A block of mass $m$ is placed on a wedge of mass $M$ and angle $\theta$. The wedge is placed on a smooth horizontal surface. If the coefficient of friction between the block and the wedge is $\mu$, what horizontal force $F$ must be applied to the wedge so that the block does not slide relative to the wedge?

Solution

1. In the non-inertial frame of the wedge, the block experiences a pseudo-force $ma$ horizontally, gravity $mg$ downwards, a normal force $N$, and friction $f=\mu N$.
2. Resolving forces parallel and perpendicular to the incline for the block, we set the net force to zero: $N\cos \theta -f\sin \theta =mg$ and $N\sin \theta +f\cos \theta =ma$.
3. Substituting $f=\mu N$ into these equations and solving for the acceleration $a$ yields $a=g\frac{\sin \theta +\mu \cos \theta }{\cos \theta -\mu \sin \theta }$.
4. Applying Newton's second law to the entire system of mass $(M+m)$, the required horizontal force is $F=(M+m)a$.
5. Hence the answer is (A).