Question

Two blocks of equal mass '$m$' are connected by massless string and pulley as shown in figure. Now the blocks are left to move. Then the force exerted by pulley on the clamp is $\frac{3\sqrt{3}mg}{k}$. Find the value $k$?

Answer 1

Let the acceleration of blocks $=a$

from (1)

$mg-T=ma$

For (2)

$T-mg\sin 30=ma$

$\Rightarrow T-mg\sin 30=mg-T$

$2T=mg+mg\sin 30=\frac{3}{2}mg$

$T=\frac{3}{4}mg$

Now FBD of clamp with, pully [ref image 2]

Now force exerted by pully on clamp $=$ resultant of the two tension [Ref. image 2]

$Fresut=\sqrt{T^2+T^2+2\times T\times T\times \cos 60}$

$=\sqrt{3}T$

$F=\frac{3\sqrt{3}mg}{4}$

$k=4$