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 $\mu$, find the horizontal force $F$ that must be applied to the wedge so that the block does not slide relative to the wedge.

Solution

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