Pully/block system on an incline

[b577phaty]

Homework Statement

A block of mass m = 10 Kg is attached to a cable that wraps around the pulley. The pulley is pivoted at the center, and the angle of incline is θ = 30°. The radius of the pulley is R = 0.5 m and its moment of inertia is I = 0.4 Kgm2. The block is released from rest and slides down the ramp, and the coefficient of friction is μk= 0.2 between the block and the ramp.
(a) Find the acceleration of the block. a = _______________________
(b) Find the tension in the cable. T = ______________________

I have no idea where to start.

 

[kuruman]

1. Draw a free body diagram for the mass on the incline.
2. Use your FBD to apply Newton's Law to the mass.
3. Draw a free body diagram for the pulley.
4. Use this FBD to apply Newton's Second Law (for rotations) to the pulley. Note that the tangential acceleration of a point on the pulley's rim is the linear acceleration of the rope and, of course, the mass.
5. Steps 2 and 4 will give two equations and two unknowns, the acceleration a and the tension T.
6. Solve the system of two equations and two unknowns to find what the problem asks.

 

[b577phaty]

For my equation, I got mgsin(theta)-I(a/r)-umgcos(theta)=ma however this does not give me the correct answer. Any ideas?

 

[kuruman]

For my equation, I got mgsin(theta)-I(a/r)-umgcos(theta)=ma however this does not give me the correct answer. Any ideas?

Unless you show what answer you got and how, I have no ideas because I don't know what you did. The equation seems correct.

 

[b577phaty]

For the block
Fx=mgsin(theta)-T-Ff=ma
Fy=-mgcos(theta)+N=0

For the pully
T=Ialpha
T=I(a/r)

10*9.8*sin30-.8a-.2*10*9.8*cos30=10a

a=2.72 ?

 

[kuruman]

For the pully
T=Ialpha
T=I(a/r)

In the above equations "T" stands for "Torque" not "Tension". You treat it as if it were tension. What is the torque τ if the tension is T and the lever arm is r?