# Pulley Problem for Engineering Dynamics Course

1. Jun 2, 2006

### VinnyCee

The cable at C is being pulled down at a velocity of $$4 \frac{ft}{s}$$. The cable at C is also decelerating at a rate of $$2 \frac{ft}{s^2}$$. At what velocity and acceleration is the block at A moving?

http://img222.imageshack.us/img222/729/engr204problem121795wl.jpg [Broken]

I have the following equations for the problem worked out:

$$\ell\,=\,2\,S_A\,+\,\left(h\,-\,S_B\right)$$

$$\frac{d}{dt}\,\left[\ell\,=\,2\,S_A\,+\,\left(h\,-\,S_B\right)\right]$$

$$0\,=\,2\,V_A\,-\,V_B$$

$$V_A\,=\,-\frac{V_C}{2}$$

$$V_A\,=\,-\frac{-4}{2}\,=\,2\,\frac{ft}{s}$$

This answer is NOT correct, it is off by a factor of two. Can anyone tell why?

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Last edited by a moderator: May 2, 2017
2. Jun 2, 2006

### Andrew Mason

Pulley C moves at half the speed of the cable around it. So C is moving down at 2 ft/sec and the cable attached to C is also moving at 2 ft/sec, which is the cable around A. Pulley A must be moving at 1/2 the rate at which the cable around it is moving.

AM

Last edited by a moderator: May 2, 2017
3. Jun 6, 2006

### VinnyCee

Thank you very much!

I guess I was just not thinking of a simple solution because I thought that I would be complicated to figure. The obvious 1/2 velocity result of the right hand pulley is what threw me!

$$V_C\,=\,\frac{1}{2}\,(-4\,\frac{ft}{s})\,=\,-2\,\frac{ft}{s}$$
$$V_A\,=\,-\frac{-2}{2}\,=\,1\,\frac{ft}{s}$$