Why is Static Friction Oriented Downwards on an Inclined Plane?

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The discussion centers on understanding the direction of static friction on a cylinder resting on an inclined plane connected to a cord. Participants explore why static friction is oriented downwards, concluding that it opposes the potential motion of the cylinder due to the tension in the cord. The conversation emphasizes analyzing forces and torques to determine the cylinder's behavior, particularly under static conditions. It is clarified that if the cylinder is at rest, static friction must act down the slope to counteract the upward force from the cord. Ultimately, the group arrives at the understanding that the direction of static friction is dictated by the need to prevent sliding, confirming its downward orientation on the incline.
  • #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.
 
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  • #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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  • #36
SammyS said:
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
Kinta said:
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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