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Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 32

Most textbook problems (including the problems in this book) have “weightless” cables and ropes. There are two reasons for this:

  1. It makes the problems less cluttered with detail and the student can concentrate on the principles.
  2. The weight of a cable or rope is often very small when compared with the weights it is carrying and taking it to be weightless is then a reasonable approximation.

Here, though, the weight of the rope is essential to the solution of the problem. Because the rope weighs 1 lb/ft, the weight on the left side of the pulley is 10 + x + y (when x and y are measured in feet), and the weight on the right side of the pulley is 20 + x. These weights must be equal to one another for moment equilibrium about the center of the pulley.

10 + x + y = 20 + x

Notice that the semicircle of rope around the top of the pulley contributes no net moment about the center of the pulley because the left and right quarter-circles balance each other out.

The x terms cancel and we are left with y = 10\,\rm{ft}. We can now find x from the length of the rope

20 = 2 x + y + \frac{1}{2} \cdot \pi \cdot 1

Plugging in our value for y and solving, we get x = 5 - \pi/4 = 4.21\,\rm{ft}.

Problem 33Problem 31

Last modified: January 22, 2009 at 8:32 PM.