# Torque and friction

1. Nov 6, 2007

### skateza

A rod of mass M rests vertically on the floor, held in place by static friction. IF the coefficient of static friction is $$\mu_{s}$$, find the maximum force F that can be applied to the rod at its midpoint before it slips.

I'm not exactly sure what i am suppose to be looking for.. Obviously there will be a FBD of the rod and friction will oppose the force, i need to find the maximum force of static friction and than anything greater than that will push the rod over. Where do i start with this?

2. Nov 6, 2007

### PhanthomJay

The rod is going to rotate but not necessarily slip if F is small enough. You must look at the torque and resulting acceleration of the rod, then apply Newton 2 for the translational motion of the rod's center of mass. You need to find I of the rod also before you begin.

3. Nov 6, 2007

### skateza

so are you saying i would use:

$$\alpha = LF/I = LF/(1/3)ML^{2}$$
than
$$a= \alpha L$$
than
$$F = m[ \alpha L + \mu_{s}g]$$

4. Nov 6, 2007

### PhanthomJay

good, you are almost there; however, the load is applied at midpoint...adjust your torque and the tangential aceleration of the center of mass accordingly.

5. Nov 6, 2007

### skateza

Ok i did that and i got as an answer

$$F = (3F/2) + \mu_{s}Mg$$

is that right?

(I'm trying to think through this verbally here:)
I can't see how its possible for a force to depend on the force itself?... or can i re-arrange the formula and than i would get:

$$F = (-2 \mu_{s}Mg)$$

But why the negative?

Last edited: Nov 6, 2007
6. Nov 7, 2007

### PhanthomJay

I don't know what you did here...your original approach was good, its just that you forgot thet the torque was FL/2 and not FL; and the acceleration was (alpha)L/2 and not (alpha)L. Redo and resubmit, please.