Question

A string of negligible mass going over a clamped pulley of mass $m$ supports a block of mass $M$ as shown in the figure. The force on the pulley by the clamp is

• A. $\sqrt{2}Mg$
• B. $\sqrt{2}mg$
• C. $g\sqrt{(M-m{)}^2+m^2}$
• D. $g\sqrt{(M+m{)}^2+M^2}$

Answer 1

The correct option is D $g\sqrt{(M+m{)}^2+M^2}$

Given,
Mass of pulley = $m$
Mass of block = $M$

Let us suppose, tension in string is $T$.


F.B.D of block -


On applying $\sum F=ma$ along vertical:
$T-Mg=0$ [ Block is at rest, $a=0$ ]
$\Rightarrow T=Mg$ ....(1)

F.B.D of pulley :


Net force on pulley,
$F_n=\sqrt{T^2+(T+mg{)}^2}$
[$\because F=\sqrt{F_1^2+F_2^2+2F_1{F}_{2}cos\theta }$]
$=\sqrt{(Mg+mg{)}^2+M^2{g}^2}$
[ from equation (1) ]
$=g\sqrt{(M+m{)}^2+M^2}$

Therefore, clamp will exert force of equal magnitude $g\sqrt{(M+m{)}^2+M^2}$.