Question

A block of mass m, attached to a fixed position O on a smooth inclined wedge of mass M, oscillates with
amplitude A and linear frequency f . The wedge is located on a rough horizontal surface. If the angle of the
wedge is 60⁰, then the force of friction acting on the wedge is given by (coefficient of static friction $=\mu$

• A. $\mu (M+m)g$
• B. $\frac{1}{2}m{\omega }^{2}Asin\omega t$
• C. $\mu (M+m){\omega }^{2}Asin\omega t$
• D. $\mu \left[\right(M+m)g+\frac{\sqrt{3}}{2}m{\omega }^{2}Asin\omega t]$

Answer 1

The correct option is B $\frac{1}{2}m{\omega }^{2}Asin\omega t$

The small block oscillates along the inclined plane with an amplitude A. As a result the centre of mass of the system undergoes SHM along the horizontal direction:
$x_{cm}=\frac{mAsin\omega t}{m+M}cos{60}^{\circ }=\frac{1}{2}\frac{m}{m+M}Asin\omega t$
The acceleration of the C.M is $a_{cm}=-{\omega }^2{x}_{cm},$along the horizontal while the net horizontal force is $=(M+m)a_{cm},$which is equal to the force of friction acting on it.