Chapter 7, Problem 49P

(III) A 3.0-m-long steel chain is stretched out along the top level of a horizontal scaffold at a construction site, in such a way that 2.0 m of the chain remains on the top level and 1.0 m hangs vertically, Fig. 7–26. At this point, the force on the hanging segment is sufficient to pull the entire chain over the edge. Once the chain is moving, the kinetic friction is so small that it can be neglected. How much work is performed on the chain by the force of gravity as the chain falls from the point where 2.0 m remains on the scaffold to the point where the entire chain has left the scaffold? (Assume that the chain has a linear weight density of 18 N/m.)

FIGURE 7-26

Problem 49.

Let y represent the length of chain hanging over the table, and let λ represent the weight per unit length of the chain. Then the force of gravity (weight) of the hanging chain is F_G = λ_y. As the next small length of chain dy comes over the table edge, gravity does an infinitesimal amount of work on the hanging chain given by the force tunes the distance. F_Gdy = λydy. To find the total amount of work that gravity does on the chain, integrate that work expression, with the limits of integration representing the amount of chain hanging over the table.

$W={\displaystyle \underset{y_{\text{initial}}}{\overset{y_{\text{final}}}{\int }}{F}_{G}dy={\displaystyle \underset{1.0\text{m}}{\overset{3.0\text{m}}{\int }}\lambda ydy=\frac{1}{2}}}\lambda y^2{|}_{1.0\text{m}}^{3.0\text{m}}=\frac{1}{2}(18\text{N}/\text{m})(9.0{\text{m}}^2-1.0{\text{m}}^2)=\boxed{72\text{J}}$