7.5 Examples of Interacting-Objects Problems

We will conclude this chapter with three extended examples. Although the mathematics will be more involved than in any of our work up to this point, we will continue to emphasize the reasoning one uses in approaching problems such as these. The solutions will be based on Problem-Solving Strategy 7.1. In fact, these problems are now reaching such a level of complexity that, for all practical purposes, it becomes impossible to work them unless you are following a well-planned strategy. Our earlier emphasis on forces and free-body diagrams will now really begin to pay off!

EXAMPLE 7.7 Placing a leg in traction

Serious fractures of the leg often need a stretching force to keep contracting leg muscles from forcing the broken bones together too hard. This is done using traction, an arrangement of a rope, a weight, and pulleys as shown in FIGURE 7.25. The rope must make the same angle on both sides of the pulley so that the net force on the leg is horizontal, but the angle can be adjusted to control the amount of traction. The doctor has specified 50 N of traction for this patient with a 4.2 kg hanging mass. What is the proper angle?

The figure illustrates an arrangement of a rope, pulley and a weight for applying traction force on a fractured leg.

FIGURE 7.25 Full Alternative Text

MODEL Model the leg and the weight as particles. The other point where forces are applied is the pulley attached to the patient’s foot, which we’ll treat as a separate object. We’ll assume massless ropes and a massless, frictionless pulley.

VISUALIZE FIGURE 7.26 shows three free-body diagrams. Forces $\vec{T}_{P 1}$ and $\vec{T}_{P 2}$ are the tension forces of the rope as it pulls on the pulley. The pulley is in equilibrium, so these forces are balanced by $\vec{F}_{L on P}$ , which forms an action/reaction pair with the 50 N traction force $\vec{F}_{P on L}$ . Our model of the rope and pulley makes the tension force constant, $T_{P 1} = T_{P 2} = T_{W}$ , so we’ll call it simply $T$ .

The figure illustrates the free body diagram of a fractured leg in traction. The free body diagrams of the leg, pulley and the weight are shown on three x y coordinate planes.

FIGURE 7.26 Full Alternative Text

SOLVE The x-component equation of Newton’s second law for the pulley is

\begin{matrix} \sum (F_{on P})_{x} & = & T_{P 1} cos θ + T_{P 2} cos θ - F_{L on P} \\ = & 2 T cos θ - F_{L on P} = 0 \end{matrix}

Thus the correct angle for the ropes is

θ = {cos}^{- 1} (\frac{F_{L on P}}{2 T})

We know, from Newton’s third law, that $F_{L on P} = F_{P on L} = 50 N$ . We can determine the tension force by analyzing the weight. It also is in equilibrium, so the upward tension force exactly balances the downward gravitational force:

T = (F_{G})_{W} = m_{W} g = (4.2 kg)(9.80 m / s^{2}) = 41 N

Thus the proper angle is

θ = {cos}^{- 1} (\frac{50 N}{2 (41 N)}) = 52 °

REVIEW The traction force would approach 82 N if angle $θ$ approached zero because the two ropes would pull in parallel. Conversely, the traction would approach 0 N if $θ$ approached $90 °$ . The desired traction is roughly midway between these two extremes, so an angle near $45 °$ seems reasonable.

EXAMPLE 7.8 The show must go on!

A 200 kg set used in a play is stored in the loft above the stage. The rope holding the set passes up and over a pulley, then is tied backstage. The director tells a 100 kg stagehand to lower the set. When he unties the rope, the set falls and the unfortunate man is hoisted into the loft. What is the stagehand’s acceleration?

MODEL The system is the stagehand M and the set S, which we will model as particles. Assume a massless rope and a massless, frictionless pulley.

VISUALIZE FIGURE 7.27 shows the pictorial representation. The man’s acceleration $a_{M y}$ is positive, while the set’s acceleration $a_{S y}$ is negative. These two accelerations have the same magnitude because the two objects are connected by a rope, but they have opposite signs. Thus the acceleration constraint is $a_{S y} = - a_{M y}$ . Forces $\vec{T}_{M on S}$ and $\vec{T}_{S on M}$ are not literally an action/reaction pair, but they act as if they are because the rope is massless and the pulley is massless and frictionless. Notice that the pulley has “turned” the tension force so that $\vec{T}_{M on S}$ and $\vec{T}_{S on M}$ are parallel to each other rather than opposite, as members of a true action/reaction pair would have to be.

The figure illustrates the sketch of a man labeled M pulling down on a rope labeled R that passes over a pulley.

The figure illustrates an Interaction diagram and Free-body diagram for a man labeled M pulling down on a rope labeled R that passes over a pulley.

Full Alternative Text

SOLVE Newton’s second law for the man and the set is

\begin{array}{l} \sum^{​} {(F_{on M})}_{y} & = & T_{S on M} - m_{M} g = m_{M} a_{M y} \\ \sum^{​} {(F_{on S})}_{y} & = & T_{M on S} - m_{S} g = m_{S} a_{S y} = - m_{S} a_{M y} \end{array}

Only the y-equations are needed. Notice that we used the acceleration constraint in the last step. Newton’s third law is

T_{M on S} = T_{S on M} = T

where we can drop the subscripts and call the tension simply T. With this substitution, the two second-law equations can be written

\begin{array}{l} T - m_{M} g = m_{M} a_{M y} \\ T - m_{S} g = - m_{S} a_{M y} \end{array}

These are simultaneous equations in the two unknowns T and $a_{M y}$ . We can eliminate T by subtracting the second equation from the first to give

(m_{S} - m_{M}) g = (m_{S} + m_{M}) a_{M y}

Finally, we can solve for the hapless stagehand’s acceleration:

a_{M y} = \frac{m_{S} - m_{M}}{m_{S} + m_{M}} g = \frac{100 kg}{300 kg} 9.80 m / s^{2} = 3.27 {m/s}^{2}

This is also the acceleration with which the set falls. If the rope’s tension was needed, we could now find it from $T = m_{M} a_{M y} + m_{M} g$ .

REVIEW If the stagehand weren’t holding on, the set would fall with free-fall acceleration g. The stagehand acts as a counterweight to reduce the acceleration.

CHAPTER 7 CHALLENGE EXAMPLE A not-so-clever bank robbery

Bank robbers have pushed a 1000 kg safe to a second-story floor-to-ceiling window. They plan to break the window, then lower the safe 3.0 m to their truck. Not being too clever, they stack up 500 kg of furniture, tie a rope between the safe and the furniture, and place the rope over a pulley. Then they push the safe out the window. What is the safe’s speed when it hits the truck? The coefficient of kinetic friction between the furniture and the floor is 0.50.

MODEL This is a continuation of the situation that we analyzed in Figures 7.15 and 7.24, which are worth reviewing. The system is the safe S and the furniture F, which we will model as particles. We will assume a massless rope and a massless, frictionless pulley.

VISUALIZE The safe and the furniture are tied together, so their accelerations have the same magnitude. The safe has a y-component of acceleration $a_{S y}$ that is negative because the safe accelerates in the negative y-direction. The furniture has an x-component $a_{F x}$ that is positive. Thus the acceleration constraint is

a_{F x} = - a_{S y}

The free-body diagrams shown in FIGURE 7.28 are modeled after Figure 7.24 but now include a kinetic friction force on the furniture. Forces $\vec{T}_{F on S}$ and $\vec{T}_{S on F}$ act as if they are an action/reaction pair, so they have been connected with a dashed line.

The figure illustrates the sketch of a safe and a furniture tied together with the help of a rope passing over a pulley. The safe is being lowered down to a truck. The sketch is drawn on a x y coordinate plane.

The figure illustrates the sketch of an interaction diagram and the free body diagram of a safe being lowered down to a truck.

Full Alternative Text

SOLVE We can write Newton’s second law directly from the free-body diagrams. For the furniture,

\begin{array}{l} \sum (F_{on F})_{x} = T_{S on F} - f_{k} = T - f_{k} = m_{F} a_{F x} = - m_{F} a_{S y} \\ \sum (F_{on F})_{y} = n - m_{F} g = 0 \end{array}

And for the safe,

\sum (F_{on S})_{y} = T - m_{S} g = m_{S} a_{S y}

Notice how we used the acceleration constraint in the first equation. We also went ahead and made use of Newton’s third law: $T_{F on S} = T_{S on F} = T$ . We have one additional piece of information, the model of kinetic friction:

f_{k} = μ_{k} n = μ_{k} m_{F} g

where we used the y-equation of the furniture to deduce that $n = m_{F} g$ . Substitute this result for $f_{k}$ into the x-equation of the furniture, then rewrite the furniture’s x-equation and the safe’s y-equation:

\begin{array}{l} T - μ_{k} m_{F} g = - m_{F} a_{S y} \\ T - m_{S} g = m_{S} a_{S y} \end{array}

We have succeeded in reducing our knowledge to two simultaneous equations in the two unknowns $a_{S y}$ and T. Subtract the second equation from the first to eliminate T:

(m_{S} - μ_{k} m_{F}) g = - (m_{S} + m_{F}) a_{S y}

Finally, solve for the safe’s acceleration:

\begin{matrix} a_{S y} & = & - (\frac{m_{S} - μ_{k} m_{F}}{m_{S} + m_{F}}) g \\ = & - (\frac{1000 kg - (0.50) (500 kg)}{1000 kg + 500 kg}) 9.80 m / s^{2} = - 4.9 m / s^{2} \end{matrix}

Now we need to calculate the kinematics of the falling safe. Because the time of the fall is not known or needed, we can use

\begin{array}{l} v_{1 y} ​^{2} = v_{0 y} ​^{2} + 2 a_{S y} Δ y = 0 + 2 a_{S y} (y_{1} - y_{0}) = - 2 a_{S y} y_{0} \\ v_{1} = \sqrt{- 2 a_{S y} y_{0}} = \sqrt{- 2 (- 4.9 {m/s}^{2}) (3.0 m)} = 5.4 m/s \end{array}

REVIEW The value of $v_{1 y}$ is negative, but we only needed to find the speed so we took the absolute value. This is about $12 mph,$ so it seems unlikely that the truck will survive the impact of the 1000 kg safe!

7.5 Examples of Interacting-Objects Problems

EXAMPLE 7.7 Placing a leg in traction

FIGURE 7.25 A leg in traction.

FIGURE 7.26 The free-body diagrams.

EXAMPLE 7.8 The show must go on!

FIGURE 7.27 Pictorial representation for Example 7.8.

CHAPTER 7 CHALLENGE EXAMPLE A not-so-clever bank robbery

FIGURE 7.28 Pictorial representation for Chapter 7 Challenge Example.