How Do Friction, Tension, and Pulley Dynamics Affect Acceleration?

[EstimatedEyes]

Homework Statement

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.

Homework Equations

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

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!

 

[Doc Al]

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)

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.

 

[kerimek]

I would first like to commend you for attempting to solve this problem using the relevant equations and concepts. However, it seems that you have overlooked the fact that there is also a tension force in the string, which is responsible for both the acceleration of block 1 and the rotation of the pulley. This tension force is equal to the weight of block 3, and therefore must be included in your equation for the net force on block 1.

The correct equation should be:

(m1 + m3)a = m3g - m1gµ - T - I*(a/r)

where T is the tension force in the string.

I would also like to point out that your equation for the moment of inertia of the pulley is incorrect. The correct equation is:

I = (1/2)m2r^2

I hope this helps you in solving the problem correctly. Keep up the good work!