# Homework Help: Boxes hanging from pulley system

1. Oct 5, 2009

### brickaney

1. The problem statement, all variables and given/known data
Three boxes are connected by cords, one of which wraps over a pulley having negligible friction on it;s axle and negligible mass. The masses are mA=30kg, mB=36kg, and mC= 12kg. When the assemblyis released from rest (a) what is the tension in the cord connecting B and C, and (b) how far does A move in the first 0.280s (assuming it does not reach the pulley)?
Box A is sitting on top of a desk while boxes B and C are being hung over the edge w/ a pulley.

2. Relevant equations
T-mg=-ma
mA-mg=-ma
a(mA+mB)=mg

3. The attempt at a solution
I substituted T into the last equation and got:
T= (mA*mB)g/(mA+mB)
but I was told this is incorrect. Also, I'm unsure as to how to apply this answer to part (b).
Thank you so much for your help!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 5, 2009

### w3390

You are overcomplicating part a. The only force acting on the rope between b and c is the weight of block c. Use this to find the tension. For part b, make a new diagram labeling all the forces except draw the two hanging blocks as a single hanging block of their combined masses. Then do the usual calculations for force and solve for acceleration. Then using the formula x=xo+vot+(1/2)at^2, find the distance block a travels in the given time.

3. Oct 5, 2009

### brickaney

Thanks for the help! I seem to be doing the problem incorrectly still. I have the values mA=32kg, mB=37kg, and mC=14kg. And time=0.230s. In the problem it says what is the tension when the assembly is released. I thought the weight of block A would effect the tension. I used the equation:
T= (14*9.8)= 137.2N & I got that answer wrong :/
I'm probably overthinking it.

4. Oct 5, 2009

### w3390

Are you sure you are using the right mass. In your first post, you said the mass of block c was 12 kg instead of 14 kg. Is that the problem?

5. Oct 5, 2009

### brickaney

Oh sorry the mass changed because I got the problem incorrect. The hw system I'm using changes the numbers every time you get an answer wrong.

6. Oct 5, 2009

### w3390

That's kinda annoying that your numbers change after each incorrect submission. Well, I am a bit stumped for the moment so I will try working it for a while and I'll get back when I figure it out. Hopefully someone else will be able to jump in and help while I can't.

7. Oct 5, 2009

### w3390

Actually, since you specify when assembly is released, I guess they want the acceleration factored in. So if you make a free body diagram for block c, you get the equation m(c)g-Ts=m(c)a, where m(c) is the mass of block c and Ts is the tension in the string between block b and c. Try solving this for tension and see if its right. I think this might be it because the tension is less than if it were just hanging there so the initial acceleration makes it so the rope is not completely taut.