- #1

AHinkle

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## Homework Statement

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.

## Homework Equations

Fnet=ma

## 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..