In the figure , masses `m_(1) , m_(2)` and M are 20 kg , 5 kg and 50 kg respectively . The coefficient of friction between M and ground is zero . The coefficient of friction between `m_(1)` and M and that between `m_(2)` and ground is`0.4` . 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 vertical between `P_(1)` and `P_(2)` . An external horizonal force F is applied to the mass M . (Take g = `10 m //s^(2))`
(a) Draw a free body diagram for mass M , clearly showing all the forces .
(b) 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 equation of motion of all the masses . Find F , tension in the string and acceleration of the masses .

Explanation

(a) The free body diagram showing all forces acting on M is shown in Fig. 7.55.

(b) The maximum static sliding frictional force on `m_(1)`,
`f_(1) = mu m_(1) g`
`= 0.3 xx 20 xx 10`
= 60 N and on `m_(2)`,
`f_(2) = mu m_(2) g = 0.3 xx 5 xx 10 = 15 N`
For given F , ` f_(1) = 2f_(2)`
`therefore` Maximum value of `f_(1)` can be 30 N as
`(f_(2))_("max") = 15 N`
This implies that `m_(1)` cannot slide on M but moves along with it with same acceleration , as pseudo force is less than maximum , static frictional force . So , all three blocks are moving with common acceleration a . The forces acting on `m_(1) , m_(2)` and M in the direction of motion are as shown in Fig. 7.56.

Applying Newton's 2nd law of motion ,
`30 - T = 20a " " ....... (i)`
`T - 15 = 5a " " ..........(ii)`
`F - 30 = 50 a " " ........ (iii)`
Adding eqns. (i) and (ii) , 15 = 25 a
`a = 0.6 m//s^(2)`
Substituting for a in eqns . (i) and (iii) ,
`T = 30 - 20 xx 0.6 = 18 N`
and `" " F = 30 + 50 xx 0.6 = 60 N`