Question

In the adjoining figure, the coefficient of friction between wedge (of mass M) and block (of mass m) is $\mu$.

Find the minimum horizontal force F required to keep the block stationary with respect to wedge.

• A. $F_{min}=2{\mu }_{s}mg\left(\frac{2m+M}{m+M}\right)$
• B. $F_{min}={\mu }_{s}mg\left(\frac{m+M}{2m+M}\right)$
• C. $F_{min}=g\left(\frac{m+M}{\mu }\right)$
• D. $F_{min}={\mu }_{s}mg\left(\frac{2m+M}{m+M}\right)$

Answer 1

The correct option is C $F_{min}=g\left(\frac{m+M}{\mu }\right)$

$FBDof{}^{\prime }{m}^{\prime }withrespecttowedge(real+onepseudoforce)withrespecttowedgeblockisstationary.\therefore \sum F_x=0=\sum F_y\therefore mg=\mu NandN=maa=\frac{g}{\mu }andF=(M+m)a=(M+m)\frac{g}{\mu }$