# Three masses suspended on pulley

## Homework Statement

A pulley is massless and frictionless. 3 kg, 2 kg, and 6 kg masses are suspended on a pulley.

(a)What is the tension T1 in the string between the two blocks on the left-hand side of the pulley?

(b)What is the magnitude of the acceleration of the lower left-hand block?

## Homework Equations

I solved for (a) using the equations:
T2-m1g=m1a
m2g+T1-T2=m2a
m3g-T1=m3a

## The Attempt at a Solution

the combining those equations:
-m1g+m2g+m3g=m1a+m2a+m3a

then I solved for a
-29.4+19.6+58.8=11a
a=4.45455 m/s^2

then plugged into T1=m3g-m3a and got T1=32.0727N (right

for part (b) I thought that I already had a since I needed it to slove for T1, but it is wrong...

So does the lower left-hand block have a different acceleration?

Also, I tried putting in -4.45455 m/s^2 since m3 is 6kg and ways more than m1 and m2 combined... also didn't work

PhanthomJay
Homework Helper
Gold Member
You have a signage error in your 2nd relevant equation....the weight acts down, so it should have a minus sign in front of it. Redo your calc for the acceleration and value of T1.

You have a signage error in your 2nd relevant equation....the weight acts down, so it should have a minus sign in front of it. Redo your calc for the acceleration and value of T1.

In front of the whole equation or just infront of all m2?

PhanthomJay
Homework Helper
Gold Member
In front of the whole equation or just infront of all m2?

Well, just in front of m2....on the left side of the equation, the weight acts down, and T2 acts down , so both those terms are minus; T1 acts up, so that's plus; on the right side of the equation, the acceleration is up, so that's a plus sign.

Well, just in front of m2....on the left side of the equation, the weight acts down, and T2 acts down , so both those terms are minus; T1 acts up, so that's plus; on the right side of the equation, the acceleration is up, so that's a plus sign.

What about m2 in front of acceleration?

PhanthomJay