IDENTIFY and SET UP:
The motion is along a curve, so we’ll use Eq. (6.14) (W=\int_{P_{1}}^{P_{2}} \overrightarrow{\boldsymbol{F}} \cdot d \overrightarrow{\boldsymbol{l}}=\int_{P_{1}}^{P_{2}} F \cos \phi d l=\int_{P_{1}}^{P_{2}} F_{\|} d l) to calculate the work done by the net force, by the tension force, and by the force \overrightarrow{\boldsymbol{F}}. Figure 6.24b shows our freebody diagram and coordinate system for some arbitrary point in Throcky’s motion. We have replaced the sum of the tensions in the two chains with a single tension T.
EXECUTE:
(a) There are two ways to find the total work done during the motion: (1) by calculating the work done by each force and then adding those quantities, and (2) by calculating the work done by the net force. The second approach is far easier here because Throcky is nearly in equilibrium at every point. Hence the net force on him is zero, the integral of the net force in Eq. (6.14) (W=\int_{P_{1}}^{P_{2}} \overrightarrow{\boldsymbol{F}} \cdot d \overrightarrow{\boldsymbol{l}}=\int_{P_{1}}^{P_{2}} F \cos \phi d l=\int_{P_{1}}^{P_{2}} F_{\|} d l) is zero, and the total work done on him is zero.
(b) It’s also easy to find the work done by the chain tension T because this force is perpendicular to the direction of motion at all points along the path. Hence at all points the angle between the chain tension and the displacement vector d \overrightarrow{\boldsymbol{l}} is 90° and the scalar product in Eq. (6.14) (W=\int_{P_{1}}^{P_{2}} \overrightarrow{\boldsymbol{F}} \cdot d \overrightarrow{\boldsymbol{l}}=\int_{P_{1}}^{P_{2}} F \cos \phi d l=\int_{P_{1}}^{P_{2}} F_{\|} d l) is zero. Thus the chain tension does zero work.
(c) To compute the work done by \overrightarrow{\boldsymbol{F}}, we need to calculate the line integral in Eq. (6.14). Inside the integral is the quantity F cos Φ dl; let’s see how to express each term in this quantity.
Figure 6.24a shows that the angle between \overrightarrow{\boldsymbol{F}} and d \overrightarrow{\boldsymbol{l}} is θ, so we replace Φ in Eq. (6.14) with θ. The value of θ changes as Throcky moves.
To find the magnitude F of force \overrightarrow{\boldsymbol{F}}, note that the net force
on Throcky is zero (he is nearly in equilibrium at all points), so \sum F_{x}=0 \text { and } \sum F_{y}=0. From Fig. 6.24b,
\sum F_{x}=F+(-T \sin \theta)=0 \quad \sum F_{y}=T \cos \theta+(-w)=0
If you eliminate T from these two equations, you can show that F = w tan θ. As the angle θ increases, the tangent increases and F increases (you have to push harder).
To find the magnitude dl of the infinitesimal displacement d \overrightarrow{\boldsymbol{l}},note that Throcky moves through a circular arc of radius R (Fig. 6.24a). The arc length s equals the radius R multiplied by the length θ (in radians): s = Rθ. Therefore the displacement d \overrightarrow{\boldsymbol{l}} corresponding to a small change of angle dθ has a magnitude dl = ds = R dθ.
When we put all the pieces together, the integral in Eq. (6.14) becomes
W=\int_{P_{1}}^{P_{2}} F \cos \phi d l=\int_{0}^{\theta_{0}}(w \tan \theta) \cos \theta(R d \theta)=\int_{0}^{\theta_{0}} w R \sin \theta d \theta
(Recall that tan θ = sin θ/cos θ, so tan θ cos θ = sin θ.) We’ve converted the line integral into an ordinary integral in terms of the angle θ. The limits of integration are from the starting position at u = 0 to the final position at θ = \theta_{0}. The final result is
W=w R \int_{0}^{\theta_{0}} \sin \theta d \theta=-\left.w R \cos \theta\right|_{0} ^{\theta_{0}}=-w R\left(\cos \theta_{0}-1\right)=w R\left(1-\cos \theta_{0}\right)
EVALUATE: If \theta_{0} = 0, there is no displacement; then cos \theta_{0} = 1 and W = 0, as we should expect. As \theta_{0} increases, cos \theta_{0} decreases and W = wR(1 – cos \theta_{0}) increases. So the farther along the arc you push Throcky, the more work you do. You can confirm that the quantity R(1 – cos \theta_{0}) is equal to h, the increase in Throcky’s height during the displacement. So the work that you do to raise Throcky is just equal to his weight multiplied by the height that you raise him.
We can check our results by calculating the work done by the force of gravity \overrightarrow{\boldsymbol{w}}. From part (a) the total work done on Throcky is zero, and from part (b) the work done by tension is zero. So gravity must do a negative amount of work that just balances the positive work done by the force \overrightarrow{\boldsymbol{F}} that we calculated in part (c).
For variety, let’s calculate the work done by gravity by using the form of Eq. (6.14) that involves the quantity \overrightarrow{\boldsymbol{F}} * d \overrightarrow{\boldsymbol{l}}, and express the force \overrightarrow{\boldsymbol{w}} and displacement d \overrightarrow{\boldsymbol{l}} in terms of their x- and y-components. The force of gravity has zero x-component and a
y-component of -w. Figure 6.24a shows that d \overrightarrow{\boldsymbol{l}} has a magnitude
of ds, an x-component of ds cos θ, and a y-component of ds sin θ. So
\vec{w}=\hat{\jmath}(-w)
d \vec{l}=\hat{\imath}(d s \cos \theta)+\hat{\jmath}(d s \sin \theta)
Use Eq. (1.19) (\overrightarrow{\boldsymbol{A}} \cdot \overrightarrow{\boldsymbol{B}}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}) to calculate the scalar product \vec{w} \cdot \vec{d} \vec{l}
\overrightarrow{\boldsymbol{w}} \cdot d \overrightarrow{\boldsymbol{l}}=(-w)(d s \sin \theta)=-w \sin \theta d s
Using ds = R dθ, we find the work done by the force of gravity:
\int_{P_{1}}^{P_{2}} \overrightarrow{\boldsymbol{w}} \cdot d \overrightarrow{\boldsymbol{l}}=\int_{0}^{\theta_{0}}(-w \sin \theta) R d \theta=-w R \int_{0}^{\theta_{0}} \sin \theta d \theta =-w R\left(1-\cos \theta_{0}\right)
The work done by gravity is indeed the negative of the work done by force \overrightarrow{\boldsymbol{F}} that we calculated in part (c). Gravity does negative work because the force pulls downward while Throcky moves upward.
As we saw earlier, R\left(1-\cos \theta_{0}\right) is equal to h, the increase in Throcky’s height during the displacement. So the work done by gravity along the curved path is -mgh, the same work that gravity would have done if Throcky had moved straight upward a distance h. This is an example of a more general result that we’ll prove in Section 7.1.
