In the figure masses `m_(1),m_(2)` and M are kg. 20 Kg, 5 kg and 50 kg respectively. The co-efficient of friction between M and ground is zero. The co-efficient of friction between `m_(1)` and M and that between `m_(2)` and ground is `0.3.` The pulleys and the string are massless. The string is perfectly horizontal between `P_(1) and m_(1)` and also between `P_(2) and m_(2).` The string is perfectly verticle between `P_(1) and P_(2).` An externam horizontal force F is applied to the mass M. Take `g=10m//s^(2).`

(i) Drew a free-body diagram for mass M, clearly showing all the forces.
(ii) Let the magnitude of the force of friction between `m_(1)` and M be `f_(1)` and that between `m,_(2)` and ground be `f_(2).` For a particular F it is found that `f_(1)=2f_(2).` Find `f_(1) and f_(2).` Write down equations of motion of all the masses. Find F, tension in the string and accelerations of the masses.

Explanation

(a)
(b) We consider the following cases:
(i) All the three blocks are moving and `m_(1)` moves relative to M.
In this case,
`(f_(1))_("max")= mu_(1)N_(1)=mu_(1)m_(1)g=0.3 xx 20xx10=60N`
`(f_(2))_("max")=mu_(2)m_(2)g=0.3xx5xx10=15N`
Given that `f_(1)=2f_(2)" or " f_(2)=f_(1)//2`
The maximum value of `f_(2)` is 15N so `f_(1)` cannot be more than 30N. This case is not possible. So, `m_(1)` remain at rest w.r.t. M.
(ii) All three blocks are at rest.
Now, `F-f_(1)=0, T=f_(1) and T=f_(2)`
So, `f_(1)=f_(2)" " ` (not possible)
(iii) All the three blocks are moving with same acceleration .a..
In this case,
Equation for `M:F-f_(1)=Ma`
Equation for `m_(1):f_(1)-T=m_(1)a`
Equation for `m_(2):T-f_(2)=m_(2)a`
`f_(1)=2f_(2)=15xx2`
Solving, we get `a=0.6m//s^(2), F=60N, T=18N, f_(2)=15N, f_(1)=30N`.