# Homework Help: Atwood's machine with 3 masses

1. Sep 23, 2010

### AHinkle

1. The problem statement, all variables and given/known data
A pulley is massless and frictionless, 2kg, 3kg, and 6kg masses are suspended. The 2 and 3 kilogram masses are suspended on the left side of the pulley (the 2 kilogram mass on the bottom) and are connected by a cord of negligible mass that does not stretch(T1), the 6kg mass is suspended on the right side and is connected around the pulley via a cord of negligible mass that does not stretch (T2). Down is the positive direction on the right side, up is positive on the left. what is the tension on the string between mass 1 and mass 2.

2. Relevant equations
Fnet=ma

3. The attempt at a solution
Sorry there's not a picture. I did not have anywhere to host it.

I drew free body diagrams for each mass and came up with the Fnet equations for each. I assume we're only working with y accelerations.
m3g-T2=m3a
T2-T1=m2a
T1-m1g=m1a

I added the equations together for m2 and m3 because they're connected by a cord and
the acceleration for the whole system should be the same.

m3g-T2+T2-T1=(m2+m3)a
m3g-T1=(m2+m3)a

m3=6kg m2=3kg g=9.8

(6kg)(9.8m/s^2)-T1 = 9a

i solved for a

T1-m1g=m1a

a = (T1-m1g)/(m1)

58.8-T1=9((T1-m1g)/(m1))

58.8-T1=9((T1-(2)(9.8))/(2))

58.8-T1=(9T1-176.4)/(2)

117.6-2(T1)=9(T1)-176.4

294-2(T1)=9(T1)

11(T1) = 294

T1 = 26.7272 N

I felt like i did this right, and it's tormenting me..

2. Sep 23, 2010

### PhanthomJay

In your second equation T2 -T1 = m2a, you forgot to include the weight, m2g, in that equation.

3. Sep 23, 2010

### ehild

This equation i wrong. Gravity acts on all masses!

ehild

4. Sep 24, 2010

### AHinkle

Thanks guys, I got it right. That small oversight cost me an hour at least.

m3g-T2 = m3a
T2-T1-m2g=m2a
T1-m1g=m1a

m3g-T1-m2g=(m2+m3)a

a=(T1-m1g)/(m1)

(6)(9.8)-T1-(3)(9.8)=9a
58.8-T1-29.4=9a
29.4-T1=9a

sub in a
29.4-T1=9((T1-m1g)/(m1))

29.4-T1=((9)(T1)-(18)g)/2
58.8-2(T1)=9(T1)-176.4

-(2)(T1)=(9)(T1)-235.2
-(11)(T1)=-235.2
T1= (-235.2)/(-11)

T1= 21.3818 N

5. Sep 24, 2010

### ehild

It is an easier job if you add up all equations. That cancels all inner forces (tensions here) and you get the acceleration as the total external force divided by the total mass. Knowing the acceleration, it is easy to find the tensions.

ehild