Question

A small disk being pulled by a constant horizontal force F on an inclined plane moves with a constant velocity v as shown in the figure. The angle of inclination of the plane from the horizontal is $\alpha$ and angle between the velocity vector and the force vector is $\beta$.Find the coefficient of kinetic friction between the disk and the plane.

Answer 1

If disk move with constant velocity then net force is zero on the disk.

Limiting value of friction is $f$

$f=\mu mg\cos \alpha$

If net force on disk is zero then

$f=\sqrt{(F\sin \beta {)}^2+(mg\sin \alpha +F\cos \beta {)}^2}$

$\mu mg\cos \alpha =\sqrt{(F\sin \beta {)}^2+(mg\sin \alpha +F\cos \beta {)}^2}$

$\mu =\frac{\sqrt{(F\sin \beta {)}^2+(mg\sin \alpha +F\cos \beta {)}^2}}{mg\cos \alpha }$