Inclined plane static friction

In summary, the problem involves a cord connecting a wall to a cylinder on an inclined plane. The cylinder has a mass of M and a radius of R and is at rest at t=0. The inclination angle is ϑ and the static and dynamic friction coefficients between cylinder and plane are μs and μd. The goal is to find the linear and angular acceleration of the cylinder. There is some confusion about the direction of static friction, but it is determined that the cylinder will roll down the slope and the direction of friction is opposite the motion. The direction of static friction can be determined by checking if it is enough to resist the motion.
  • #1
DottZakapa
239
17

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

 
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  • #2
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
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
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
Kinta said:
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
DottZakapa said:
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
DottZakapa said:
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
Kinta said:
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 realized 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
PeroK said:
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
DottZakapa said:
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 realized 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
DottZakapa said:
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
Kinta said:
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
DottZakapa said:
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
PeroK said:
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
DottZakapa said:
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
Kinta said:
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?
Kinta said:
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
Kinta said:
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
DottZakapa said:
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
DottZakapa said:
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
Kinta said:
Let me put it this way: If theI'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
PeroK said:
You have got the friction in the wrong direction. Both equations.
the x axes is downwards
 
  • #22
PeroK said:
You have got the friction in the wrong direction. Both equations.
Let's not simply state that.

DottZakapa said:
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
DottZakapa said:
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
Kinta said:
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
Does your rotation equation reflect this?
 
  • #26
yep, in fact:
to check the static friction you think that avery point of the system is at rest, hence the center of mass is at rest, which implies zero velocity of center of mass and zero angular velocity.
both equations will be equated to zero, in which from the last you get that Fs=-T from there you substitute in the other, from which you'll get Fs at this point you solve the inequality of Fs max
 
  • #27
DottZakapa said:
yep, in fact:
to check the static friction you think that avery point of the system is at rest, hence the center of mass is at rest, which implies zero velocity of center of mass and zero angular velocity.
both equations will be equated to zero, in which from the last you get that Fs=-T from there you substitute in the other, from which you'll get Fs at this point you solve the inequality of Fs max
Ok, I'm starting to see what you're saying. What I was unaware of was that you were making ##F_s## have the opposite sign of ##T##, which is a messy way of putting the two vectors parallel to each other (both up the plane, not with ##\vec{F}_s## down the plane). Having learned torques, you know of cross products, right?
 
  • #28
Kinta said:
Ok, I'm starting to see what you're saying. What I was unaware of was that you were making ##F_s## have the opposite sign of ##T##, which is a messy way of putting the two vectors parallel to each other (both up the plane, not with ##\vec{F}_s## down the plane). Having learned torques, you know of cross products, right?
no problems with that
 
  • #29
No problems with just cross products or both that and ##\vec{F}_s## being up the plane, not down?
 
  • #30
Kinta said:
No problems with just cross products or both that and ##\vec{F}_s## being up the plane, not down?

I suggest to review :
torque theorem,
I and II law of Newton,
system of points cardinal equations,
you will surely come to the same conclusion.
If the system is at rest which implies static friction (time t=0), we surely have zero total external forces ⇒ total torque zero and velocity zero.
Go through all the equations and you i'll see.
STATIC friction (rolling with no sliding of the body or at rest) downhill and Tension uphill (Opposite isn't possible because tension goes uphill without any doubt ).
The friction uphill is only Dynamical friction.
Is physically and mathematically impossible the other way, unless other assumption like pulling the cord, in that case static friction goes uphill.
 
  • #31
DottZakapa said:
I suggest to review :
torque theorem,
I and II law of Newton,
system of points cardinal equations,
you will surely come to the same conclusion.
If the system is at rest which implies static friction (time t=0), we surely have zero total external forces ⇒ total torque zero and velocity zero.
Go through all the equations and you i'll see.
STATIC friction (rolling with no sliding of the body or at rest) downhill and Tension uphill (Opposite isn't possible because tension goes uphill without any doubt ).
The friction uphill is only Dynamical friction.
Is physically and mathematically impossible the other way, unless other assumption like pulling the cord, in that case static friction goes uphill.

This is what you're claiming for the static case:
BallStringInclinedPlane.jpg

If these force vectors are correct, I just cannot see how that cylinder is going to roll down the inclined plane without "slipping". I also cannot see how this picture could ever possibly imply that the ball would remain stationary if ##|\vec{T}| = |\vec{F}_s|##. Please help me see what I'm missing. When I take the sum of the cross products to get the net torque, ##\vec{\tau}^{net} = \vec{r}_1 \times \vec{T} + \vec{r}_2 \times \vec{F}_s##, I don't see any way to arrive at a zero net torque because the two cross products result in vectors with equivalent directions.

Edit: ##\vec{r}_1## points from the cylinder's radial center to the point at which the cord is no longer in contact with the cylinder. ##\vec{r}_2## points from the radial center of the cylinder to the line along which the cylinder is in contact with the inclined plane.
 
Last edited:
  • #32
Can someone please clarify this issue? I feel like I'm going bonkers over this.
 
  • #33
Kinta said:
Can someone please clarify this issue? I feel like I'm going bonkers over this.
Which issue?

By the way (to the OP): It's so much more inviting to respond to a thread when the actual image is posted rather than a link to some image, especially when a download is necessary.

upload_2015-6-25_14-7-49.png
 
  • #34
SammyS said:
Which issue?

By the way: It's so much more inviting to respond to a thread when the actual image is posted rather than a link to some image, especially when a download is necessary.

View attachment 85205

DottZakapa has come to the conclusion that, if the situation calls for a static frictional force, it is directed down the plane as I tried to illustrate in my previous post. This doesn't make sense to me. I could see how this would be the case in the absence of the rope, but not with its presence. Did my illustration not display properly in my previous post?
 
  • #35
Kinta said:
This is what you're claiming for the static case:
[ ATTACH=full]85184[/ATTACH]
If these force vectors are correct, I just cannot see how that cylinder is going to roll down the inclined plane without "slipping". I also cannot see how this picture could ever possibly imply that the ball would remain stationary if ##|\vec{T}| = |\vec{F}_s|##. Please help me see what I'm missing. When I take the sum of the cross products to get the net torque, ##\vec{\tau}^{net} = \vec{r}_1 \times \vec{T} + \vec{r}_2 \times \vec{F}_s##, I don't see any way to arrive at a zero net torque because the two cross products result in vectors with equivalent directions.

Edit: ##\vec{r}_1## points from the cylinder's radial center to the point at which the cord is no longer in contact with the cylinder. ##\vec{r}_2## points from the radial center of the cylinder to the line along which the cylinder is in contact with the inclined plane.
Yes. You are correct.

With the forces in the directions indicated there is no way for the torque on the ball to be zero. It will have angular acceleration causing rotation in a counter-clockwise sense.

(Sorry. I was a bit late in sorting out the issues! I see them now.)
 
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