# Homework Help: Friction force and maximum torque

1. Aug 11, 2012

### Acut

Hi! I'm struggling with this problem from my Mechanics book. It's actually an Engineering book, but this problem actually seems like an elementary Physics problem - if it's not, someone please move it to the appropriate forum. I tried my best to translate it clearly, but if it's ambiguous or unclear, let me know. Please just indicate what principles/ideas I'm missing here - I really want to work on this by myself as much as I can. I'll come back if I still get stuck and need more help.

1. The problem statement, all variables and given/known data

An horizontal bar, length 2L, is supported by a homogeneous disk, of weight P and radius R. The friction coefficient between the bar and the disk and between the disk and the ground is μ. (I've redrawn the problem from my textbook. Sorry, it's poorly done, but I hope you can still get what the problem is about. The picture is here https://docs.google.com/document/d/1yOec6udxJKYkh58MUUIY1ktfeKcu0OAR5jPHjXKxi4U/edit). The force 2P on the bar is applied in its mid-point.

1) What is the maximum torque that can be applied to the disk, clockwise, so that it still maintains the equilibrium?
2) What is the maximum horizontal force F that can be applied from right to left at the center of the disk, so that it still maintains equilibrium?
3) Applying a force F=μP (still from right to left) at the center of the disk, what is the maximum (clockwise) torque that can be applied to the disk so that it still maintains equilibrium?

1) T = 2RμP
2) F = 2μP
3) T = 3RμP

2. Relevant equations
F = μN and Archimedes' principle.

3. The attempt at a solution
I'm struggling with all three items, which is a shame for me, who used to perform well in Physics. The bar is obviously applying a force P on the disk, so that the normal force between the disk and the ground is 2P. I know that the maximum static friction force that may occur between two surfaces is F = μN, but I really don't see how I can determine if this maximum force is reached in either the top or the bottom of the disk. If one simply supposes that friction is maximum on both surfaces, we would get F = 3μP (μP at the top and 2μP at the bottom of the disk) for the second item, so that clearly isn't the case. What should I do to solve this? I've tried putting some "dummy" horizontal +G and -G at the top and at the bottom, so I had a binary acting on the system (whose torque would be 2RG) and no net extra force on the system, but I reached nowhere in parts 1 and 3.

2. Aug 11, 2012

### pgardn

The bar is massless and the disk touches the bar at the very end, I am assuming.
Would the red not work for the first question also? only you are going to have to multiply by R for torque?

3. Aug 11, 2012

### Acut

Yes, I also assumed that the bar is massless when trying to solve it. And in the original drawing, the bar extends a little bit beyond the disk, but I also guess it should be touching the cylinder on its very end, since there are no indications of the length of the bar extending beyond the disk.

Yup, the red would work for the first question too if my reasoning (maximum friction both at top and bottom) was valid. But, given the answers, it isn't.

4. Aug 12, 2012

### AGNuke

The force on the disc due to rod is not 2P, but P/2 (HINT: Write the relation between the force and Moment of Inertia for rod from the hinge and from its centre). Therefore, the friction on upper side is 1/2μP and on bottom side is 3/2μP. Note that if we are to rotate the disc clockwise, the friction due to rod will act towards left, while the friction due to ground will act towards side.

When balancing torque on the disc along the centre of the disc, T=2μPR.

Last edited: Aug 12, 2012
5. Aug 12, 2012

### Acut

Wouldn't it be simply P instead of P/2? It's analogous to a lever, and we're applying a force 2P right on the middle of the lever. So the force on the disk would be simply P, wouldn't it? There would be a force P upwards both on the hinge and on the disk so that the sum of the torques on the bar is zero and the net force on it is also zero.

6. Aug 13, 2012

### AGNuke

No. You see, torque IS dependent on the moment of inertia of the object. If we are talking from the hinge of the rod (assume its mass to be negligible, but consider it has mass)$$(2P)L=\frac{M(2L)^2}{3}\alpha$$
Now, finding out the torque on the edge of the rod, by finding the angular acceleration about the centre, which is equal to α we found above. I hope you know the moment of inertia of rod from its edge, centre, etc.$$\left (\frac{P}{2} \right )L=\frac{M(2L)^2}{12}\alpha$$
Barring L (the distance where force is applied) we get our force P/2.

7. Aug 13, 2012

### pgardn

Ahhhh....

I see what nuke did.

You could just pretend like 2P is the weight of the rod as well, that it does have mass, and that the earth is applying the force.

This brings up an some interesting stuff about rotation. It can be counter intuitive. This must mean that the vertical reaction force on the pivot is 3/2P which is not apparent if you took the whole thing to act like a table... If instead of a pivot, you had a vertical leg, and the same thing at the disk.

Its as counter intuitive as the acceleration:
The end of the rod actually would accelerate at a rate greater than g also which makes sense in that it is also changing direction while falling, its just not clearly apparent what the acceleration is until you do the math.