A horizontal force F is applied on a uniform sphere of radius r and weight W and as a result, the sphere slides over the horizontal table. If the coefficient of friction between the sphere and the table is mu, then show that h =r (1- (muW)/(F)).

A horizontal force F is applied on a uniform sphere of radius r and we

1.	A horizontal force F is applied on a uniform sphere of radius r and weight W and as a result, the sphere slides over the horizontal table. If the coefficient of friction between the sphere and the table is mu, then show that h =r (1- (muW)/(F)).
Answer» Solution :The point of contact between the table and the sphere is Q. Here, the FORCE of friction f acts along the surface of the table through the point Q opposite to the force F. The weight of the sphere W acts vertically DOWNWARDS through the CENTRE of the sphere. Since the sphere slides over the horizontal table it has no rotational motion. So, with respect to the centre O, `fxxr = F(r-h)` or, `mu WXXR =F(r-h)` `h= r(1-(MUW)/(F))`.