Pulley Problem for Engineering Dynamics Course

[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

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?

 

[Andrew Mason]

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

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?

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

 

[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!

Anyways, the correct answer is:

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

V_A\,=\,-\frac{-2}{2}\,=\,1\,\frac{ft}{s}