1. Nov 17, 2007

### Johnny87

Uniform sphere of radius a and weight W is resting on a horizontal ground in contact with a step of height a/5.

coefficient of friction between sphere and the ground is 3/4
coefficient of friction between sphere and the step is µ

A gradually increasing force P is applied to the highest point of the sphere in a direction perpendicular to the edge of the step.

If equilibrium is broken by the sphere rotating about the step (rather than by slipping against the step and the ground) show that this happens when P=W/3.

If, on the other hand, equilibrium is broken by slipping, show that this happens when P=1/6W(3+µ)/(2-µ)

For what range of values of µ is the equilibrium broken by slipping?

Have no clue at all.

2. Nov 17, 2007

### Shooting Star

Write down what clues you do have about friction and the forces in equilibrium in general..

3. Nov 17, 2007

### Johnny87

Ok, I've tried like this:

To find P rotating:
I took the moment about the centre of the sphere and obtained:
P=mu Reaction(step)

To find P slipping:
Balance:
Horiz.: P + mu (4/5) Reaction(step) + (3/4) Reaction(ground) = (3/5) Reaction(step)
Vertical: W = Reaction(ground) + (4/5) Reaction(step) + mu (3/5) Reaction(step)
(Should I take also the moment? If yes, about which point?)

Then I sketched the graph, and equated the two pulls to see where they intersecate (to find the range of mu) but the result wasn't the one I expected.

4. Nov 17, 2007

### Shooting Star

Draw a diagram first. Suppose the circle touches the ground at G, and intersects the edge of the step at E. Notice that the vertical side of the step is not a tangent to the circle. It’s best to take moment about all the forces about E. We have to find the horizontal dist between the E and the line of action of W. (You calculate it to verify.) When P is max and the sphere is just about to rotate about E, then,

P(a+4a/5) = W*3a/5, where 3a/5 is the horizontal dist between G and the bottom of the step => P = W/3.

Can you do the next part?

5. Nov 17, 2007

### Johnny87

I'll try. I'm supposed to take moments and balance, right?
(Sorry for these very stupid questions, but I'm not very good at statics..)

6. Nov 17, 2007

### Shooting Star

That's right. Also, since this is statics, the net force on the body is zero. So, the sum of the horizontal forces and the sum of the vertical forces must individually be zero. Actually, it's valid for the sum in any direction, but generally it's found to be convenient to resolve the forces horizontally or vertically.