# Pulley problem

1. Nov 8, 2007

### chiurox

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

The problem seems like this: on the left, closer to the floor, is block3 with mass 1.0kg, there is a string above it which connects it to block2 with mass 3.0kg which has a string above it that circles around a pulley and comes back down connecting it to block1 with mass 5.0kg

The pulley is frictionless and massless

2. Relevant equations
a. Derive an equation for the acceleration of the block m1=5.0kg and calculate the value of the acceleration.

b. What is the tensionin the string connected between the 1.0kg block (mass3) and the 3.0kg (mass2) block?

c. What is the tension in the string connected between the 3.0kg block (mass2) and the 5.0kg block (mass1)?

d. After the 5.0kg block (mass1) comes to rest on a table surface, what is the force the table exerts on the block?
3. The attempt at a solution

a. I'm not sure how I would derive an equation for it, but I just solved it using the Fnet-y of the larger block and the second block with mass 3.0k. I added both equations:
T1 - m1g = m1a ---------- T1 - 50 = 5a (plugged)
-T1 + (m2g + m3g) = m2a ---------- -T1 + 40 = 3a
After adding both equations I found a to be -1.25m/s^2 (because I established counter-clockwise movement of the pulley as being positive) Is this right?

b. 10-T2 = -1.25 so T2 = 11N I think it makes sense...

c. T1 - 50 = 5(-1.25) so T1 = 44N

d. N = m1g - T1 = 50-44 = N = 6N Well, I think it makes sense, but i'm not sure though.

Could anyone help check these please? because at first I was really lost and still am.

2. Nov 8, 2007

### Staff: Mentor

It's almost right. In your second equation, what you really did was combine m2 & m3 as a single system--perfectly OK. But then you need to use the mass of that system:
-T1 + (m2g + m3g) = (m2 + m3)a

And then you'd solve for both a and T1.

3. Nov 8, 2007

### chiurox

Alright, I understand it, thank you very much. Good thing it didn't change much (because of sig figs) on the other answers.