Note: On this last problem they change the orientation of the distance. We have been using x = 0 is the ground and the max height is our upper limit. This next two problem has a couple of differences. First, the top of the building is x = 0 and the ground is *=h. They do this type of thing (swapping which is x=0 and x=h all the time in physics so it is good to get some practice with it). The second difference in this problem is we are only lifting half the chain in this problem. You may need to type Ax = Delta x (capital D important) and T; = X_i A chain, with length h = 160 feet and weighing 7 pounds per foot, hangs over the side of a building. We are going to determine the work required to lift half the chain to the top of the building. X= 0 Xi di IAX x= h This problem depends on the vertical coordinate system to the left. The force, Fi, on the it strip of the chain = weight of the it strip of the chain. Fi = pounds The distance, di, we are moving the it strip is di = feet (Must relate back to I; ). The work, Wi, required to lift the it" strip of the chain is W, = Fi . di. Wi = ft-lbs Adding up the work on all the subintervals and allowing n -> oo, where n represents the number of subintervals, gives the integral for total work, 1, required to lift half the chain to the top of the building. W = Finally, W = foot-pounds