Cylinder rolling down a wedge - wedge should not slip

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Homework Help Overview

The problem involves a solid cylinder rolling down an inclined wedge without slipping. The wedge has specific dimensions, and the task is to determine the minimum coefficient of friction required to prevent the wedge from slipping on the ground.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to balance forces acting on the wedge but struggles to reach the correct answer despite having solved a similar problem. They express confusion regarding the equations used.
  • Some participants question the number of friction forces acting on the wedge, indicating a need for clarification on the forces involved.
  • Others note that the cylinder must experience torque as it rolls, which introduces additional forces acting on the wedge due to Newton's third law.

Discussion Status

The discussion is ongoing, with participants exploring the forces involved and questioning assumptions about the problem setup. There is an acknowledgment of the complexity of the interactions between the cylinder and the wedge, but no consensus has been reached yet.

Contextual Notes

The problem is constrained by the requirement that the wedge does not slip, which influences the analysis of forces and friction involved. The original poster's reference to a similar problem suggests a potential gap in understanding the specific dynamics at play in this scenario.

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Homework Statement


The height of the slope shown in the figure is 30 cm, and its base is 40 cm. A solid cylinder of uniform density, which has the same mass as the slope, rolls down along the slope without slipping. What is the least value of the coefficient of friction between the inclined plane and the horizontal ground if the slope does not slip?


Homework Equations





The Attempt at a Solution


I understand that this is a simple problem but I cannot reach the specified answer though I did a similar problem recently.

The forces acting on the wedge are the normal reaction from cylinder (N), normal reaction from ground (N'), weight (mg) and the frictional force (f).

Balancing forces on the wedge in the vertical direction:
$$N'=mg+N\cos\theta$$
where ##\theta## is the angle made by the slope with the horizontal.
Since the wedge doesn't slip,
$$f \geq N\sin\theta \Rightarrow \mu N' \geq N\sin\theta$$
Using ##N=mg\cos\theta## and solving the equations, I don't get the right answer.

Any help is appreciated. Thanks!
 

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Pranav-Arora said:
The forces acting on the wedge are the normal reaction from cylinder (N), normal reaction from ground (N'), weight (mg) and the frictional force (f).

How many friction forces act on the wedge?
 
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The cylinder has to spin up as it rolls down without slipping, which means there is some torque on it, which is provided by a force. By Newton's third law, the wedge will experience the opposite force.
 
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TSny said:
How many friction forces act on the wedge?

voko said:
The cylinder has to spin up as it rolls down without slipping, which means there is some torque on it, which is provided by a force. By Newton's third law, the wedge will experience the opposite force.

Silly me, thanks a lot both of you! :)
 

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