Is the normal force constant when lowering a bowling ball on a scale?

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Homework Help Overview

The discussion revolves around the dynamics of a block on a frictionless incline and the implications of lowering a bowling ball on a scale. Participants explore concepts related to momentum conservation and the effects of external forces on the system.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants question the conservation of momentum in the vertical direction and the role of external forces acting on the system. There is a focus on the normal force and its relationship to the weight of the system, with some participants suggesting that the normal force may not remain constant when lowering the bowling ball.

Discussion Status

The discussion is active, with participants raising questions about the assumptions made regarding the normal force and momentum. Some guidance has been offered regarding the implications of lowering the bowling ball, but there is no explicit consensus on the correct interpretation of the forces involved.

Contextual Notes

Participants are considering the effects of quickly lowering an object on the normal force measured by a scale, which introduces complexities in the analysis of the system's dynamics.

ritwik06
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Let there be an incline which is placed on a frictionless surface. A block is placed at th top of the incline. The surface of th incline is also frictionless. The block starts falling down due to gravity. Consider only the incline and the block in the system. Is th momentum conserved in y direction.
 
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What external forces act on this system?
 
Doc Al said:
What external forces act on this system?

Well, if the mass of the incline is M and that of block is m.
The gravitational force =(M+m)g
and the normal reaction from the ground which is equal to the weight of the to bodies. Hence the net force being zero.
 
ritwik06 said:
and the normal reaction from the ground which is equal to the weight of the to bodies.
Sure about that? (Consider what happens to the center of mass of the system.)
 
Doc Al said:
Sure about that? (Consider what happens to the center of mass of the system.)

Thats exactly the problem, The block comes down so it will have momentum in y direction. The incline is sure not to go up. That means momentum is created in the y direction. and the there is no net force in the y direction. How is this possible?
Yes, I think that the normal will also be equal to (M+m)g . Is it wrong?
 
ritwik06 said:
That means momentum is created in the y direction. and the there is no net force in the y direction. How is this possible?
It's not.
Yes, I think that the normal will also be equal to (M+m)g . Is it wrong?
Yes, that's wrong.

Example: Stand on a scale while you hold a bowling ball. The normal force (scale reading) will be (M + m)g. But what if you quickly lower the ball? Will the normal force remain constant?
 
Doc Al said:
It's not.

Yes, that's wrong.
ok.

Doc Al said:
Example: Stand on a scale while you hold a bowling ball. The normal force (scale reading) will be (M + m)g. But what if you quickly lower the ball? Will the normal force remain constant?
Nope the reading must not be constant. This would mean that I can't take the incline and block as one system. Then how will I figure out the magnitude of th normal?
 

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