# Torque: stick resting on a bowling ball

1. Mar 3, 2015

### henry3369

1. The problem statement, all variables and given/known data
http://imgur.com/d0kW2Kr [Broken]

2. Relevant equations
Net torque = 0

3. The attempt at a solution
The work is done in the image. I only included the forces that were relevant to the answer.
The answer is N = 0.983 newtons.

I have a feeling that my incorrect answer is due to the incorrect distance from the rotation of axis (the left end of the side) and the force N, but I can't see where I went wrong.

Last edited by a moderator: May 7, 2017
2. Mar 3, 2015

### haruspex

Not sure I understand the question. If there's no friction between stick and ball then the stick exerts a torque about the point of contact of ball with ground. That would seem to ensure the ball will roll, yet it says the stick 'rests'.
Your diagram is strange. The horizontal line for the floor should be tangent to the ball at its lowest point.

3. Mar 3, 2015

### henry3369

It is tangent, I drew it quickly, sorry.

4. Mar 3, 2015

### henry3369

Wouldn't the friction between the ball and the ground prevent it from rolling? Either way, can't the ball be ignored since I'm looking at the forces on the stick? I assume the ball was only used to find the lever arm for the force of the stick on the ball (N in the picture).

5. Mar 3, 2015

### CWatters

Yes I think it's irrelevant if the ball moves. If it does then you can assume that the force you are being asked to calculate is the initial force just before the ball starts moving.

Redraw it again. This error means your calculation includes "0.0892 * 2" which I think is incorrect. I believe it should be "0.0892 + 0.103"

6. Mar 3, 2015

### haruspex

No. If it accelerates the force will be less because of the stick's inertia.

7. Mar 3, 2015

### henry3369

Yeah that is correct. I'm confused why it is 0.0892 + 0.103 though. Isn't 0.0892 the top half so the entire side of the triangle would be 0.0892 *2. I'm starting at my picture and I can't see where I've gone wrong.

8. Mar 3, 2015

### henry3369

Never mind, I see where the 0.103 comes from. Thanks!

9. Mar 3, 2015

### henry3369

I'm trying to do a problem that follows the one given in the problem which says:

Find the horizontal and vertical component of the force exerted on the stick by the floor.

I was able to solve for the vertical component to be 1.25, but I don't know how the normal force can have a horizontal component in this situation. The normal force points straight up (which is the vertical component), and then there is no component of the normal force pointing along the ground. I originally thought it it wanted the force along the stick, but in order for the vertical component to be correct, the x-axis has to be parallel to the ground, while the y-axis perpendicular.

10. Mar 3, 2015

### BvU

N is the normal force. I see N pointing up $\perp$ the stick. It definitely has a component $\perp \vec g$ and the only one that can prevent the stick from moving to the left is the floor!

11. Mar 3, 2015

### haruspex

No. It can prevent sliding, but not rolling.
No. If the ball accelerates then so does the stick (rotationally), and the stick's acceleration implies a net torque on it.
However, since no mass is given for the bowling ball there is no way to solve the question correctly. Either the ball is effectively screwed down or the question is screwed up.
For what it's worth, taking the ball's mass to be 1kg instead of infinite reduces the normal force on the stick by about 3%.

12. Mar 3, 2015

### haruspex

Quite so. But it doesn't ask for the horizontal component of the normal force. It asks for the horizontal component of the total force:
The floor is rough and horizontal. The normal force is therefore vertical. What do we call the horizontal force from such a floor?