Difficult problem involving friction and torque

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The discussion revolves around a physics problem involving a block sliding down an inclined plane connected to a cylinder via a cord. The key challenge is to analyze the forces acting on both the block and the cylinder, including tension, friction, and gravitational forces. Participants express uncertainty about the direction of the normal force on the cylinder and the nature of the frictional force. The equations derived from the forces lead to confusion, particularly regarding the assumptions about the cylinder's contact with the inclined plane. Ultimately, the problem requires a clear understanding of the dynamics involved to determine the block's speed after traveling a specified distance.
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Homework Statement


A cord connected at one end to a block which can slide on an inclined plane has its other end wrapped around a cylinder resting in a depression at the top of the plane as shown in the figure:
http://img35.imageshack.us/img35/6743/giancolich10p098.jpg
Determine the speed of the block after it has traveled 1.40 m along the plane, starting from rest.

Assume the coefficient of friction between all surfaces is μ = 3.00×10^-2. [Hint: First determine the normal force on the cylinder, and make any reasonable assumptions needed.]


Homework Equations


Newton's second law in linear and angular form

The Attempt at a Solution


The forces on the block are fairly obvious: tension T, friction f, weight mg, and normal F.
However, I am not so sure about the forces on the cylinder. First there is T and Mg. I know there is a normal force N, but I wasn't sure of the direction so I drew it pointing to the upper left, an angle θ above the horizontal. I also was not sure about the frictional force. Is it perpendicular to N? If so, I got the following 4 equations in 4 unknowns:

<br /> \]<br /> \\<br /> -Ncos\theta+Tcos27+\mu Nsin\theta=0 \\<br /> <br /> \mu Ncos\theta+Nsin\theta-Tsin27-Mg=0 \\<br /> <br /> T-f=.5Ma \\<br /> <br /> -T-mg cos27+mg sin27=ma<br /> \[<br />

I can't seem to get the right answer from these though.
 
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From the diagram, I get the impression that the cylinder is not actually touching the triangle and is instead attached to a pivot point (which I will assume frictionless).

If this is the case, the forces on the block are the gravitational force, kinetic friction force, tension force, and normal force. The only force on the cylinder is the tension force.

You can use these to find that:
\Sigma F = ma = mg \sin \theta - T - \mu_k N
and
\Sigma \tau = I\alpha = R \times T
 
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