Normal and tangential forces on a cylinder

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Homework Help Overview

The problem involves analyzing the forces acting on a cylinder in contact with two walls, one smooth and one rough, with a specific weight and friction coefficient. Participants are tasked with calculating the normal and tangential forces while considering the equilibrium conditions of the system.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the initial calculations of normal forces and the implications of introducing friction. Questions arise about the direction of frictional forces and the conditions under which they act. Some participants express confusion regarding the equilibrium state of the cylinder when friction is introduced.

Discussion Status

The discussion is ongoing, with participants reviewing their calculations and questioning the assumptions about friction in equilibrium. Some guidance has been offered regarding the nature of static friction and its relationship to the forces acting on the cylinder, but no consensus has been reached.

Contextual Notes

There is a focus on the conditions of equilibrium and the role of friction in this context. Participants are considering the implications of adding a friction coefficient to a previously frictionless scenario, and the discussion includes varying interpretations of how forces may adjust in response to this change.

Karol
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Homework Statement


The walls are smooth, calculate the reactions. the cylinder weighs 100[N].
Now the 600 wall is rough with a friction coefficient 0.4. what are the reactions

Homework Equations


Friction: F=μN

The Attempt at a Solution


I made the first step and it was good: N1=287.9, N2=253.2
But with friction there is another force, tangential, but the results in the book are the same.
 

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You got the first one correct? Because I'm getting different answers.

As for the second part, what I think, is that the cylinder is stationary, in equilibrium. There is no resultant force on it. So which direction will you consider the friction to act in?
 
siddharth23 said:
You got the first one correct? Because I'm getting different answers.
I agree with Karol's answers.
Karol said:
But with friction there is another force, tangential, but the results in the book are the same.
If the cylinder is in equilibrium when there is no friction, and then the only change that is made is that a coefficient of friction is added, then why would a force of friction act?

Edit:
Karol said:
Friction: F=μN
For the case of static friction, that is only the maximum, not the applied force.
 
Last edited:
Ok I'll check the first part again.
 
Nathanael said:
If the cylinder is in equilibrium when there is no friction, and then the only change that is made is that a coefficient of friction is added, then why would a force of friction act?
I may agree if μ is added later. but the forces can arrange themselves differently (only magnitudes) to accommodate the changes.
Maybe because there is no relative movement without the friction force anyway it isn't created.
 
Last edited:
See the image I've uploaded. The surface is rough and has a coeffecient of friction 'μ'. A force 'F' pushes the block towards the right side but its motion is hindered by the wall. Now even though a force of friction would act on the block, in this case it won't as the system is in equilibrium. There is no scope for the block to move.
 

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Karol said:
Maybe because there is no relative movement without the friction force anyway it isn't created.
More accurately, it would be because there is no tendency to move.
However, the question is not really correct. While there need not be a frictional force, there could be one. You could find a solution to the equations in which the walls exert a greater normal force and, to balance that, a small frictional force. In the real world, this corresponds to the cylinder being jammed in.
 
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