Why is Static Friction Oriented Downwards on an Inclined Plane?

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Homework Help Overview

The discussion revolves around a problem involving a cylinder on an inclined plane connected to a cord. The original poster questions the orientation of static friction, particularly why it is considered to act downwards along the incline.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the reasoning behind the direction of static friction, questioning how it relates to the forces acting on the cylinder and the role of the cord. Some express uncertainty about the setup and seek clarification on the forces involved.

Discussion Status

There is an ongoing exploration of the forces at play, with some participants suggesting that the cord constrains the motion of the cylinder. Several have acknowledged the need to analyze the forces to understand the direction of static friction better, while others are still grappling with the implications of the cord's tension and the conditions for static versus dynamic friction.

Contextual Notes

Participants note that the problem lacks certain details, such as the state of the cord at time t=0 and whether it is in tension. There is also a discussion about the conditions under which static friction would hold versus when it would transition to dynamic friction.

  • #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:
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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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