Question

The force F to be applied on the triangular block of mass M so that the block of mass m placed on it stationary with respect to wedge is

• A. $mgtan\theta$
• B. $(M+m)gtan\theta$
• C. $(M+m)gcos\theta$
• D. $(M+m)gsin\theta$

Answer 1

Answer: (B)

$(M+m)gtan\theta$

$\begin{array}{c}\text{Correct option is}\left(B\right)\\ \text{If block is stationary with respect to}\\ \text{wedge acceleration of both}a=\frac{F}{M+m}\\ \text{Applying pseudo force:}\\ \begin{array}{c}\left(\frac{F}{M+m}\right)m\cos \theta =mg\sin \theta \\ a\cos \theta =\frac{m}{m}g\sin \theta \\ a=g\tan \theta \end{array}\\ \begin{array}{cc}\text{so;}& a=\\ F& =(M+m)g\tan \theta \end{array}\end{array}$