- #1

albertrichardf

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## Homework Statement

A differential pulley carries a weight W. The chain used has N links per foot. The bigger pulley has n notches, which can hold two chain links each. The smaller pulley contains n - 1 notches, which can also hold 2 chain links. The friction is such that the ratio of the force required to lift the weight up to the force required to lower it is R. Assuming that the friction is the same in both directions, find the force to lift the weight and find the force to lower the weight. Use the principle of virtual work.

## Homework Equations

[tex] ∆E = 0 [/tex] for virtual displacement

[tex] F_{dn} = \frac {W} {n(R - 1) + 1} [/tex]

[tex] F_{up} = RF_{dn} [/tex]

The solutions

## The Attempt at a Solution

I set up the two following equations:

Call the downwards force F, the radius of the big pulley X, and the radius of the smaller pulley x. I then imagine a virtual rotation of the pulleys of dø. I yank on the chain with F, and the chain moves a distance Xdø, So I do a work of FXdø. However, at the same time, the smaller pulley moves a distance xdø opposite the force, so I do work of F(Xdø - xdø). Furthermore, the weight moves down a distance Xdø/2, which is a work of WXdø/2. There is friction, f, which acts along both pulleys and opposite the movement each time, so it does a total work of -f(X + x)dø. Virtual work states:

[tex] \frac{WXdø}{2} + F(X - x)dø = f(X + x)dø. [/tex]

The situation is the same to move the weight upwards, except that the work done by gravity is negative and the force applied is RF instead of F. Thus:

[tex] \frac{-WXdø}{2} + RF(X - x)dø = f(X + x)dø [/tex]

Equating them, and moving the weight to the right, dividing by dø:

[tex] RF(X - x) = WX + F(X - x) [/tex]

To determine the radius X and x, I note that they can hold 2n chain links and 2(n - 1) chain links respectively. There are N links per foot, so the circumference is 2n/N. Dividing by 2π to obtain the radius I get: n/πN = X. I get something similar for x, except that the numerator is (n-1). I also note that because everything in my equation contains a term in X or x, I can eliminate any factors in front of n and n - 1.

Therefore X - x = n - n + 1 = 1. The equation reduces to

[tex] RF = Wn + F [/tex]

Solving for F I get:

[tex] F = \frac{Wn}{(R -1).} [/tex]

Which is not the answer. I think I might have made a mistake in setting up my equations but I have no idea what it could be. Thank you for helping.