A mass of mass `60 kg` is pulling a mass `M` by an inextensible light rope passing through a smooth and massless pulley as shown in figure. The coefficient of friction between the mass and ground is `mu = 1//2` find the maximum value of `M` that can be pulled on the ground

Explanation

`T = Mg (T =` tension in the rope)
`N = 60g - T sin 60^(@)(N` is normal reaction between the man and the ground) and `T cos 60^(@) = mu N`
Solving these three equations with proper substitutions, we get
`M = (120)/(sqrt(3) + 2) kg`