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.

Answer

(i) 30 (ii) 15 (iii) 60 (iv) 18 (v) 0.6

Explanation

Free body diagram of M is shown in the figure below.

(i) Supposing all the blocks are in motion
`f_(1_("max")) = mu_(1) N_(1) = mu_(1) m_(1) g = 0.3 xx 20 xx 10 = 60 N` and `f_(2_("max")) = mu_(2) N_(2) = mu_(2) m_(2) g = 0.3 xx 5 xx 10 = 15 N`
Friction between `m_(2)` and ground will be maximum which is 15 N. Given that
`f_(1) = 2 f_(2)`, so `f_(1) = 2 xx 15 = 30 N`
The block `m_(1)` can not move on M.
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 given condition
Since `m_(1)` cannot move over the block M so `m_(2)` can't move relative to M, therefore all the blocks move together and `T = f_(1) = 30 N` and `f_(2) = 15 N`

for `m_(1) : 30 - T = 20 a`
for `M : F - 30 = 50 a`
and for `m_(2) : T - 15 - 5a`
After solving these equations we get , `a = (3)/(5) m//s^(2), T = 18 N` and F = 60 N