Difficult problem involving friction and torque

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SUMMARY

The discussion focuses on a physics problem involving a block sliding down an inclined plane connected to a cylinder via a cord. The coefficient of friction is given as μ = 3.00×10^-2. Key equations include Newton's second law in both linear and angular forms, leading to a system of equations that describe the forces acting on both the block and the cylinder. The challenge lies in correctly identifying the forces and their directions, particularly the normal force and frictional force acting on the cylinder.

PREREQUISITES
  • Understanding of Newton's second law in linear and angular forms
  • Knowledge of frictional forces and coefficients of friction
  • Familiarity with inclined plane dynamics
  • Basic concepts of torque and rotational motion
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  • Study the derivation of equations for inclined plane motion with friction
  • Learn about the relationship between linear and angular acceleration in rotational systems
  • Explore the concept of normal force in different scenarios, particularly with inclined surfaces
  • Investigate the effects of varying coefficients of friction on motion dynamics
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of inclined plane problems involving friction and torque.

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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:

[tex] \]<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 /> \[[/tex]

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:
[tex]\Sigma F = ma = mg \sin \theta - T - \mu_k N[/tex]
and
[tex]\Sigma \tau = I\alpha = R \times T[/tex]
 

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