Torque: stick resting on a bowling ball

AI Thread Summary
The discussion centers on a physics problem involving a stick resting on a bowling ball and the calculation of forces acting on the stick. Participants express confusion about the torque and forces involved, particularly regarding the role of friction and the correct calculation of distances in the torque equation. There is debate over whether the ball's movement affects the stick and how to accurately represent the forces in the diagram. The conversation highlights the importance of understanding the relationship between the stick and the ball, as well as the need for precise calculations to determine the horizontal and vertical components of the force exerted by the floor on the stick. Ultimately, clarification on the problem's parameters is needed to arrive at a correct solution.
henry3369
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Homework Statement


http://imgur.com/d0kW2Kr

Homework Equations


Net torque = 0

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.
 
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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.
 
haruspex said:
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.
It is tangent, I drew it quickly, sorry.
 
haruspex said:
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.
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).
 
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.

henry3369 said:
It is tangent, I drew it quickly, sorry.

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"
 
CWatters said:
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.
No. If it accelerates the force will be less because of the stick's inertia.
 
CWatters said:
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"
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.
 
CWatters said:
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"
Never mind, I see where the 0.103 comes from. Thanks!
 
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
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
henry3369 said:
Wouldn't the friction between the ball and the ground prevent it from rolling?
No. It can prevent sliding, but not rolling.
henry3369 said:
Either way, can't the ball be ignored since I'm looking at the forces on the stick?
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
henry3369 said:
there is no component of the normal force pointing along the ground
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:
henry3369 said:
Find the horizontal and vertical component of the force exerted on the stick by the floor.
The floor is rough and horizontal. The normal force is therefore vertical. What do we call the horizontal force from such a floor?
 
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