed and sent sliding on a horizontal surface. The suitcase slides without any rotation. A and B are the only contact points of the suitcase with the ground. If the coefficient of friction between the suitcase and the ground is $\mu$ , find all the forces applied by the ground on the suitcase. Discuss the results obtained for normal forces.

Solution As usual, we first draw a free-body diagram of the suitcase. The FBD is shown in Fig. 14.50.

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Figure 14.50: FBD of the suitcase.

Assuming Coulomb’s law of friction holds, we can write

{\accentset{\rightharpoonup}{\boldsymbol{F}}}_{1}=-\mu N_{1}\hat{\boldsymbol{% \imath}}\hbox to36.135pt{\hfil and\hfil}{\accentset{\rightharpoonup}{% \boldsymbol{F}}}_{2}=-\mu N_{2}\hat{\boldsymbol{\imath}}.

(14.29)

Now we write the balance of linear momentum for the suitcase:

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}}_{\mathrm{cm}}

\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad-(F_{1}+F_{2})\hat{\boldsymbol{% \imath}}+(N_{1}+N_{2}-mg)\hat{\boldsymbol{\jmath}}=ma_{C}\hat{\boldsymbol{% \imath}}

(14.30)

where ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{C}=a_{C}\hat{\boldsymbol{\imath}}$ is the unknown acceleration. Dotting eqn. (14.30) with $\hat{\boldsymbol{\imath}}$ and $\hat{\boldsymbol{\jmath}}$ and substituting for $F_{1}$ and $F_{2}$ from eqn. (14.29) we get

	$\displaystyle-\mu(N_{1}+N_{2})$	$\displaystyle=$	$\displaystyle ma_{C}$		(14.31)
	$\displaystyle N_{1}+N_{2}$	$\displaystyle=$	$\displaystyle mg.$		(14.32)

Equations (14.31) and (14.32) represent 2 scalar equations in three unknowns $N_{1},\;N_{2}$ and $a$ . Obviously, we need another equation to solve for these unknowns.

We can write the balance of angular momentum about any point. Points A or B are good choices because they each eliminate some reaction components. Let us write the balance of angular momentum about point A:

\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\rm A}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{\rm A}

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\rm A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{B/A}\times N_{2}% \hat{\boldsymbol{\jmath}}+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{D/A}% \times(-mg)\hat{\boldsymbol{\jmath}}$	(14.33)
	$\displaystyle=$	$\displaystyle\ell\hat{\boldsymbol{\imath}}\times N_{2}\hat{\boldsymbol{\jmath}% }+\frac{\ell}{2}\hat{\boldsymbol{\imath}}\times(-mg)\hat{\boldsymbol{\jmath}}$
	$\displaystyle=$	$\displaystyle(\ell N_{2}-mg\frac{\ell}{2})\hat{\boldsymbol{k}}$

and

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{\rm A}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C/A}\times m{% \accentset{\rightharpoonup}{\boldsymbol{a}}}_{C}$	(14.34)
	$\displaystyle=$	$\displaystyle(\frac{\ell}{2}\hat{\boldsymbol{\imath}}+h\hat{\boldsymbol{\jmath% }})\times ma_{C}\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle-ma_{C}h\hat{\boldsymbol{k}}.$

Equating (14.33) and (14.34) and dotting both sides with $\hat{\boldsymbol{k}}$ we get the following third scalar equation:

\ell N_{2}-mg\frac{\ell}{2}=-ma_{C}h.

(14.35)

Solving eqns. (14.31) and (14.32) for $a$ we get

a_{C}=-\mu g

and substituting this value of $a_{C}$ in eqn. (14.35) we get

	$\displaystyle N_{2}$	$\displaystyle=$	$\displaystyle\frac{m\mu gh+mg\ell/2}{\ell}$
		$\displaystyle=$	$\displaystyle mg\left(\frac{1}{2}+\frac{h}{\ell}\mu\right).$

Substituting the value of $N_{2}$ in either of the equations (14.31) or (14.32) we get

N_{1}=mg\left(\frac{1}{2}-\frac{h}{\ell}\mu\right).

Answer: $N_{1}=mg(\frac{1}{2}-\frac{h}{\ell}\mu),\;N_{2}=mg(\frac{1}{2}+\frac{h}{\ell}% \mu),\;f_{1}=\mu N_{1},\;f_{2}=\mu N_{2}.$

Discussion: From the expressions for $N_{1}$ and $N_{2}$ we see that

1.

$N_{1}=N_{2}=\frac{1}{2}mg\;$ if $\mu=0$ because without friction there is no deceleration. The problem becomes equivalent to a statics problem.
2.

$N_{1}=N_{2}\approx\frac{1}{2}mg\;$ if $\ell>>h$ . In this case, the moment produced by the friction forces is too small to cause a significant difference in the magnitudes of the normal forces. For example, take $\ell=20h$ and calculate moment about the center-of-mass to convince yourself.

Graphically, $N_{1}$ , $N_{2}$ and their difference $N_{1}-N_{2}$ are shown in the plot below as a function of $h/\ell$ for a particular value of $\mu$ and $m g$ . As the equations indicate, $N_{1}-N_{2}$ increases steadily as $h/\ell$ increases, showing how the moment produced by the friction forces makes a bigger and bigger difference between $N_{1}$ and $N_{2}$ as this moment gets bigger.

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Figure 14.51: The normal forces

N_{1}

and

N_{2}

differ from each other more and more as

h/\ell

increases.

SAMPLE 14.8

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Figure 14.52: An accelerating board in 3-D

Uniform acceleration of a board in 3-D. A uniform sign-board of mass $m=20\,\mathrm{kg}$ sits in the back of an accelerating flatbed truck. The board is supported with a ball-and-socket joint at $O$ and a hinge at $G$ . A light rod from $H$ to $I$ keeps the board from falling over. The truck is on level ground and has forward acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=0.6\,\mathrm{m}/\mathrm{s}^{2}% \hat{\boldsymbol{\imath}}$ . The relevant dimensions are $b=1.5\,\mathrm{m},\;c=1.5\,\mathrm{m},\;d=3\,\mathrm{m},\;e=0.5\,\mathrm{m}.$ There is gravity ( $g=10\,\mathrm{m}/\mathrm{s}^{2}$ ).

1.

Draw a free-body diagram of the board.
2.

Set up equations to solve for all the unknown forces shown on the FBD.
3.

Use the balance of angular momentum about an axis to find the tension in the rod.

Solution

1.

The free-body diagram of the board is shown in Fig. 14.53.

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Figure 14.53: FBD of the board

Linear momentum balance for the board:

\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}=m{\accentset{\rightharpoonup% }{\boldsymbol{a}}},\quad\mbox{or}

(G_{x}+O_{x})\hat{\boldsymbol{\imath}}+(G_{y}+O_{y})\hat{\boldsymbol{\jmath}}+% (G_{z}+O_{z}-mg)\hat{\boldsymbol{k}}+T\hat{\boldsymbol{\lambda}}_{HI}=ma\hat{% \boldsymbol{\imath}}

(14.36)

where

\hat{\boldsymbol{\lambda}}_{HI}=\frac{d\hat{\boldsymbol{\imath}}+b\hat{% \boldsymbol{\jmath}}+e\hat{\boldsymbol{k}}}{\sqrt{d^{2}+b^{2}+e^{2}}}=\frac{d% \hat{\boldsymbol{\imath}}+b\hat{\boldsymbol{\jmath}}+e\hat{\boldsymbol{k}}}{% \ell},

and $\ell$ is the length of the rod HI.

Dotting eqn. (14.36) with $\hat{\boldsymbol{\imath}},\;\hat{\boldsymbol{\jmath}}$ and $\hat{\boldsymbol{k}}$ we get the following three scalar equations:

$\displaystyle G_{x}+O_{x}+T\frac{d}{\ell}$	$\displaystyle=$	$\displaystyle ma$	(14.37)
$\displaystyle G_{y}+O_{y}+T\frac{b}{\ell}$	$\displaystyle=$	$\displaystyle 0$	(14.38)
$\displaystyle G_{z}+O_{z}+T\frac{e}{\ell}$	$\displaystyle=$	$\displaystyle mg$	(14.39)

Angular momentum balance about point G:

\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{G}=\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{G}

$\displaystyle\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{G}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C/G}\times(-mg\hat% {\boldsymbol{k}})+{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{O/G}\times(O_% {x}\hat{\boldsymbol{\imath}}+O_{z}\hat{\boldsymbol{k}})+{\accentset{% \rightharpoonup}{\boldsymbol{r}}}_{H/G}\times T\hat{\boldsymbol{\lambda}}_{HI}$	(14.40)
	$\displaystyle=$	$\displaystyle(-\frac{b}{2}\hat{\boldsymbol{\jmath}}+\frac{c-e}{2}\hat{% \boldsymbol{k}})\times(-mg\hat{\boldsymbol{k}})-b\hat{\boldsymbol{\jmath}}% \times(O_{x}\hat{\boldsymbol{\imath}}+O_{z}\hat{\boldsymbol{k}})$
		$\displaystyle+[-b\hat{\boldsymbol{\jmath}}+(c-e)\hat{\boldsymbol{k}}]\times% \frac{T}{\ell}(d\hat{\boldsymbol{\imath}}+b\hat{\boldsymbol{\jmath}}+e\hat{% \boldsymbol{k}})$
	$\displaystyle=$	$\displaystyle\left(\frac{b}{2}mg-bO_{z}-be\frac{T}{\ell}-(c-e)b\frac{T}{\ell}% \right)\hat{\boldsymbol{\imath}}$
		$\displaystyle+(c-e)d\frac{T}{\ell}\hat{\boldsymbol{\jmath}}+\left(bO_{x}+bd% \frac{T}{\ell}\right)\hat{\boldsymbol{k}}$

and

$\displaystyle\dot{\accentset{\rightharpoonup}{\boldsymbol{H}}}_{G}$	$\displaystyle=$	$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{r}}}_{C/G}\times ma\hat{% \boldsymbol{\imath}}$	(14.41)
	$\displaystyle=$	$\displaystyle(-\frac{b}{2}\hat{\boldsymbol{\jmath}}+\frac{c-e}{2}\hat{% \boldsymbol{k}})\times ma\hat{\boldsymbol{\imath}}$
	$\displaystyle=$	$\displaystyle\frac{b}{2}ma\hat{\boldsymbol{k}}+\frac{c-e}{2}ma\hat{\boldsymbol% {\jmath}}.$

Equating (14.40) and (14.41) and dotting both sides with $\hat{\boldsymbol{\imath}},\;\hat{\boldsymbol{\jmath}}$ and $\hat{\boldsymbol{k}}$ we get the following three additional scalar equations:

$\displaystyle O_{z}+\frac{c}{\ell}T$	$\displaystyle=$	$\displaystyle\frac{1}{2}mg$	(14.42)
$\displaystyle\frac{d}{\ell}T$	$\displaystyle=$	$\displaystyle\frac{1}{2}ma$	(14.43)
$\displaystyle O_{x}+\frac{d}{\ell}T$	$\displaystyle=$	$\displaystyle\frac{1}{2}ma$	(14.44)

Now we have six scalar equations in seven unknowns — $O_{x},\,O_{y},\,O_{z}$ , $G_{x},\,G_{y},G_{z},$ and $T$ . From basic linear algebra, we know that we cannot find unique solutions for all these unknowns from the given equations. A closer inspection of eqns. (14.37–14.39) and (14.42–14.44) shows that we can easily solve for $O_{x},\,O_{z}$ , $G_{x},G_{z},$ and $T$ , but $O_{y}$ and $G_{y}$ cannot be determined uniquely because they appear together as the sum $G_{y}+O_{y}$ . ^†^†margin: Note that $G_{y}$ and $O_{y}$ will always appear together as the sum $G_{y}+O_{y}$ even if you took the angular momentum balance about some other point. This is because they have the same line of action. Thus, they cannot be found independently. This mathematical problem corresponds to the physical reality that the supports at points $O$ and $G$ could be squeezing the plate along the line $O G$ with, say, $O_{y}=1000\,{\mbox{N}}$ and $G_{y}=-1000\,{\mbox{N}}$ even if there were no gravity, and the truck was not accelerating. To make prestress problems like this tractable, people often make assumptions like, ‘Assume $G_{y}=0$ ’, that is, they try to get rid of the redundancy in supports to make the problem statically determinate. Fortunately, we can find the tension in the wire $H I$ without worrying about the values of $O_{y}$ and $G_{y}$ as we show below.

3.

Balance of angular momentum about axis OG gives:

$\displaystyle\hat{\boldsymbol{\lambda}}_{OG}\cdot\sum{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{G}$ $\displaystyle=$ $\displaystyle\hat{\boldsymbol{\lambda}}_{OG}\cdot\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{G}$ (14.45)

$\displaystyle=$ $\displaystyle\hat{\boldsymbol{\lambda}}_{OG}\cdot({\accentset{\rightharpoonup}% {\boldsymbol{r}}}_{C/G}\times ma\hat{\boldsymbol{\imath}}).$

Since all reaction forces and the weight go through axis OG, they do not produce any moment about this axis (convince yourself that the forces from the reactions have no torque about the axis by calculation or geometry). Therefore,

$\displaystyle\hat{\boldsymbol{\lambda}}_{OG}\cdot\sum{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{G}$ $\displaystyle=$ $\displaystyle\hat{\boldsymbol{\jmath}}\cdot({\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{H/G}\times T\hat{\boldsymbol{\lambda}}_{HI})$ (14.46)

$\displaystyle=$ $\displaystyle T\frac{d(c-e)}{\ell}.$

$\displaystyle\hat{\boldsymbol{\lambda}}_{OG}\cdot({\accentset{\rightharpoonup}% {\boldsymbol{r}}}_{C/G}\times ma\hat{\boldsymbol{\imath}})$ $\displaystyle=$ $\displaystyle\hat{\boldsymbol{\jmath}}\cdot\left[\left(\frac{b}{2}\hat{% \boldsymbol{\jmath}}+\frac{c-e}{2}\hat{\boldsymbol{k}}\right)\times ma\hat{% \boldsymbol{\imath}}\right]$ (14.47)

$\displaystyle=$ $\displaystyle ma\frac{(c-e)}{2}.$

Equating (14.46) and (14.47), as required by eqn. (14.45), we get

$\displaystyle T$ $\displaystyle=$ $\displaystyle\frac{ma\ell}{2d}$

$\displaystyle=$ $\displaystyle\frac{20\,\mathrm{kg}\cdot 0.6\,\mathrm{m}/\mathrm{s}^{2}\cdot 3.% 39\,\mathrm{m}}{2\cdot 3\,\mathrm{m}}$

$\displaystyle=$ $\displaystyle 6.78\,{\mbox{N}}.$

Answer: $T_{HI}=6.78\,{\mbox{N}}$

SAMPLE 14.9 Computer solution of algebraic equations. In the previous sample problem (Sample 14.51), six equations were obtained to solve for the six unknown forces (assuming $G_{y}=0)$ . (i) Set up the six equations in matrix form and (ii) solve the matrix equation on a computer. Check the solution by substituting the values obtained in one or two equations.

Solution

1.

The six scalar equations — (14.37), (14.38), (14.39), (14.42), (14.43), and (14.44) are amenable to hand calculations. We, however, set up these equations in matrix form and solve the matrix equation on the computer. The matrix form of the equations is:

$\left[\begin{array}[]{cccccc}1&0&0&1&0&\frac{d}{\ell}\\ 0&1&0&0&0&\frac{b}{\ell}\\ 0&0&1&0&1&\frac{e}{\ell}\\ 0&0&1&0&0&\frac{c}{\ell}\\ 0&0&0&0&0&\frac{d}{\ell}\\ 1&0&0&0&0&\frac{d}{\ell}\end{array}\right]\left\{\begin{array}[]{c}O_{x}\\ O_{y}\\ O_{z}\\ G_{x}\\ G_{z}\\ T\end{array}\right\}=\left\{\begin{array}[]{c}ma\\ 0\\ mg\\ mg/2\\ ma/2\\ ma/2\end{array}\right\}.$ (14.48)

The above equation can be written, in matrix notation, as

${\bf A\,x}={\bf b}$

where A is the coefficient matrix, x is the vector of the unknown forces, and b is the vector on the right hand side of the equation. Now we are ready to solve the system of equations on the computer.
2.

We use the following pseudo-code to solve the above matrix equation. ^†^†margin: Be careful with units. Most computer programs will not take care of your units. They only deal with numerical input and output. You should, therefore, make sure that your variables have proper units for the required calculations. Either do dimensionless calculations or use consistent units for all quantities.
```
     m = 20, a = 0.6,
     b = 1.5, c = 1.5, d = 3, e = 0.5, g = 10,
     l = sqrt(b^2 + d^2 + e^2),

     A = [1 0 0 1 0 d/l
          0 1 0 0 0 b/l
          0 0 1 0 1 e/l
          0 0 1 0 0 c/l
          0 0 0 0 0 d/l
          1 0 0 0 0 d/l]
     b = [m*a, 0, m*g, m*g/2, m*a/2, m*a/2]’

    {Solve A x = b for x}
    x =               % this is the computer output
             0
       -3.0000
       97.0000
        6.0000
      102.0000
        6.7823
```
The solution obtained from the computer means:

Answer: $O_{x}=0,\;O_{y}=-3\,{\mbox{N}},\;O_{z}=97\,{\mbox{N}},\;G_{x}=6\,{\mbox{N}},\;% G_{z}=102\,{\mbox{N}},\;T=6.78\,{\mbox{N}}.$

We now hand-check the solution by substituting the values obtained in, say, Eqns. (14.38) and (14.43). Before we substitute the values of forces, we need to calculate the length $\ell$ .

	$\displaystyle\ell$	$\displaystyle=$	$\displaystyle\sqrt{d^{2}+b^{2}+e^{2}}$
		$\displaystyle=$	$\displaystyle 3.3912\,\mathrm{m}.$

Therefore,

$\displaystyle\mbox{Eqn.~{}(\ref{eq12.3}):}\quad\qquad O_{y}+T\frac{b}{\ell}$	$\displaystyle=$	$\displaystyle-3\,{\mbox{N}}+6.78\,{\mbox{N}}\cdot\frac{1.5\,\mathrm{m}}{3.3912% \,\mathrm{m}}$
	$\displaystyle\mathrel{\mathop{=}\limits^{\surd}}$	$\displaystyle 0,$
$\displaystyle\mbox{Eqn.~{}(\ref{eq12.8}):}\qquad\frac{d}{\ell}T-\frac{1}{2}ma$	$\displaystyle=$	$\displaystyle\frac{3\,\mathrm{m}}{3.3912\,\mathrm{m}}6.78\,{\mbox{N}}-\frac{1}% {2}20\,\mathrm{kg}\,0.6\,\mathrm{m}/\mathrm{s}^{2}$
	$\displaystyle\mathrel{\mathop{=}\limits^{\surd}}$	$\displaystyle 0.$

Thus, the computer solution agrees with our equations.

Comments: We could have solved the six equations for seven unknowns without assuming $G_{y}=0$ if our computer program or package allows us to do so. We will, of course, not get a unique solution. For example, by taking the following A, a $6\times 7$ matrix, and solving A x = b for x $=[O_{x}\;O_{y}\;O_{z};G_{x}\;G_{y}\;G_{z}\;T]^{T}$ with the same b as input above, we get the solution as shown below.

     A = [1 0 0 1 0 0 d/l
          0 1 0 0 1 0 b/l
          0 0 1 0 0 1 e/l
          0 0 1 0 0 0 c/l
          0 0 0 0 0 0 d/l
          1 0 0 0 0 0 d/l]
     b = [m*a, 0, m*g, m*g/2, m*a/2, m*a/2]’

    {Solve A x = b for x}
    x =               % this is the computer output
             0
       -3.0000
       97.0000
        6.0000
             0
      102.0000
        6.7823

This is the same solution as we got before except that it includes $G_{y}=0$ in the solution. Now, if we add a vector $\Delta$ x $=[0\;\alpha\;0\;0\;-\alpha\;0\;0]^{T}$ to x where $\alpha$ is any number, and compute A (x+ $\Delta$ x) , we get back b. That is, the six equilibrium conditions are satisfied irrespective of the actual values of $O_{y}$ and $G_{y}$ as long as the value of $O_{y}+G_{y}$ remains the same.

Problems for 14.2 1D motion with 2D and 3D forces

Preparatory Problems

14.2.1 Mass pulled by two strings. $F_{1}$ and $F_{2}$ are applied so that the system shown accelerates to the right at $5\,\mathrm{m}/\mathrm{s}^{2}$ (i.e., $\boldsymbol{a}=5\,\mathrm{m}/\mathrm{s}^{2}\hat{\boldsymbol{\imath}}+0\hat{% \boldsymbol{\jmath}}$ ) and has no rotation. The mass of D and forces $F_{1}$ and $F_{2}$ are unknown. What is the tension in string AB?

Filename:pfigure-blue-127-2

Figure 14.54:

14.2.2 The two blocks, $m_{1}=m_{2}=m$ , are connected by an inextensible string $A B$ . The string can only withstand a tension $T_{cr}$ . Find the maximum value of the applied force $P$ so that the string does not break. The sliding coefficient of friction between the blocks and the ground is $\mu$ .

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Figure 14.55:

14.2.3 A point mass $m$ is attached to a piston by two inextensible cables. The piston has upwards acceleration of $a_{y}\hat{\boldsymbol{\jmath}}$ . There is gravity. In terms of some or all of $m, g, d$ , and $a_{y}$ find the tension in cable $A B$ . Answer: $T_{AB}=\frac{5\sqrt{13}}{28}m(a_{y}+g)$

Filename:pfigure-blue-90-2

Figure 14.56:

14.2.4 A point mass of mass $m$ moves on a frictional surface with coefficient of friction $\mu$ and is connected to a spring with constant $k$ and unstretched length $\ell$ . There is gravity. At the instant of interest, the mass is at a distance $x$ to the right from its position where the spring is unstretched and is moving with $\dot{x}>0$ to the right.

(a)

Draw a free-body diagram of the mass at the instant of interest.
(b)

At the instant of interest, write the equation of linear momentum balance for the block evaluating the left hand side as explicitly as possible. Let the acceleration of the block be ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=\ddot{x}\hat{\boldsymbol{\imath}}$ .

Filename:s92f1p7

Figure 14.57:

14.2.5 Consider the mass at B ( $2\,\mathrm{kg}$ ) supported by two strings in the back of a truck which has acceleration of $3\,\mathrm{m}/\mathrm{s}^{2}$ . Use $g=10\,\mathrm{m}/\mathrm{s}^{2}$ . What is the tension $T_{AB}$ in the string AB in Newtons?

Filename:twodisks-ang-mom

Figure 14.58:

14.2.6 At the instant shown, the mass is moving to the right at speed $v=3\,\mathrm{m}/\mathrm{s}$ . Find the rate of work done on the mass.

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Figure 14.59:

14.2.7 A point mass ‘ $m$ ’ is pulled straight up by two strings. The two strings pull the mass symmetrically about the vertical axis with constant and equal force $T$ . At an instant in time $t$ , the position and the velocity of the mass are $y(t)\hat{\boldsymbol{\jmath}}$ and $\dot{y}(t)\hat{\boldsymbol{\jmath}}$ , respectively. Find the power input to the moving mass.

Filename:pfigure-blue-110-1

Figure 14.60:

More-Involved Problems

14.2.8 Two blocks, each of mass $m$ , are connected by a rod of length $S$ . They slide down a slope of angle $\theta$ . Do not neglect gravity but do neglect friction.

(a)

Draw separate free-body diagrams of each block, the rod, and the system of the two blocks and rod.
(b)

Write separate equations for linear momentum balance for each block, the rod, and the system of blocks and rod.
(c)

What is the acceleration of the center of mass of the two blocks? Answer: $a_{1}=a_{2}=g\sin\theta$
(d)

What is the force in the rod? Answer: $T=0$
(e)

What is the speed of the center of mass for the two blocks after they have traveled a distance $d$ down the slope, having started from rest. [Hint: Dot your momentum balance equations with a unit vector along the ramp in order to reduce this problem to a problem in one dimensional mechanics.] Answer: $v=\sqrt{2dg\sin\theta}$

Filename:pfigure-blue-107-1

Figure 14.61:

14.2.9 Two blocks, each of mass $m$ , are connected by a massless rod of length $S$ ; the blocks’ dimensions are small compared to $S$ . They slide down a slope of angle $\theta$ . The coefficient of friction of the top block is $\mu$ and of the bottom block is $\mu/2$ .

(a)

Draw separate free-body diagrams of each block, the string, and the system of the two blocks and rod.
(b)

Write separate equations for linear momentum balance for each block, the string, and the system of blocks and rod.
(c)

What is the acceleration of the center of mass of the two blocks? Answer: $a_{com}=a_{1}=a_{2}=g(sin(\theta)-\frac{3}{4}\mu cos(\theta))$
(d)

What is the force in the rod? Answer: $T=\frac{\mu}{4}mgcos(\theta)$
(e)

What is the speed of the center of mass for the two blocks after they have traveled a distance $d$ down the slope, having started from rest. Answer: $v=\sqrt{2dg(sin(\theta)-\frac{3}{4}\mu cos(\theta))}$
(f)

How would your solutions to parts (a) and (c) differ if the two blocks were interchanged with the slippery one on top? Answer: The upper block would push the lower one down the ramp so the rod tension would be rod compression. But the acceleration would be unchanged.

Filename:pfigure-s94h14p4

Figure 14.62:

14.2.10 Coin on a car on a ramp. A student engineering design course asked students to build a cart (mass $=m_{c}$ ) that rolls down a ramp with angle $\theta$ . A small weight (mass $m_{w}\ll m_{c}$ ) is placed on top of the cart on a surface tipped with respect to the cart (angle $\phi$ ). Assume the small mass does not slide. Assume massless wheels with frictionless bearings. $\hat{\boldsymbol{\imath}}$ is horizontal and $\hat{\boldsymbol{\jmath}}$ is vertical up.

(a)

Find the acceleration of the cart. Answer in terms of some or all of $m_{c},g,\hat{\boldsymbol{\imath}},\theta$ and $\hat{\boldsymbol{\jmath}}$ .
(b)

What coefficient of friction $\mu$ is required (the smallest that will work) to keep the small mass from sliding as the cart rolls down the slope? Answer in terms of some or all of $m_{c},m_{w},g,\theta,$ and $\phi$ .
(c)

What angle $\phi$ will allow a small mass to ride on the cart with the smallest coefficient of friction? Answer in terms of some or all of $m_{c},m_{w},g,$ and $\theta$ .

14.2.11 Guyed plate on a cart A uniform rectangular plate $A B C D$ of mass $m$ is supported by a rod $D E$ and a hinge joint at point $B$ . The dimensions are as shown. There is gravity. What must the acceleration of the cart be in order for massless rod $D E$ to be in tension? Answer: $a_{x}>\frac{3}{2}g$

Filename:pfigure-blue-112-1

Figure 14.63:

Uniform plate supported by a hinge and a cable on an accelerating cart.

14.2.12 A uniform rectangular plate of mass $m$ is supported by two inextensible cables $A B$ and $C D$ and by a hinge at point $E$ on the cart as shown. The cart has acceleration $a_{x}\hat{\boldsymbol{\imath}}$ due to a force not shown. There is gravity.

(a)

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

Write the equation of linear momentum balance for the plate and evaluate the left hand side as explicitly as possible.
(c)

Write the equation for angular momentum balance about point $E$ and evaluate the left hand side as explicitly as possible.

Filename:tfigure8-alt-app2c

Figure 14.64:

14.2.13 A uniform rectangular plate of mass $m$ is supported by an inextensible cable $C D$ and a hinge joint at point $E$ on the cart as shown. The hinge joint is attached to a rigid column welded to the floor of the cart. The cart is at rest. There is gravity. Find the tension in cable $C D$ .

Filename:sfig4-6-4a

Figure 14.65:

14.2.14 A uniform rectangular plate of mass $m$ is supported by an inextensible cable $A B$ and a hinge joint at point $E$ on the cart as shown. The hinge joint is attached to a rigid column welded to the floor of the cart. The cart has acceleration $a_{x}\hat{\boldsymbol{\imath}}$ . There is gravity. Find the tension in cable $A B$ . (What’s ‘wrong’ with this problem? What if instead point $B$ were at the bottom left hand corner of the plate?) Answer: Can’t solve for $T_{AB}$ .

Filename:sfig4-6-3

Figure 14.66:

14.2.15 A block of mass $m$ is sitting on a frictionless surface and acted upon at point $E$ by the horizontal force $P$ through the center of mass. Draw a free-body diagram of the block. There is gravity. Find a) the acceleration of the block and b) reactions on the block at points $A$ and $B$ . Answer: a) ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}=\frac{P}{m}\hat{\boldsymbol{% \imath}}$ with $\hat{\boldsymbol{\imath}}$ to the right. b) $R_{A}=R_{B}=\frac{mg}{2}\hat{\boldsymbol{\jmath}}$ , with $\hat{\boldsymbol{\jmath}}$ upwards

Filename:sfig4-6-3a

Figure 14.67:

14.2.16 Reconsider the block in problem 14.66. This time, find the acceleration of the block and the reactions at $A$ and $B$ if the force $P$ is applied instead at point $D$ . Are the acceleration and the reactions on the block different from those found when $P$ is applied at point $E$ ?

Answer: The acceleration is still ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{G}=\frac{P}{m}\hat{\boldsymbol{% \imath}}$ . But the reactions are changed to $R_{A}=\frac{mg}{2}-\frac{d}{b}\frac{P}{2}$ and $R_{B}=\frac{mg}{2}+\frac{d}{b}\frac{P}{2}$

14.2.17 A block of mass $m$ is sitting on a frictional surface and acted upon at point $D$ by the horizontal force $P$ . The block is resting on a sharp edge at point $B$ and is supported by an ideal wheel at point $A$ . There is gravity. Assuming the block is sliding with coefficient of friction $\mu$ at point $B$ , find the acceleration of the block and the reactions on the block at points $A$ and $B$ .

Filename:pfigure4-4-rp12

Figure 14.68:

14.2.18 A force $F_{C}$ is applied to the corner $C$ of a box of weight $W$ with dimensions and center of gravity at $G$ as shown in the figure. The coefficient of sliding friction between the floor and the points of contact $A$ and $B$ is $\mu$ . Assuming that the box slides when $F_{C}$ is applied, find the acceleration of the box and the reactions at $A$ and $B$ in terms of $W$ , $F_{C}$ , $\theta$ , $b$ , and $d$ .

Filename:pfigure-blue-38-2

Figure 14.69:

14.2.19 A uniform rod with mass $m_{r}$ rests on a cart (mass $m_{c}$ ) which is being pulled to the right. The rod is hinged at one end (with a frictionless hinge) and has no friction at the contact with the cart. The cart is rolling on wheels that are modeled as having no mass and no bearing friction (ideal massless wheels). Answer in terms of $g$ , $m_{r}$ , $m_{c}$ , $\theta$ and $F$ . Find:

(a)

The force on the rod from the cart at point B.
(b)

The force on the rod from the cart at point A.

Filename:pfigure-blue-49-2

Figure 14.70:

14.2.20 The box shown in the figure is dragged in the $x$ -direction with a constant acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=0.5\,\mathrm{m}/\mathrm{s}^{2}% \hat{\boldsymbol{\imath}}$ . At the instant shown, the velocity of (every point on) the box is ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=0.8\,\mathrm{m}/\mathrm{s}\hat{% \boldsymbol{\imath}}$ .

(a)

Find the linear momentum of the box.
(b)

Find the rate of change of linear momentum of the box.
(c)

Find the angular momentum of the box about the contact point $O$ .
(d)

Find the rate of change of angular momentum of the box about the contact point $O$ .

Filename:pfigure4-4-rp13

Figure 14.71:

14.2.21 The groove and disk accelerate upwards, ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=a\hat{\boldsymbol{\jmath}}$ . Neglecting gravity, what are the forces on the disk due to the groove?

Filename:f92h7p1

Figure 14.72:

14.2.22 The following problems concern a box that is in the back of a pickup truck. The pickup truck is moving forward with acceleration of $a_{t}$ . The truck’s speed is $v_{t}$ . The box has sharp feet at the front and back ends so the only place it contacts the truck is at the feet. The center of mass of the box is at the geometric center of the box. The box has height $h$ , length $\ell$ and depth $w$ (into the paper.) Its mass is $m$ . There is gravity. The friction coefficient between the truck and the box edges is $\mu$ .

In the problems below you should express your solutions in terms of the variables given in the figure, $\ell$ , $h$ , $\mu$ , $m$ , $g$ , $a_{t}$ , and $v_{t}$ . If any variables do not enter the expressions comment on why they do not.

In all cases you may assume that the box does not rotate (though it might be on the verge of doing so).

(a)

Assuming the box does not slide, what is the total force that the truck exerts on the box (i.e. the sum of the reactions at A and B)?
(b)

Assuming the box does not slide what are the reactions at A and B? [Note: You cannot find both of them without additional assumptions.]
(c)

Assuming the box does slide, what is the total force that the truck exerts on the box?
(d)

Assuming the box does slide, what are the reactions at A and B?
(e)

Assuming the box does not slide, what is the maximum acceleration of the truck for which the box will not tip over (hint: just at that critical acceleration what is the vertical reaction at B?)?
(f)

What is the maximum acceleration of the truck for which the block will not slide?
(g)

The truck hits a brick wall and stops instantly. Does the block tip over?

Assuming the block does not tip over, how far does it slide on the truck before stopping (assume the bed of the truck is sufficiently long)?

Filename:pfigure-blue-52-2

Figure 14.73:

14.2.23 A collection of uniform boxes with various heights $h$ and widths $w$ and masses $m$ sit on a horizontal conveyer belt. The acceleration $a(t)$ of the conveyer belt gets extremely large sometimes due to an erratic over-powered motor. Assume the boxes touch the belt at their left and right edges only and that the coefficient of friction there is $\mu$ . It is observed that some boxes never tip over. What is true about $\mu$ , $g$ , $w$ , $h$ , and $m$ for the boxes that always maintain contact at both the right and left bottom edges? (Write an inequality that involves some or all of these variables.)

Filename:pfigure4-4-rp16

Figure 14.74:

14.2.24 After failure of her normal brakes, a driver pulls the emergency brake of her old car. This action locks the rear wheels (friction coefficient $=\mu$ ) but leaves the well lubricated and light front wheels spinning freely. The car, braking inadequately as is the case for rear wheel braking, hits a stiff and slippery phone pole which compresses the car bumper. The car bumper is modeled here as a linear spring (constant $=k$ , rest length $=l_{0}$ , present length $=l_{s}$ ). The car is still traveling forward at the instant of interest. The bumper is at a height $h_{b}$ above the ground. Assume that the car, excepting the bumper, is a non-rotating rigid body and that the wheels remain on the ground (that is, the bumper is compliant but the suspension is stiff).

•

What is the acceleration of the car in terms of $g$ , $m$ , $\mu$ , $l_{f}$ , $l_{r}$ , $k$ , $h_{b}$ , $h_{cm}$ , $l_{0}$ , and $l_{s}$ (and any other parameters if needed)?

Filename:pfigure4-4-rp14

Figure 14.75:

14.2.25

Car braking: front brakes versus rear brakes versus all four brakes. What is the peak deceleration of a car when you apply: the front brakes till they skid, the rear brakes till they skid, and all four brakes till they skid? Assume that the coefficient of friction between rubber and road is $\mu=1$ (about right, the coefficient of friction between rubber and road varies between about $.7$ and $1.3$ ) and that $g=10\,\mathrm{m}/\mathrm{s}^{2}$ (2% error). Pick the dimensions and mass of the car, but assume the center of mass height $h$ is greater than zero but is less than half the wheel base $w$ , the distance between the front and rear wheel. Also assume that the $C M$ is halfway between the front and back wheels (i.e., $l_{f}=l_{r}=w/2$ ). The car has a stiff suspension so the car does not move up or down or tip appreciably during braking. Neglect the mass of the rotating wheels in the linear and angular momentum balance equations. Treat this problem as two-dimensional problem; i.e., the car is symmetric left to right, does not turn left or right, and that the left and right wheels carry the same loads. To organize your work, here are some steps to follow.

(a)

Draw a FBD of the car assuming rear wheel is skidding. The FBD should show the dimensions, the gravity force, what you know a priori about the forces on the wheels from the ground (i.e., that the friction force $F_{r}=\mu N_{r}$ , and that there is no friction at the front wheels), and the coordinate directions. Label points of interest that you will use in your momentum balance equations. (Hint: also draw a free-body diagram of the rear wheel.)
(b)

Write the equation of linear momentum balance.
(c)
Write the equation of angular momentum balance relative to a point of your choosing. Some particularly useful points to use are:
- •
  
  the point above the front wheel and at the height of the center of mass;
- •
  
  the point at the height of the center of mass, behind the rear wheel that makes a $45$ degree angle line down to the rear wheel ground contact point; and
- •
  
  the point on the ground straight under the front wheel that is as far below ground as the wheel base is long.
(d)

Solve the momentum balance equations for the wheel contact forces and the deceleration of the car. If you have used any or all of the recommendations from part (c) you will have the pleasure of only solving one equation in one unknown at a time. Answer: Normal reaction at rear wheel: $N_{r}=\frac{mgw}{2(h\mu+w)}$ , normal reaction at front wheel: $N_{f}=mg-\frac{mgw}{2(h\mu+w)}$ , deceleration of car: $a_{car}=-\frac{\mu gw}{2(h\mu+w)}$ .
(e)

Repeat steps (a) to (d) for front-wheel skidding. Note that the advantageous points to use for angular momentum balance are now different. Does a car stop faster or slower or the same by skidding the front instead of the rear wheels? Would your solution to (e) be different if the center of mass of the car were at ground level( $h$ =0)? Answer: Normal reaction at rear wheel: $N_{r}=mg-\frac{mgw}{2(w-\mu h)}$ , normal reaction at front wheel: $N_{f}=\frac{mgw}{2(w-\mu h)}$ , deceleration of car: $a_{car}=-\frac{\mu gw}{2(w-\mu h)}$ . Car stops more quickly for front wheel skidding. Car stops at same rate for front or rear wheel skidding if $h=0$ .
(f)

Repeat steps (a) to (d) for all-wheel skidding. There are some shortcuts here. You determine the car deceleration without ever knowing the wheel reactions (or using angular momentum balance) if you look at the linear momentum balance equations carefully. Answer: Normal reaction at rear wheel: $N_{r}=\frac{mg(w/2-\mu h)}{w}$ , normal reaction at front wheel: $N_{f}=\frac{mg(w/2+\mu h)}{w}$ , deceleration of car: $a_{car}=-\mu g$ .
(g)

Does the deceleration in (f) equal the sum of the decelerations in (d) and (e)? Why or why not? Answer: No. Simple superposition just doesn’t work.
(h)

What peculiarity occurs in the solution for front-wheel skidding if the wheel base is twice the height of the CM above ground and $\mu=1$ ? Answer: No reaction at rear wheel.
(i)

What impossibility does the solution predict if the wheel base is shorter than twice the CM height? What wrong assumption gives rise to this impossibility? What would really happen if one tried to skid a car this way? Answer: Reaction at rear wheel is negative. Not allowing for rotation of the car in the $x y$ -plane gives rise to this impossibility. In actuality, the rear of the car would flip over the front.

Filename:pfigure4-4-rp15

Figure 14.76:

14.2.26 Assuming massless wheels, an infinitely powerful engine, a stiff suspension (i.e., no rotation of the car) and a coefficient of friction $\mu$ between tires and road,

(a)

what is the maximum forward acceleration of this front wheel drive car? Answer: Hint: the answer reduces to $a=\ell_{r}g/h$ in the limit $\mu\rightarrow\infty$ .]
(b)

what is the force of the ground on the rear wheels during this acceleration?
(c)

what is the force of the ground on the front wheels?

Filename:tfigure4-spherical-rotaxis

Figure 14.77:

14.2.27 At time $t=0$ , the block of mass $m$ is released from rest on the slope of angle $\phi$ . The coefficient of friction between the block and slope is $\mu$ .

(a)

What is the acceleration of the block for $\mu>0$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=g(\sin\phi-\mu\cos\phi)\hat{% \boldsymbol{\imath}}$ , where $\hat{\boldsymbol{\imath}}$ is parallel to the slope and pointing downwards
(b)

What is the acceleration of the block for $\mu=0$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=g\sin\phi$
(c)

Find the position and velocity of the block as a function of time for $\mu>0$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=g(\sin\phi-\mu\cos\phi)t\hat{% \boldsymbol{\imath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=g(\sin\phi-\mu\cos\phi)\frac{t^{% 2}}{2}$
(d)

Find the position and velocity of the block as a function of time for $\mu=0$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=g\sin\phi t\hat{\boldsymbol{% \imath}}$ , ${\accentset{\rightharpoonup}{\boldsymbol{r}}}=g\sin\phi\frac{t^{2}}{2}\hat{% \boldsymbol{\imath}}$

Filename:tfigure5-7

Figure 14.78:

14.2.28 A small block of mass $m_{1}$ is released from rest at altitude $h$ on a frictionless slope of angle $\alpha$ . At the instant of release, another small block of mass $m_{2}$ is dropped vertically from rest at the same altitude. The second block does not interact with the ramp. What is the velocity of the first block relative to the second block after $t$ seconds have passed?

Filename:tfigure5-gen-rigid-body

Figure 14.79:

14.2.29 Block sliding on a ramp with friction. A square box is sliding down a ramp of angle $\theta$ with instantaneous velocity $v\hat{\boldsymbol{\imath}}^{\prime}$ . Assume it does not tip over.

(a)

What is the force on the block from the ramp at point $A$ ? Answer in terms of any or all of $\theta$ , $\ell$ , $m$ , $g$ , $\mu$ , $v$ , $\hat{\boldsymbol{\imath}}^{\prime}$ , and $\hat{\boldsymbol{\jmath}}^{\prime}$ . As a check, your answer should reduce to ${mg\over 2}\hat{\boldsymbol{\jmath}}^{\prime}$ when $\theta=\mu=0$ . Answer: ${\accentset{\rightharpoonup}{\boldsymbol{R}}}_{A}=\frac{(1-\mu)mg\cos\theta}{2% }(\hat{\boldsymbol{\jmath}}^{\prime}-\mu\hat{\boldsymbol{\imath}}^{\prime})$ .
(b)

In addition to solving the problem by hand, see if you can write a set of computer commands that, if $\theta$ , $\mu$ , $\ell$ , $m$ , $v$ and $g$ were specified, would give the correct answer.
(c)

Assuming $\theta=80^{\circ}$ and $\mu=0.9$ , can the box slide this way or would it tip over? Why? Answer: No tipping if $N_{A}=\frac{(1-\mu)mg\cos\theta}{2}>0$ ; i.e., no tipping if $\mu<1$ since $\cos\theta>0$ for $0<\theta<\frac{\pi}{2}$ .(Here $\mu=0.9$ )

Filename:tfigure5-term1-a

Figure 14.80:

14.2.30 A coin is given a sliding shove up a ramp with angle $\phi$ with the horizontal. It takes twice as long to slide down as it does to slide up. What is the coefficient of friction $\mu$ between the coin and the ramp. Answer in terms of some or all of $m,g,\phi$ and the initial sliding velocity $v$ .

14.2.31 A skidding car. What is the braking acceleration of the front-wheel braked car as it slides down hill. Express your answer as a function of any or all of the following variables: the slope $\theta$ of the hill, the mass of the car $m$ , the wheel base $\ell$ , and the gravitational constant $g$ . Use $\mu=1$ . Answer: braking acceleration $=g(\frac{1}{2}\cos\theta-\sin\theta)$ .

Filename:tfigure5-term1-b

Figure 14.81:

A car skidding downhill on a slope of angle $\theta$

14.2.32 Two blocks A and B are pushed up a frictionless inclined plane by an external force $F$ as shown in the figure. The coefficient of friction between the two blocks is $\mu=0.2$ . The masses of the two blocks are $m_{A}=5\,\mathrm{kg}$ and $m_{B}=2\,\mathrm{kg}$ . Find the magnitude of the maximum allowable force such that no relative slip occurs between the two blocks.

Filename:sfig1-2-12

Figure 14.82:

14.2.33 A bead slides on a frictionless rod. The spring has constant $k$ and rest length $\ell_{0}$ . The bead has mass $m$ .

(a)

Given $x$ and $\dot{x}$ find the acceleration of the bead (in terms of some or all of $D,\ell_{0},x,\dot{x},m,k$ and any base vectors that you define).
(b)

If the bead is allowed to move, as constrained by the slippery rod and the spring, find a differential equation that must be satisfied by the variable $x$ . (Do not try to solve this somewhat ugly non-linear equation.)
(c)

In the special case that $\ell_{0}=0^{*}$ find how long it takes for the block to return to its starting position after release with no initial velocity at $x=x_{0}$ .

Filename:sfig4-6-5

Figure 14.83:

14.2.34 A bead oscillates on a straight frictionless wire. The spring obeys the equation $F=k$ ( $\ell-\ell_{o})$ , where $\ell$ = length of the spring and $\ell_{0}$ is the ’rest’ length. Assume

x(t=0)=x_{0},\ \dot{x}(t=0)=0.

(a)

Write a differential equation satisfied by $x(t)$ .
(b)

What is ${\dot{x}}$ when $x=0$ ? [hint: Don’t try to solve the equation in (a)!]
(c)

What is the simplification in (a) if $\ell_{o}=0$ (spring is then a so-called “zero–length” spring).
(d)

For this special case ( $\ell_{o}=0$ ) solve the equation in (a) and show the result agrees with (b) in this special case.

Filename:sfig4-6-5a

Figure 14.84:

14.2.35 A cart on an elastic leash. A cart $B$ (mass $m$ ) rolls on a frictionless level floor. One end of an inextensible string is attached to the cart. The string wraps around a pulley at point $A$ and the other end is attached to a spring with constant $k$ . When the cart is at point $O$ , it is in static equilibrium. The spring relaxed length, rope length, and room height $h$ are such that the spring would be relaxed if the end of rope at $B$ were disconnected from the cart and brought up to point $A$ . The gravitational constant is $g$ . The cart is pulled a horizontal distance $d$ from the center of the room (at $O$ ) and released.

(a)

Assuming that the cart never leaves the floor, what is the speed of the cart when it passes through the center of the room, in terms of $m$ , $h$ , $g$ and $d$ . Answer: $v=d\sqrt{\frac{k}{m}}$ .
(b)

Does the cart undergo simple harmonic motion for small or large oscillations (specify which if either)? (Simple harmonic motion occurs when position varies sinusoidally with time.) Answer: The cart undergoes simple harmonic motion for any size oscillation.

Filename:sfig4-6-5b

Figure 14.85:

14.2.36 The cart moves to the right with constant acceleration $a$ . The ball has mass $m$ . The spring has unstretched length $\ell_{0}$ and spring constant $k$ . Assuming the ball is stationary with respect to the cart find the distance from $O$ to $A$ in terms of $k$ , $\ell_{0}$ , and $a$ . [Hint: find $\theta$ first.]

Filename:sfig4-6-5c

Figure 14.86:

14.2.37 Consider a person, modeled as a rigid body, riding an accelerating motorcycle (2-D). The person is sitting on the seat and cannot slide fore or aft, but is free to rock in the plane of the motorcycle (as if there were a hinge connecting the motorcycle to the rider at the seat). The person’s feet are off the pegs and the legs are sticking down and not touching anything. The person’s arms are like cables (they are massless and only carry tension). Assume all dimensions and masses are known (you have to define them carefully with a sketch and words). Assume the forward acceleration of the motorcycle is known. You may use numbers and/or variables to describe the quantities of interest.

(a)

Draw a clear sketch of the problem showing needed dimensional information and the coordinate system you will use.
(b)

Draw a Free-Body Diagram of the rider.
(c)

Write the equations of linear and angular momentum balance for the rider.
(d)

Find all forces on the rider from the motorcycle (i.e., at the hands and the seat).
(e)

What are the forces on the motorcycle from the rider?

14.2.38 Acceleration of a bicycle on level ground. 2-D . A very compact bicycler (modeled as a point mass $M$ at the bicycle seat $C$ with height $h$ , and distance $b$ behind the front wheel contact), rides a very light old-fashioned bicycle (all components have negligible mass) that is well maintained (all bearings have no frictional torque) and streamlined (neglect air resistance). The rider applies a force $F_{p}$ to the pedal perpendicular to the pedal crank (with length $L_{c}$ ). No force is applied to the other pedal. The radius of the front wheel is $R_{f}$ .

(a)

Assuming no slip, what is the forward acceleration of the bicycle? [ Hint: draw a FBD of the front wheel and crank, and another FBD of the whole bicycle-rider system.] Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{bike}=\frac{F_{p}L_{c}}{MR_{f}}$ .
(b)

(Harder) Assuming the rider can push arbitrarily hard but that $\mu=1$ , what is the maximum possible forward acceleration of the bicycle. Answer: max( ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{bike})$ = $\frac{ga}{a+b+2R_{f}}$ .

Filename:sfig4-6-5d

Figure 14.87:

14.2.39 A $320\,{\mathrm{lbm}}$ mass is attached at the corner $C$ of a light rigid piece of pipe bent as shown. The pipe is supported by ball-and-socket joints at $A$ and $D$ and by cable $E F$ . The points $A$ , $D$ , and $E$ are fastened to the floor and vertical sidewall of a pick-up truck which is accelerating in the $z$ -direction. The acceleration of the truck is ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=5\,{\mathrm{ft}}/{\mathrm{s}}^{2% }\hat{\boldsymbol{k}}$ . There is gravity. Find the tension in cable $E F$ . Answer: $T_{EF}=640\sqrt{2}\,{\rm lbf}$ .

Filename:sfig4-5-6

Figure 14.88:

14.2.40 A $5\,\mathrm{ft}$ by $8\,\mathrm{ft}$ rectangular plate of uniform density has mass $m=10\,{\mathrm{lbm}}$ and is supported by a ball-and-socket joint at point $A$ and the light rods $C E$ , $B D$ , and $G H$ . The entire system is attached to a truck which is moving with acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{T}$ . The plate is moving without rotation or angular acceleration relative to the truck. Thus, the center of mass acceleration of the plate is the same as the truck’s. Dimensions are as shown. Points $A$ , $C$ , and $D$ are fixed to the truck but the truck is not touching the plate at any other points. Find the tension in rod $B D$ .

(a)

If the truck’s acceleration is $\boldsymbol{a}_{cm}=(5\,{\mathrm{ft}}/{\mathrm{s}}^{2})\boldsymbol{k}$ , what is the tension or compression in rod $B D$ ? Answer: $T_{BD}=92.6\,{\mathrm{lbm}}\cdot\,{\mathrm{ft}}/{\mathrm{s}}^{2}$ .
(b)

If the truck’s acceleration is ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{\rm cm}=(5\,{\mathrm{ft}}/{% \mathrm{s}}^{2})\hat{\boldsymbol{\jmath}}+(6\,{\mathrm{ft}}/{\mathrm{s}}^{2})% \hat{\boldsymbol{k}}$ , what is the tension or compression in rod $G H$ ? Answer: $T_{GH}=5\sqrt{61}\,{\mathrm{lbm}}\cdot\,{\mathrm{ft}}/{\mathrm{s}}^{2}$ .

Filename:sfig4-5-6a

Figure 14.89:

14.2.41 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 at the other three corners with strings. At the instant of interest the shelf is in a rocket in outer space and accelerating at $10\,\mathrm{m}/\mathrm{s}^{2}$ in the $\boldsymbol{k}$ direction. The shelf is in the $x y$ plane.

(a)

Draw a FBD of the shelf.
(b)

Challenge: without doing any calculations on paper can you find one of the reaction force components or the tension in any of the cables? Give yourself a few minutes of staring to try this approach. If you can’t, then come back to this question after you have done all the calculations. Answer: $T_{EH}=0$
(c)

Write down the linear momentum balance equation (a vector equation). Answer: $(R_{C_{x}}-T_{AB})\hat{\boldsymbol{\imath}}+(R_{C_{y}}-\frac{T_{GD}}{\sqrt{2}}% )\hat{\boldsymbol{\jmath}}+(T_{HE}+R_{C_{z}}+\frac{T_{GD}}{\sqrt{2}})\hat{% \boldsymbol{k}}=m{\accentset{\rightharpoonup}{\boldsymbol{a}}}=10\,{\mbox{N}}% \hat{\boldsymbol{k}}$ .
(d)

Write down the angular momentum balance equation using the center of mass as a reference point. Answer: $\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathrm{cm}}=(\frac{T_{GD}}% {\sqrt{2}}-T_{HE}-R_{C_{z}})\hat{\boldsymbol{\imath}}+(R_{C_{z}}-\frac{T_{GD}}% {\sqrt{2}}-T_{HE})\hat{\boldsymbol{\jmath}}+(T_{AB}+R_{C_{x}}-R_{C_{y}}-\frac{% T_{GD}}{\sqrt{2}})\hat{\boldsymbol{k}}={\accentset{\rightharpoonup}{% \boldsymbol{0}}}$
(e)

By taking components, turn (b) and (c) into six scalar equations in six unknowns. Answer:

$\displaystyle R_{C_{x}}-T_{AB}=0$

$\displaystyle R_{C_{y}}-\frac{T_{GD}}{\sqrt{2}}=0$

$\displaystyle R_{C_{z}}+\frac{T_{GD}}{\sqrt{2}}+T_{EH}=5\,{\mbox{N}}$

$\displaystyle-T_{EH}+\frac{T_{GD}}{\sqrt{2}}-R_{C_{z}}=0$

$\displaystyle-T_{EH}-\frac{T_{GD}}{\sqrt{2}}+R_{C_{z}}=0$

$\displaystyle T_{AB}-\frac{T_{GD}}{\sqrt{2}}+R_{C_{x}}-R_{C_{y}}=0$
(f)

Solve these equations by hand or on the computer. Answer: $R_{C_{x}}=5\,{\mbox{N}}$ , $R_{C_{y}}=5\,{\mbox{N}}$ , $R_{C_{z}}=5\,{\mbox{N}}$ , $T_{GD}=\frac{10}{\sqrt{2}}\,{\mbox{N}}$ , $T_{EH}=0\,{\mbox{N}}$ , $T_{AB}=5\,{\mbox{N}}$ .
(g)

Instead of using a system of equations try to find a single equation which can be solved for $T_{EH}$ . Solve it and compare to your result from before. Answer: Find moment about $C D$ axis; e.g., $(\sum{\accentset{\rightharpoonup}{\boldsymbol{M}}}_{\mathchoice{\lower 1.0pt% \hbox{$\displaystyle\mathrm{C}$}}{\lower 1.0pt\hbox{$\textstyle\mathrm{C}$}}{% \lower 1.0pt\hbox{$\scriptstyle\mathrm{C}$}}{\lower 1.0pt\hbox{$% \scriptscriptstyle\mathrm{C}$}}}={\accentset{\rightharpoonup}{\boldsymbol{r}}}% _{cm/C}\times\,\mathrm{m}{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{cm})% \cdot\hat{\boldsymbol{\lambda}}_{CD}$ , where $\hat{\boldsymbol{\lambda}}_{CD}$ is a unit vector in the direction of axis $C D$ .
(h)

Challenge: For how many of the reactions can you find one equation which will tell you that particular reaction without knowing any of the other reactions? [Hint, try angular momentum balance about various axes as well as linear momentum balance in an appropriate direction. It is possible to find five of the six unknown reaction components this way.] Must these solutions agree with (d)? Do they?

Filename:sfig4-6-8

Figure 14.90:

14.2.42 A uniform rectangular plate of mass $m$ is supported by an inextensible cable $C D$ and a hinge joint at point $E$ on the cart as shown. The hinge joint is attached to a rigid column welded to the floor of the cart. The cart has acceleration $a_{x}\hat{\boldsymbol{\imath}}$ . There is gravity. Find the tension in cable $C D$ .

Filename:sfig4-6-8a

Figure 14.91:

14.2.43 The uniform $2\,\mathrm{kg}$ plate DBFH is held by six massless rods (AF, CB, CF, GH, ED, and EH) which are hinged at their ends. The support points A, C, G, and E are all accelerating in the $x$ -direction with acceleration $\boldsymbol{a}=3\,\mathrm{m}/\mathrm{s}^{2}\boldsymbol{\imath}$ . There is no gravity.

(a)

What is $\{\sum{\accentset{\rightharpoonup}{\boldsymbol{F}}}\}\cdot\hat{\boldsymbol{% \imath}}$ for the forces acting on the plate?
(b)

What is the tension in bar CB?

Filename:sfig4-6-8b

Figure 14.92:

14.2.44 A massless triangular plate rests against a frictionless wall of a pick-up truck at point $D$ and is rigidly attached to a massless rod supported by two ideal bearings fixed to the floor of the pick-up truck. A ball of mass $m$ is fixed to the centroid of the plate. There is gravity. The pick-up truck skids across a road with acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=a_{x}\hat{\boldsymbol{\imath}}+a% _{z}\hat{\boldsymbol{k}}$ . What is the reaction at point $D$ on the plate?

Filename:sfig5-5-2

Figure 14.93:

14.2.45 Towing a bicycle. A bicycle on the level $x y$ plane is steered straight ahead and is being towed by a rope. The bicycle and rider are modeled as a uniform plate with mass $m$ (for the convenience of the artist). The tow force $F$ applied at C has no $z$ component and makes an angle $\alpha$ with the $x$ axis. The rolling wheel contacts are at A and B. The bike is tipped an angle $\phi$ from the vertical. The towing force $F$ is the magnitude needed to keep the bike accelerating in a straight line (along the $y$ axis) without tipping any more or less than the angle $\phi$ . What is the acceleration of the bicycle? Answer in terms of some or all of $b,h,\alpha,\phi,m,g$ and $\hat{\boldsymbol{\jmath}}$ (Note: $F$ should not appear in your final answer.)

Filename:sfig5-5-2a

Figure 14.94:

14.2.46 An airplane is in straight level flight but is accelerating in the forward direction. In terms of some or all of the following parameters,

•

$m_{tot}\equiv$ the total mass of the plane (including the wings),
•

$D$ = the drag force on the fuselage,
•

$F_{D}$ = the drag force on each wing,
•

$g$ = gravitational constant, and,
•

$T=$ the thrust of one engine.

(a)

What is the lift on each wing $F_{L}$ ? Answer: $F_{L}=\frac{1}{2}{m_{\mathrm{tot}}}g$ .
(b)

What is the acceleration of the plane ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}$ ? Answer: ${\accentset{\rightharpoonup}{\boldsymbol{a}}}_{P}=\frac{1}{m_{\mathrm{tot}}}% \left[2(T-F_{D})-D\right]\hat{\boldsymbol{\imath}}$ .
(c)

A free-body diagram of one wing is shown. The mass of one wing is $m_{w}$ . What, in terms of $m_{tot}$ , $m_{w}$ , $F_{L}$ , $F_{D}$ , $g$ , $a$ , $b$ , $c$ , and $\ell$ are the reactions at the base of the wing (where it is attached to the plane), ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=F_{x}\hat{\boldsymbol{\imath}}+F% _{y}\hat{\boldsymbol{\jmath}}+F_{z}\hat{\boldsymbol{k}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}=M_{x}\hat{\boldsymbol{\imath}}+M% _{y}\hat{\boldsymbol{\jmath}}+M_{z}\hat{\boldsymbol{k}}$ ?

Answer: ${\accentset{\rightharpoonup}{\boldsymbol{F}}}=\left[\frac{m_{w}}{m_{\mathrm{% tot}}}(2T-D-2F_{D})-T+F_{D}\right]\hat{\boldsymbol{\imath}}+(m_{w}g-F_{L})\hat% {\boldsymbol{\jmath}}$ and ${\accentset{\rightharpoonup}{\boldsymbol{M}}}=(bF_{L}-am_{w}g)\hat{\boldsymbol% {\imath}}+\left[(bF_{D}-cT)+a\frac{m_{w}}{m_{\mathrm{tot}}}(2T-D-2F_{D})\right% ]\hat{\boldsymbol{\jmath}}$ .

Filename:sfig5-5-2b

Figure 14.95:

14.2.47 A rear-wheel drive car on level ground. The two left wheels are on perfectly slippery ice. The right wheels are on dry pavement. The negligible-mass front right wheel at $B$ is steered straight ahead and rolls without slip. The right rear wheel at $C$ also rolls without slip and drives the car forward with velocity ${\accentset{\rightharpoonup}{\boldsymbol{v}}}=v\hat{\boldsymbol{\jmath}}$ and acceleration ${\accentset{\rightharpoonup}{\boldsymbol{a}}}=a\hat{\boldsymbol{\jmath}}$ . Dimensions are as shown and the car has mass $m$ . What is the sideways force from the ground on the right front wheel at $B$ ? Answer in terms of any or all of $m$ , $g$ , $a$ , $b$ , $\ell$ , $w$ , and $\hat{\boldsymbol{\imath}}$ . Answer: sideways force = $F_{B}\hat{\boldsymbol{\imath}}=\frac{wma}{2\ell}\hat{\boldsymbol{\imath}}$ .

Filename:sfig7-4-2

Figure 14.96:

The left wheels of this car are on ice.

14.2.48 A somewhat crippled car slams on the brakes. The suspension springs at A, B, and C are frozen and keep the car level and at constant height. The normal force at D is kept equal to $N_{D}$ by the only working suspension spring which is on the left rear wheel at D. The only brake which is working is that of the right rear wheel at C which slides on the ground with friction coefficient $\mu$ . Wheels A, B, and D roll freely without slip. Dimensions are as shown.

(a)

Find the acceleration of the car in terms of some or all of $m,w,\ell,b,h,g,\mu,\hat{\boldsymbol{\jmath}}$ , and $N_{D}$ .
(b)

From the information given could you also find all of the reaction forces at all of the wheels? If so, why? If not, what can’t you find and why? (No credit for correct answer. Credit depends on clear explanation.)

Filename:pfigure-blue-99-1

Figure 14.97:

Given normal force at D, skidding at C.

14.2.49 Speeding tricycle gets a branch caught in the right rear wheel. A scared-stiff tricyclist riding on level ground gets a branch stuck in the right rear wheel so the wheel skids with friction coefficient $\mu$ . Assume that the center of mass of the tricycle-person system is directly above the rear axle. Assume that the left rear wheel and the front wheel have negligible mass, good bearings, and have sufficient friction that they roll in the $\hat{\boldsymbol{\jmath}}$ direction without slip, thus constraining the overall motion of the tricycle. Dimensions are shown in the lower sketch. Find the acceleration of the tricycle (in terms of some or all of $\ell,\ h,\ b,\ m,\ [I^{cm}],\mu,\ g,\ \hat{\boldsymbol{\imath}},\ \hat{% \boldsymbol{\jmath}},$ and $\hat{\boldsymbol{k}}$ ). [Hint: check your answer against special cases for which you might guess the answer, such as when $\mu=0$ or when $h=0$ .]

Filename:pfigure-blue-47-2

Figure 14.98:

Right rear wheel skids.

14.2.50 A 3-wheeled robot. A 3-wheeled robot with mass $m$ is being transported on a level flatbed trailer also with mass $m$ . The trailer is being pushed with a force $F\hat{\boldsymbol{\jmath}}$ . The ideal massless trailer wheels roll without slip. The ideal massless robot wheels also roll without slip. The robot steering mechanism has turned the wheels so that wheels at A and C are free to roll in the $\hat{\boldsymbol{\jmath}}$ direction and the wheel at B is free to roll in the $\hat{\boldsymbol{\imath}}$ direction. The center of mass of the robot at G is $h$ above the trailer bed and symmetrically above the axle connecting wheels A and B. The wheels A and B are a distance $b$ apart. The length of the robot is $\ell$ .

Find the force vector ${\accentset{\rightharpoonup}{\boldsymbol{F}}}_{A}$ of the trailer on the robot at A in terms of some or all of $m,g,\ell,F,b,h,\hat{\boldsymbol{\imath}},\hat{\boldsymbol{\jmath}},$ and $\hat{\boldsymbol{k}}$ . [Hints: Use a free-body diagram of the cart with robot to find their acceleration. With reference to a free-body diagram of the robot, use angular momentum balance about axis BC to find $F_{Az}$ .]

Filename:pfigure-blue-89-1

Figure 14.99:

	$\displaystyle m\ddot{x}_{1}$	$\displaystyle=$	$\displaystyle T$		(14.9)
	$\displaystyle F-T$	$\displaystyle=$	$\displaystyle m\ddot{x}_{2}$		(14.10)

	$\displaystyle m\ddot{x}+m\ddot{x}$	$\displaystyle=$	$\displaystyle F$
	$\displaystyle\ddot{x}$	$\displaystyle=$	$\displaystyle F/2m$

$\displaystyle(2T-F)\hat{\boldsymbol{\imath}}+(2N_{A}-mg)\hat{\boldsymbol{% \jmath}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}=ma_{A}\hat{% \boldsymbol{\imath}}$
$\displaystyle\mbox{(dotting with $\hat{\boldsymbol{\jmath}}$)}\hbox to36.135pt% {\hfil$\Rightarrow$\hfil}\qquad 2N_{A}$	$\displaystyle=$	$\displaystyle mg$
$\displaystyle\mbox{(dotting with $\hat{\boldsymbol{\imath}}$)}\hbox to36.135pt% {\hfil$\Rightarrow$\hfil}2T-F$	$\displaystyle=$	$\displaystyle ma_{A}$	(14.14)

$\displaystyle-3T\hat{\boldsymbol{\imath}}+(2N_{B}-mg)\hat{\boldsymbol{\jmath}}$	$\displaystyle=$	$\displaystyle m{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}=ma_{B}\hat{% \boldsymbol{\imath}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\qquad 2N_{B}$	$\displaystyle=$	$\displaystyle mg$
$\displaystyle\hbox to36.135pt{and\hfil}-3T$	$\displaystyle=$	$\displaystyle ma_{B}.$	(14.15)

$\displaystyle\ell_{tot}$	$\displaystyle\equiv$	length of the string connecting A and B
	$\displaystyle=$	$\displaystyle 3(x_{B}-X_{C})+2(X_{C}-x_{A})$
	$\displaystyle=$	$\displaystyle 3x_{B}-X_{C}-2x_{A}$

	$\displaystyle\overbrace{\frac{d}{dt}(\ell_{tot})}^{0}\;$	$\displaystyle=$	$\displaystyle 3{\dot{x}}_{B}-0-2{\dot{x}}_{A}$
	$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad{\ddot{x}}_{B}$	$\displaystyle=$	$\displaystyle\frac{2}{3}{\ddot{x}}_{A}$		(14.16)

$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{A}$	$\displaystyle=$	$\displaystyle v_{A}\hat{\boldsymbol{\imath}}={\dot{x}}_{A}\hat{\boldsymbol{% \imath}},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{A}$	$\displaystyle=$	$\displaystyle a_{A}\hat{\boldsymbol{\imath}}={\ddot{x}}_{A}\hat{\boldsymbol{% \imath}},$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{v}}}_{B}$	$\displaystyle=$	$\displaystyle v_{B}\hat{\boldsymbol{\imath}}=\ {\dot{x}}_{B}\hat{\boldsymbol{% \imath}},\mbox{ and}$
$\displaystyle{\accentset{\rightharpoonup}{\boldsymbol{a}}}_{B}$	$\displaystyle=$	$\displaystyle a_{B}\hat{\boldsymbol{\imath}}=\ {\ddot{x}}_{B}\hat{\boldsymbol{% \imath}},$

	$\displaystyle a_{A}$	$\displaystyle=$	$\displaystyle-\frac{9T}{2m}=-\frac{9\cdot 20\,{\mbox{N}}}{2\cdot 40\,\mathrm{% kg}}=-2.25\,\mathrm{m}/\mathrm{s}^{2}$
	$\displaystyle a_{B}$	$\displaystyle=$	$\displaystyle\frac{2}{3}a_{A}=-1.5\,\mathrm{m}/\mathrm{s}^{2}$

$\displaystyle-Fv_{A}-T\overbrace{v_{q}}^{=v_{c}=0}$	$\displaystyle=$	$\displaystyle m_{A}v_{A}a_{A}+m_{B}v_{B}a_{B}$
$\displaystyle\hbox to36.135pt{or\hfil}F$	$\displaystyle=$	$\displaystyle-ma_{A}-m\frac{v_{B}}{v_{A}}a_{B}$
	$\displaystyle=$	$\displaystyle-a_{A}(m+m\frac{v_{B}}{v_{A}}\frac{a_{B}}{a_{A}}).$

$\displaystyle a_{A}$	$\displaystyle=$	$\displaystyle\frac{-F}{m+\frac{4}{9}m}$
	$\displaystyle=$	$\displaystyle\frac{-130\,{\mbox{N}}}{40\,\mathrm{kg}(1+\frac{4}{9})}$
	$\displaystyle=$	$\displaystyle-2.25\,\mathrm{m}/\mathrm{s}^{2},$

	$\displaystyle-ky=(m+M)\ddot{y}$
	$\displaystyle\mbox{or}\quad\ddot{y}+\frac{k}{(m+M)}y=0.$		(14.23)

	$\displaystyle y(t)$	$\displaystyle=$	$\displaystyle A\sin\lambda t+B\cos\lambda t$		(14.24)
	$\displaystyle\mbox{where }\quad\lambda$	$\displaystyle=$	$\displaystyle\sqrt{\frac{k}{m+M}}.$		(14.25)

	$\displaystyle\overbrace{y(0)}^{y_{0}}$	$\displaystyle=A\sin(\lambda\cdot 0)+B\cos(\lambda\cdot 0)\hbox to36.135pt{% \hfil$\Rightarrow$\hfil}$	$\displaystyle B=y_{0}$
	$\displaystyle\underbrace{\dot{y}(0)}_{0}$	$\displaystyle=A\lambda\cos(\lambda\cdot 0)-B\lambda\sin(\lambda\cdot 0)\hbox to% 36.135pt{\hfil$\Rightarrow$\hfil}$	$\displaystyle A=0.$

$\displaystyle mg-N$	$\displaystyle=$	$\displaystyle m\overbrace{(-y_{0}\lambda^{2}\cos\lambda t)}^{\ddot{y}}$
$\displaystyle\hbox to36.135pt{\hfil$\Rightarrow$\hfil}\quad N$	$\displaystyle=$	$\displaystyle mg+my_{0}\lambda^{2}\cos\lambda t$	(14.28)
	$\displaystyle=$	$\displaystyle mg\left(1+\frac{y_{0}\lambda^{2}}{g}\cos\lambda t\right)$	(14.28)

$\displaystyle W$	$\displaystyle=$	$\displaystyle\Delta{E_{\mathrm{K}}}$
$\displaystyle\int_{t_{1}}^{t_{2}}P_{\rm ave}dt$	$\displaystyle=$	$\displaystyle\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})$
$\displaystyle P_{\rm ave}(t_{2}-t_{1})$	$\displaystyle=$	$\displaystyle\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})$
$\displaystyle P_{\rm ave}$	$\displaystyle=$	$\displaystyle\frac{m}{2\Delta t}(v_{2}^{2}-v_{1}^{2})$

	$\displaystyle\hat{\boldsymbol{\lambda}}_{OG}\cdot\sum{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{G}$	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\lambda}}_{OG}\cdot\dot{\accentset{% \rightharpoonup}{\boldsymbol{H}}}_{G}$		(14.45)
		$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\lambda}}_{OG}\cdot({\accentset{\rightharpoonup}% {\boldsymbol{r}}}_{C/G}\times ma\hat{\boldsymbol{\imath}}).$		(14.45)

	$\displaystyle\hat{\boldsymbol{\lambda}}_{OG}\cdot\sum{\accentset{% \rightharpoonup}{\boldsymbol{M}}}_{G}$	$\displaystyle=$	$\displaystyle\hat{\boldsymbol{\jmath}}\cdot({\accentset{\rightharpoonup}{% \boldsymbol{r}}}_{H/G}\times T\hat{\boldsymbol{\lambda}}_{HI})$		(14.46)
		$\displaystyle=$	$\displaystyle T\frac{d(c-e)}{\ell}.$		(14.46)

$\displaystyle T$	$\displaystyle=$	$\displaystyle\frac{ma\ell}{2d}$
	$\displaystyle=$	$\displaystyle\frac{20\,\mathrm{kg}\cdot 0.6\,\mathrm{m}/\mathrm{s}^{2}\cdot 3.% 39\,\mathrm{m}}{2\cdot 3\,\mathrm{m}}$
	$\displaystyle=$	$\displaystyle 6.78\,{\mbox{N}}.$

	$\displaystyle R_{C_{x}}-T_{AB}=0$
	$\displaystyle R_{C_{y}}-\frac{T_{GD}}{\sqrt{2}}=0$
	$\displaystyle R_{C_{z}}+\frac{T_{GD}}{\sqrt{2}}+T_{EH}=5\,{\mbox{N}}$
	$\displaystyle-T_{EH}+\frac{T_{GD}}{\sqrt{2}}-R_{C_{z}}=0$
	$\displaystyle-T_{EH}-\frac{T_{GD}}{\sqrt{2}}+R_{C_{z}}=0$
	$\displaystyle T_{AB}-\frac{T_{GD}}{\sqrt{2}}+R_{C_{x}}-R_{C_{y}}=0$

Chapter 14 Constrained straight-line motion

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Kinematic constraints

14.1 1-D constrained motion and pulleys

Kinematic constraint: two approaches

Method 1: Finding the accelerations and the constraint forces together

Method 2: Finesse having to find the constraint force

Kinematic constraints

Pulleys

Constant length and constant tension

The classic simple uses of pulleys

Force amplification is motion attenuation

Power balance

Problems for 14.1 1D constrained motion and pulleys

Preparatory Problems

More-Involved Problems

14.2 1D motion with 2D and 3D forces

Highly constrained bodies

Constraint forces are of interest

1D mechanics with 1D forces, moving on

Kinematics of parallel motion and straight line motion

Velocity of a point

Acceleration of a point

General results

Linear momentum 𝑳⇀ and its rate of change 𝑳⇀˙

Angular momentum 𝑯⇀/C and its rate of change, 𝑯⇀˙C for straight-line motion

Kinetic energy

Approach

Sliding and pseudo-sliding objects

Wheels

3-D forces in straight-line motion

Angular momentum balance about an axis

Problems for 14.2 1D motion with 2D and 3D forces

Preparatory Problems

More-Involved Problems

Linear momentum ${\accentset{\rightharpoonup}{\boldsymbol{L}}}$ and its rate of change $\dot{\accentset{\rightharpoonup}{\boldsymbol{L}}}$