Question

Magnitude of $F$ such that block remains stationary with respect to the wedge. All surfaces are smooth.

• A. $2g(M+m)\tan \theta$
• B. $g(m-M)\tan \theta$
• C. $g(M-m)\tan \theta$
• D. $g(M+m)\tan \theta$

Answer 1

The correct option is D $g(M+m)\tan \theta$

We will work from ground frame of reference.
Let us suppose acceleration of system is $a$.


Acceleration, $a=\frac{F}{M+m}$ .............(1)
$[\sum F=m_{total}a]$

F.B.D of block:


On applying, $\sum F=ma$ along horizontal.
$N\sin \theta =ma$ .............(2)
On applying, $\sum F=ma$ along vertical.
$N\cos \theta -mg=0$
$[$ No vertical acceleration, $a=0$$]$
$N\cos \theta =mg$ .............(3)
On dividing equation (2) by (3).
$\tan \theta =\frac{a}{g}$
$\Rightarrow a=g\tan \theta$
$\Rightarrow \frac{F}{M+m}=g\tan \theta$ $[$ from (1) $]$
$\therefore F=g(M+m)\tan \theta$
Magnitude of $F$ to keep block stationary with respect to wedge is $F=g(M+m)\tan \theta$