# Friction generating torque

1. May 2, 2009

### fobos3

http://en.wikipedia.org/wiki/File:Friction_alt.svg" [Broken]

Look at the picture. If the friction force acts as it is shown, it would create torque. The force $$\textbf{F}$$ acts at angle $$0^\circ$$ so it doesn't generate torque. A simple thought experiment shows that if this is true, then the cube would topple over as soon as we apply any force at the centre of mass (which is obviously wrong). Can you clarify this for me, please.

Last edited by a moderator: May 4, 2017
2. May 2, 2009

### Staff: Mentor

The diagram is misleading as it implies that the normal force acts through the center of mass. But it doesn't. As you push with force F, the normal force shifts to exert a counter-torque.

3. May 2, 2009

### Staff: Mentor

I'm not sure I would call it misleading, but definitely a simplification. You'll note that it does not give any dimensions anywhere. That's an indication to you that you should ignore any implications of those dimensions - they aren't part of what is trying to be explained.

Perhaps it would be helpful in trying to avoid the confusion if they applied the force at the base of the square/box to eliminate the torque, but I wouldn't typically consider that necessary for the example to be useful.

 Meh - perhaps that isn't even necessary: a varying normal force due to the torque wouldn't change the basic form of the friction equation at all - you'd still integrate it to get a single normal force and the friction would remain unchanged.

4. May 2, 2009

### elect_eng

I think you are essentially correct, but in real life gravity would exert a stabilizing torque once the block starts to topple (as per DocAl's explanation). Under the right conditions, the stabilizing torque can be insufficient, and the block will actually topple. I just did a simple test pushing my coffee cup with my finger. I can push even below the center of gravity (but not at the base - as russ_watters says, this eliminates the torque) and still seem to see a toppling effect, if the friction is strong enough to keep the cup from sliding. My cup is very tall and narrow which tends to make the effect more noticeable since the moment arm for the gravity counter torque is smaller if the base is shorter.

Last edited: May 2, 2009
5. May 2, 2009

### fobos3

Ok, I get this. But when we apply a force at the top corrner the friction moves tocounter the torque. And the normal force acts at the centre of gravity. I don't understand when the normal force moves and when the friction force moves.

6. May 2, 2009

### elect_eng

If I understand you correctly, I think russ_watters basically answered this in his last edit. The application of a force at the top corner induces a torque both from the applied force and from the reaction (friction force). Assuming that the force is not strong enough to topple the block, there will be a new equilibrium condition. The picture looks the same, but the distribution of the normal force and the friction force is no longer constant over the bottom surface. The far side of the block now has greater normal force and greater friction force. You can now think of the far bottom corner as a fulcrum about which the block wants to turn. It's easier to think about it if you consider the case where the near bottom corner actually comes off the ground. Then, the entire friction force and normal force is at the far bottom corner. Now you can see how gravity can create the counter torque. Gravity pulls the block down through the center of gravity, but the turning point is at the far bottom corner.

Wow, that sounds confusing, but I think it should give you the intuition you need.

Last edited: May 2, 2009
7. May 2, 2009

### Staff: Mentor

Gravity acts at the center of mass, thus exerts no torque about the center of mass. But friction and the normal force do exert torques about the center of mass. (I know what you mean, but I'm trying to be precise.)

8. May 2, 2009

### Staff: Mentor

The line of action of the friction force doesn't change, thus the torque it produces remains the same.
The normal force is a "reaction" force. It shifts to prevent acceleration, as best it can. (Just like when you push a wall, the wall pushes back--up to the point that it crumbles.) It does not simply act through the center of mass. That's why I said your diagram was misleading for understanding torques. (It's perfectly OK if all you care about is translational motion.)

9. May 2, 2009

### Staff: Mentor

Bottom line is you have to draw a free body diagram to see where, exactly, the forces and moments are and where they work. Ie, if the force pushing on the block is applied to the middle, it obviously can't apply a torque about the center of mass. If applied above or below the center of mass of the block, it can.

10. May 2, 2009

### elect_eng

Yes, I agree. Once you start pushing, the normal force redistributes on the bottom surface. So gravity pulls through the center, but normal force does not. But, I'm trying to create the intuitive explanation that the turning point for torque is not the center of gravity, but the far lower corner of the block. I think this only is strictly correct if the near bottom corner lifts off the ground, but at that point it's clear that gravity, acting through the center of gravity of the block, can create the counter torque. Hopefully, this mental picture makes the understanding easier.

11. May 2, 2009

### Staff: Mentor

Exactly. (The shifting line of action of the normal force is a subtle point not generally discussed in intro textbooks.)
Right, but the friction and normal force can certainly exert such torques.

12. May 2, 2009

### Staff: Mentor

Choosing the center of mass as the pivot point for calculating torque is always legitimate. Choosing the lower edge as the pivot is also perfectly fine, as long as the cube is not accelerating. (If it accelerates, the relationship between torques about the edge and the subsequent rotational acceleration is a bit more complicated.)

Since the diagram showed Ff = μN, I assume the block is moving (kinetic friction) and perhaps accelerating.

13. May 2, 2009

### elect_eng

I wasn't implying that it is not legitimate to choose the center of mass as the pivot point (i said intuitive explanation to mean not a strictly correct one). I'm just trying to give some type of explanation that the OP finds helpful. As you say, the lower edge as the pivot point is also perfectly fine, so with that thought, one can view gravity as inducing a stabilizing torque.

I'm not trying to argue about semantics or perfect explanations. I know the experts here can handle that. However, sometimes certain visualization ideas can help someone come to grips with a confusing concept.

Last edited: May 2, 2009
14. May 2, 2009

### Staff: Mentor

We sure know how to beat a topic into the ground, don't we?

15. May 2, 2009

### fobos3

Thanks for the replies. I've found it explained in the next chapter in my textbook (I'm studying for A-levels. This is before uni in the UK).