1
$\begingroup$

enter image description here Adapted from JEE advanced paper-1 of 2020

If you see the left side of the Planck which the ball touches, it seems so that as we vary $\theta$, the contribution of the force from that point of contact drops to zero. I have marked the point of interest in paint:

enter image description here

As we reduce $\theta$ , we see both the edges touching the sphere contribute to supporting it's weight. But, as we increase $\theta$ it drops off. It is intuitive to understand for me, but I can't give a precise reason why as to this happens. Hence, the question:

For what exact reason does the normal force shift to the right edge as we vary $\theta$?

Improve this question
$\endgroup$

2 Answers 2

Reset to default
1
$\begingroup$

If we assume that the two sides can exert a force only normal to ball, and that they are the only forces on the ball other than gravity, then this becomes just a linear vector solution.

The two forces must sum to zero out gravity and simultaneously have net zero horizontal force. This can only happen if the "downhill" side exerts more force than the "uphill" side.

Another way to say the same thing is that as gravity pulls the ball down, the forces from the downhill side most directly oppose this force, while the uphill side is not able to act in that direction. So the pull downward results in strong reaction forces on the low side and weak ones on the high side. In the limit where the bottom force acts vertically through the center of mass, there will be no force pulling the ball into the plank and no force from the uphill edge on the ball.

Improve this answer
$\endgroup$
0
$\begingroup$

To simplify first you can visualize what happens with the cavity when incline has an angle zero. The normal reaction on both the edges is same. Direction of normal reaction is vertical upwards opposite to gravity.

When there is incline the sphere is actually also falling towards the right edge. More the θ, more the sphere falls towards the right edge. There will be two components of weight of sphere. One is mgcosθ, other is mgsinθ. mgcosθ is supported by normal components at both the edges. mgsinθ is supported just by the right edge.

So on the right edge you can think of two normal forces - One stopping sphere from falling into cavity and other is just the plank surface stopping forward motion on incline. When θ is 90, sphere is actually not falling into cavity its just hitting the right edge plank

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.