Ok sir. On equating the forces in the vertical direction, I get:
$$\mu N_1+N_2\cos\theta=W$$
Similarly for the horizontal direction:
$$N_1=N_2\sin\theta$$
From these two, we get:
$$N_1=\frac{W}{\mu+\cot\theta}$$
Or in other words, the normal contact force exerted by the wall depends on the coefficient of friction. As $\mu$ increases, $N_1$ decreases and vice versa.
We know that the coefficient of friction is independent of the normal contact force. But here the normal contact force depends on coefficient of friction. What is the reason for this?
But the effect of the frictional force is to reduce the weight of the ball because it applies an upwards force that counteracts the downwards force due to gravity.
But if the (net) weight of the ball is reduced then the force with which it presses against the wall is also reduced.
Ok sir. I understand the apparent weight is less than the true weight due to the upward frictional force. But I think here we need to compare the degree of roughness of the wall and the normal contact force.
Or in other words, how can the wall repel away the ball when the roughness (coefficient of friction) increases?
Suppose the friction was very high then as soon as the rotating ball touched the wall it would fly upwards so the (net) weight would be zero. If the ball is not in contact with the slope then there is nothing pressing it against the wall so the normal force is zero.
I think the problem with my comparison of the normal force and coefficient of friction is here there are other factors disturbing these two quantities where under normal conditions only these two are present.
Particle Accelerator
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