Question

In the arrangement shown in the figure, there is friction between the blocks of masses m and 2m which are in contact. The ground is smooth. The mass of the suspended block is m. The block of mass m which is kept on mass 2m is stationary with respect to a block of mass 2m. The force of friction between m and 2m is (pulleys and string are lig frictionless):

• A. $\frac{mg}{2}$
• B. $\frac{mg}{\sqrt{2}}$
• C. $\frac{mg}{4}$
• D. $\frac{mg}{3}$

Answer 1

Answer: (C)

$\frac{mg}{4}$

Let $a$ be the downwards acceleration of the suspended block. Then, according to Newton's 2nd law:

$ma=mg-T$

And for the $(2m+m)$ system:

$T=(2m+m)a$

From the above two equations, we can easily calculate that $a=\frac{g}{4}$.

Now consider the non-inertial reference frame of the $2m$ block. It accelerates towards the right with acceleration $a$, so objects in that frame experience a psuedo force equal to their mass times the acceleration of the frame, in the opposite direction. Thus, according to the FBD given:

$f-ma=0$, since the mass is at rest in this frame.

$\therefore f=ma=\frac{mg}{4}$

$\to QED.$