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 .

Answer

A, B, C

Explanation

Given `m_1=20kg`, `m_2=5kg`, `M=50kg`, `mu=0.3` and `g=10m//s^2`
(A) Free body diagram of mass M is

(B) The maximum value of `f_1` is
`(f_1)_(max)=(0.3)(20)(10)=60N`
The maximum value of `f_2` is
`(f_2)_(max)=(0.3)(5)(10)=15N`
Forces on `m_1` and `m_2` in horizontal direction are as follows:

(1) Either both `m_1` and `m_2` will remain stationary (w.r.t. ground) or (2) both `m_1` and `m_2` will move (w.r.t. ground). First case is possible when.
`Tle(f_1)_(max)` or `Tle60N`
and `Tle(f_2)_(max)` or `Tle15N`
These conditions will be satisfied when `Tle15N` say `T=14` then `f_1=f_2=14N`.
Therefore the condition `f_1=2f_2` will not be satisfied.
Thus `m_1` and `m_2` both can't remain stationary.
In the second case, when `m_1` and `m_2` both move
`f_2=(f_2)max=15N`
Therefore, `f_1=2f_2=30N`
Free body diagrams and equations of motion are as follows:

For `m_1:30-T=20a` ...(i)
For `m_2:T-15=5a` ...(ii)
For `M:F-30=50a` ...(iii)
Solving these three equations, we get,
`F=60N`, `T=18N` and `a=3/5m//s^2`.