Inclined plane static friction

  • Thread starter DottZakapa
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  • #26
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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
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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
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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
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No problems with just cross products or both that and ##\vec{F}_s## being up the plane, not down?
 
  • #30
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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
71
5
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
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Can someone please clarify this issue? I feel like I'm going bonkers over this.
 
  • #33
SammyS
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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
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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
SammyS
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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.)
 
  • #36
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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.

Ok...good. I about lost my marbles over that.
 
  • #37
SammyS
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Ok...good. I about lost my marbles over that.
Were they on an incline?
Which way were they rolling? :wink:

Anyway, save those marbles!
 

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