Two Blocks and Pulley

1. Oct 30, 2009

EstimatedEyes

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

A block of mass m1 = 3 kg rests on a table with which it has a coefficient of friction µ = 0.54. A string attached to the block passes over a pulley to a block of mass m3 = 5 kg. The pulley is a uniform disk of mass m2 = 0.4 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.

2. Relevant equations

F=ma
tau=I*alpha
I(disc)=(1/2)mr^2
alpha*r=a

3. The attempt at a solution
I determined what I thought were all of the forces acting in the direction of acceleration (friction on block 1, gravity on block 3, and the torque causing the angular acceleration of the pulley. From this I derived the equation:
(m1 + m3)a = m3g -m1gµ - I*(a/r)
and then solved for a
a = (m3g-m1gµ)/(m1 + m2 + .5*m2r)
but that did not give me the right answer; where did I go wrong? Thanks!

2. Oct 30, 2009

Staff: Mentor

Redo your derivation. For one thing, torque and force have different units and thus cannot be added together.

I recommend that you separately apply Newton's 2nd law to each mass and the pulley. By combining those three equations, you'll derive the equation that you want.