Boxes hanging from pulley system

[brickaney]

Homework Statement

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.

Homework Equations

T-mg=-ma
mA-mg=-ma
a(mA+mB)=mg

The Attempt at a Solution

I substituted T into the last equation and got:
T= (mA*mB)g/(mA+mB)
my answer was T= 178.5N
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!

 

[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+volt+(1/2)at^2, find the distance block a travels in the given time.

 

[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.

 

[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?

 

[brickaney]

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

 

[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.

 

[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.