# Homework Help: Pulley question

1. Nov 1, 2007

### bungie77

1. The problem statement, all variables and given/known data

http://img214.imageshack.us/img214/6531/pulleysystemph6.th.jpg [Broken]

2. Relevant equations

sum of forces

3. The attempt at a solution

So I tried summing the forces in the x-direction where

m*a (block a) = -Ff + T = m(block a)*g*u + m*a(block b) + m*g(block b)

where u is the coefficient of friction, T is the tension in the rope and Ff is the frictional force.

I'm not, however, getting an answer on the multi choice list. Is there something I'm missing?

Last edited by a moderator: May 3, 2017
2. Nov 1, 2007

3. Nov 1, 2007

### PhanthomJay

You've got to look at each block separately when doing these problems. This is essential. Draw a free body diagram for each. Identify the forces acting on each block and use newton 2. You'll have 2 equations with 2 unknowns that you can solve. Note that the magnitudes of the tension and acceleration for each block are the same.

4. Nov 1, 2007

### bungie77

well i've been doing that but i cant seem to get the right answer.... i was wondering if maybe i was screwing something up with the directions...

so far ive come to the following;

for block A;

Fx = T - Uk*N = mA*ax
Fy = N - mA*g = 0

So essentially, Fx = T - Uk*mA*g = mA*ax

For block C;

Fy = T - mC*g = mC*ay

I understand ay = ax but i cant seem to get the right answer... i'm getting something around 1.6 :(

5. Nov 2, 2007

### PhanthomJay

Oh, good, you are on track; only error you made was for block C, since it accelerates down, its weight must be greater than T, so your plus/minus signs are incorrect. It should be mCg - T = mC*ay. Do you see why?

6. Nov 2, 2007

### bungie77

ahhh yeah of course. and since the tension is obviously greater in block A the signs are opposite to block C. thanks :)