In the figure masses `m_1`, `m_2` and M are 20 kg, 5kg and 50kg 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 0.3. The pulleys and the strings 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 horizontal force F is applied to the mass M. Take `g=10m//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 equations of motion of all the masses. Find F, tension in the spring and acceleration of masses.

Explanation

a. Free- body diagram of `M` is shown in Fig

b. Supposing all the blocks are in motion
i. `f_(max) = mu_(2) N_(1) = mu_(1) m_(2) g = 0.3 xx 20 xx 10 = 60N`
and `f_(max) = mu_(2) N_(2) = mu_(2) m_(2) g = 0.3 xx 5 xx 10 = 15N`
Friction between `m^(2)` and ground will be maximum , which is `15 N` given than `f_(1) = 2f_(2)` so `f_(1) = 2 xx 15 xx 30N`
The blocks `m_(1)` connot move on `M`.
ii. Let all the blocks are at rest , then

For `M :F - f_(1) = 0 `
For `m_(1) :T - f_(1) = 0` and
For `m_(2) :T - f_(2) = 0`
which gives ` f_(1) = f_(2)` which does not satisfy the gives condition
iii. Since `m_(1)` cannot move over the blocks `M , m_(2)` cannot move relative to `M : ` therefore all the blocks move togather and `T = f_(1) = 30N` and `f_(2) = 15 N`

For `m_(1): 30- T = 20a`....(i)
For `M:F - 30 = 50a`....(ii)
For `m_(2) :T - 15= 5a`....(iii)
After solving these equation, we get
`a = (3)/(5)ms^(-2) , T = 18 N , F = 60N`