# Pulley question

1. Nov 1, 2007

### bungie77

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

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?

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 :)