# Simple Pulley Problem With Acceleration

1. Feb 26, 2015

### Goofball Randy

1. The problem statement, all variables and given/known data
Block A in the figure has a mass of 5.00kg , and block B has mass 12.0kg . The coefficient of kinetic friction between block B and the horizontal surface is 0.25.

What is the mass of block C if block B is moving to the right and speeding up with an acceleration 3.00m/s2 ?

2. Relevant equations

Friction = coefficient * normal force
F = ma

3. The attempt at a solution

I found the frictional force heading left (since block is going right). 0.25 * (12 * 9.81) = 29.43 N
The tension force on B is then just F = ma + friction force, so 12 * 3 + 29.43 = 65.43 N
And the acceleration of block C, which should also be 3 m/s2, gives the equation Ca = Cg - T, since the acceleration downwards is just gravitational force - tension force. So 3C = 9.81C - 65.43, C = 9.608kg

But this is wrong!!! And I don't know why :(
The textbook sucks too...pulleys aren't even so much as mentioned in the chapter we're in.

Edit: Tried accounting for A.

A is moving up at 3 m/s2.
So -3A = Ag - T
-3(5) = (9.81)(5) - T
T = 34.05 N
So then the total forces pulling left are 34.05 + 29.43 = 63.48N (A and friction). So the forces pulling right must equal that + 36 (F = (12)(3)) = 99.48N. So 3C = 9.81C - 99.48, C = 14.608kg

Still doesn't work :(

Last edited: Feb 26, 2015
2. Feb 26, 2015

### Dishsoap

Maybe I'm misunderstanding your work, but did you consider that the gravitational force of blocks A and C also act on B?

3. Feb 26, 2015

### Goofball Randy

Still doesn't work :(

Last edited: Feb 26, 2015
4. Feb 26, 2015

### haruspex

I don't see where you took into account the tension on the A side.

5. Feb 26, 2015

### Goofball Randy

Yeah, realized that from the other post. Edited my post above, but it still is wrong. I was so hoping it would be right too, since it made so much sense...

6. Feb 26, 2015

### Goofball Randy

Ahhhh never mind. I'm just bad at math. :D