Question

In the arrangement shown in 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 block of mass 2m. The force of friction between m and 2m is (pulley and string are
ig frictionless)​

Answer 1

$\huge\fcolorbox{black}{teal}{Solution:-}$

★ Answer:-

→ μ = 1/4

★ Explanation:-

→ Let a be the downwards acceleration of the suspended block. Then, as per the Newton's second law of motion, we get,

ma = mg - T

And for the (2m + m) system:

=) T = (2m + m)a

From the above two equations, we can easily calculate that a= 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 pseudo 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 = mg/4

=) μmg = mg/4

=) μ = 1/4

_______________________________________

I hope this helps! :)