# Newton's Law- 3 objects on a pulley

1. Oct 11, 2007

### johnsonandrew

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

Three objects are connected by light strings as shown in Figure P4.62. The string connecting the m1 = 5.00 kg mass and the m2 = 4.00 kg mass passes over a light frictionless pulley.

a) find the acceleration of each object and
b) the tension in the two strings

2. Relevant equations

sumF = m*a

sumF1= m1*a = T1 - m1*g
sumF2= m2*a = T1 - m2*g - T2
sumF3= m3*a = T2 - m3*g

3. The attempt at a solution

I combined equations sumF1 and sumF2:

m1*a = T1 - m1*g
-m2*a = -T1 + m2g +T2

to get rid of one variable, T1, and I got

m2*g - m1*g + T2 = m1a - m2a

then in the sumF3 equation I solved for T2,
T2 = m3*a + m3*g

and plugged that into the above equation to get
m2*g - m1*g + (m3*a + m3*g) =m1*a - m2*a

I then plugged in my knowns to get 9.8 m/s/s, while the answer is supposed to be 1.63 m/s/s.

Last edited: Oct 11, 2007
2. Oct 11, 2007

### PhanthomJay

you're on track, but the stuff on the left accelerates up, and the stuff on the right acceleartes with the same magnitude, except it accelerates down, You've slipped up on your plus and minus signs.

3. Oct 11, 2007

### johnsonandrew

Ohh. I was wondering about that. So on the right side down would be positive if I took up to be positive on the left?

4. Oct 11, 2007

### PhanthomJay

Yes, that should work.

5. Oct 11, 2007

### johnsonandrew

Ahhh and it did. It gave me 1.6! Thanks a lot!