Inclined plane static friction

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Homework Statement


An ideal cord is connected to a wall on the top of an inclined plane at one end and is wound around a thin homogeneous cylinder of mass M and radius R. The inclination angle is ϑ and the static and dynamic friction coefficients between cylinder and plane are μs, μd , and the viscosity of the air is
negligible. The cylinder is at rest at the time t=0. Find the linear and angular acceleration of the cylinder.

- I didn't really get why the static friction is oriented downwards along the inclined.


Homework Equations




The Attempt at a Solution

 

Answers and Replies

  • #2
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I don't think I understand the entirety of the problem with the details you've given, but I'll ask an important question you should always ask yourself in situations such as the one you're experiencing. What makes you question the direction of static friction in this problem?

Edit: Is there any way you could upload a picture of the problem including all of the forces acting on the cylinder? If you want, you could draw the problem using this site's "whiteboard" feature bellow the reply box, save it, then upload it.
 
  • #3
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You right, i didn't make my self clear! Sorry about.
Basically i need to know the direction of the friction in order to evaluate the torque
 

Attachments

  • exercise 2.pdf
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  • #4
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If a friction force is present, in any situation, it always opposes the direction of the net force. That is, friction always does negative work.
 
  • #5
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If a friction force is present, in any situation, it always opposes the direction of the net force. That is, friction always does negative work.
So the cylinder, in presence of static friction is supposed to go upwards due to the pulling force applied by the cord and goes downwards if the friction is dynamic due to the cord holding the cylinder from the top.
Am picturing it right?
 
  • #6
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So the cylinder, in presence of static friction is supposed to go upwards due to the pulling force applied by the cord and goes downwards if the friction is dynamic due to the cord holding the cylinder from the top.
Am picturing it right?

I'm unsure because I don't know exactly what that cord is doing. Have you left out any information about the problem? From what you've given me, I see no reason for the cylinder to roll up the inclined plane. Is the cord in a state of extension at t=0? Is there any mechanism pulling the rope closer to the wall?
 
  • #7
PeroK
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So the cylinder, in presence of static friction is supposed to go upwards due to the pulling force applied by the cord and goes downwards if the friction is dynamic due to the cord holding the cylinder from the top.
Am picturing it right?

Not quite. The cord is effectively constraining the motion of the cylinder forcing it to roll the "wrong" way. I suggest you work out the direction of all the forces and see what you get.

The cylinder must go down the slope, surely?
 
  • #8
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I'm unsure because I don't know exactly what that cord is doing. Have you left out any information about the problem? From what you've given me, I see no reason for the cylinder to roll up the inclined plane. Is the cord in a state of extension at t=0? Is there any mechanism pulling the rope closer to the wall?
That is also my problem, I can't picture what the motion exactly would be.
What you see in the attachment is all I have, I also got the solution in which uses the static friction going down, that I don't get why.
In my view at time t=0 is all at rest, hence you check if the static situation will hold for ever or not; if the cord is ideal then it has no elasticity (extension ) so i think that the cord contact point with the cylinder will always be at rest, seen that hasn't been specified any other kind of actions.
As I am writing I've just realised why the friction goes down.
If the contact point of the cylinder with the plane slides, the only direction of the sliding will be downwards because the cord holds the cylinder from the top, it then follows that the direction of dynamic friction is opposite the motion, in conclusion if the system is at rest the direction of static friction is the opposite.
Some times just speaking helps figure out things :)
 
  • #9
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Not quite. The cord is effectively constraining the motion of the cylinder forcing it to roll the "wrong" way. I suggest you work out the direction of all the forces and see what you get.

The cylinder must go down the slope, surely?

If the contact point of the cylinder with the plane slides, the only direction of the sliding will be downwards because the cord holds the cylinder from the top, it then follows that the direction of dynamic friction is opposite the motion, in conclusion if the system is at rest the direction of static friction is the opposite.

In order to analyse if slides or not you first check if static friction is enough checking that:

|Fs| ≤ |Fs max|
Fs max= μd*N

If the inequality is satisfied the static will hold, if not then in the equations you swap static with dynamic.
 
  • #10
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That is also my problem, I can't picture what the motion exactly would be.
What you see in the attachment is all I have, I also got the solution in which uses the static friction going down, that I don't get why.
In my view at time t=0 is all at rest, hence you check if the static situation will hold for ever or not; if the cord is ideal then it has no elasticity (extension ) so i think that the cord contact point with the cylinder will always be at rest, seen that hasn't been specified any other kind of actions.
As I am writing I've just realised why the friction goes down.
If the contact point of the cylinder with the plane slides, the only direction of the sliding will be downwards because the cord holds the cylinder from the top, it then follows that the direction of dynamic friction is opposite the motion, in conclusion if the system is at rest the direction of static friction is the opposite.
Some times just speaking helps figure out things :)
Just to be clear, are you saying that, when the system is at rest, the force on the cylinder from static friction points down the plane?
 
  • #11
PeroK
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If the contact point of the cylinder with the plane slides, the only direction of the sliding will be downwards because the cord holds the cylinder from the top, it then follows that the direction of dynamic friction is opposite the motion, in conclusion if the system is at rest the direction of static friction is the opposite.

In order to analyse if slides or not you first check if static friction is enough checking that:

|Fs| ≤ |Fs max|
Fs max= μd*N

If the inequality is satisfied the static will hold, if not then in the equations you swap static with dynamic.

Yes. You're getting there. How are you going to calculate the force needed to stop the cylinder moving?
 
  • #12
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Just to be clear, are you saying that, when the system is at rest, the force on the cylinder from static friction points down the plane?
Exactly, picture this:
The cord constrains the cylinder motion, it can roll up if the cord is pulled up and if static friction is enough to avoid the sliding, but considering that the cord has no motion and it is in tension the only possible state of the cylinder is, at rest if static is enough or slide down wards.
 
  • #13
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Exactly, picture this:
The cord constrains the cylinder motion, it can roll up if the cord is pulled up and if static friction is enough to avoid the sliding, but considering that the cord has no motion and it is in tension the only possible state of the cylinder is, at rest if static is enough or slide down wards.

It's possible that we have different meanings of "down the plane". If we don't, then you're implying that the static friction force is antiparallel to that provided by the rope. What then would stop the cylinder from rotating counterclockwise in the "static" case?
 
  • #14
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Yes. You're getting there. How are you going to calculate the force needed to stop the cylinder moving?
I solved as follows:

M*d2X/dt= Mgsinϑ+Fs-T
N=Mgcosϑ
I*dω/dt= R(-T-Fs)

Assuming as positive reference a clockwise rotation.
Then from here you get everything.
My only doubt was on the torque sign for the static force.
 
  • #15
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I solved as follows:

M*d2X/dt= Mgsinϑ+Fs-T
N=Mgcosϑ
I*dω/dt= R(-T-Fs)

Assuming as positive reference a clockwise rotation.
Then from here you get everything.
My only doubt was on the torque sign for the static force.

Look at your rotation equation. Imagine what that equation would imply if the static friction coefficient is very large. Does this reflect what you would expect?
 
  • #16
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It's possible that we have different meanings of "down the plane". If we don't, then you're implying that the static friction force is antiparallel to that provided by the rope. What then would stop the cylinder from rotating counterclockwise in the "static" case?
It's possible that we have different meanings of "down the plane". If we don't, then you're implying that the static friction force is antiparallel to that provided by the rope. What then would stop the cylinder from rotating counterclockwise in the "static" case?
The fact that in the problem hasn't been mentioned that the cord would be pulled up
 
  • #17
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Look at your rotation equation. Imagine what that equation would imply if the static friction coefficient is very large. Does this reflect what you would expect?
Sorry I made a mistake reporting the second cardinal equation, both torques are positive not negative as I have written
 
  • #18
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Sorry I made a mistake reporting the second cardinal equation, both torques are positive not negative as I have written

I'm still having trouble seeing how that rotation equation reflects reality. Disregarding equations for a moment, what would happen if the torque from static friction were equivalent to that provided by the cord in terms of magnitude?

Edit: Deleted something above the quote that I forgot to erase.
 
  • #19
PeroK
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I solved as follows:

M*d2X/dt= Mgsinϑ+Fs-T
N=Mgcosϑ
I*dω/dt= R(-T-Fs)

Assuming as positive reference a clockwise rotation.
Then from here you get everything.
My only doubt was on the torque sign for the static force.

You have got the friction in the wrong direction. Both equations.
 
  • #20
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Let me put it this way: If the


I'm still having trouble seeing how that rotation equation reflects reality. Disregarding equations for a moment, what would happen if the torque from static friction were equivalent to that provided by the cord in terms of magnitude?
It all depends from the inclination of the plane, if too high it will slide if not it stays where it is, the sum of forces is equal to zero hence in such case the requirements for static situation is achieved.
 
  • #21
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You have got the friction in the wrong direction. Both equations.
the x axes is downwards
 
  • #22
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You have got the friction in the wrong direction. Both equations.
Let's not simply state that.

It all depends from the inclination of the plane, if too high it will slide if not it stays where it is, the sum of forces is equal to zero hence in such case the requirements for static situation is achieved.
Ok, so you agree that if the magnitude of the static friction torque equals the magnitude of the torque from the rope, the cylinder will not rotate, yes?
 
  • #23
PeroK
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the x axes is downwards
You mean ##F_s## is negative? In that case, okay.

You just need to relate the two accelerations and you've cracked it. Subject to some messy algebra.
 
  • #24
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Let's not simply state that.


Ok, so you agree that if the magnitude of the static friction torque equals the magnitude of the torque from the rope, the cylinder will not rotate, yes?
absolutely
 
  • #25
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Does your rotation equation reflect this?
 

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