mechanics question

http://www.goiit.com/posts/list/mechanics-problem-laws-of-motion-6585.htm#35994

Before solving the first question let?s us get a feel for the situation given:

Now here if a force is exerted on the huge cart it feel some acceleration given by the variable a_x. Now what the first question says is that the block m2 shouldn?t slip relative to the huge block m. Now what it means is that the acceleration of the cart and the block in the x direction has to equal. So both the block m2 and the cart have to move with an acceleration a_x in the x direction

Now let?s choose the system to be all the masses in the problem. The acceleration of the system as whole is given by the following equation:

F = (M+m1+m2+m3) a_x------------1

This for the system as a whole here we don?t consider the tension in the strings and the frictional forces as they become internal forces. So the only external force on the system is F
And it gives the acceleration a_x is the x direction.

Now as going by the above explanation the system consisting of mass m2 and m1 have to have the acceleration a_x. Now what is the direction of the frictional force?? Well u have a tension force in the forward direction therefore the frictional should be in ?x direction
Since the blocks are at relatively rest with respect to the cart, the acceleration in the y direction is 0. Therefore

T =m3g

T - mu*(m1+m2)g =(m1+m2)a_x-----2

from 1AND 2 UGET THE answer as
m3g-mu*(m1+m2)g/(m1+m2) *(M+m1+m2+m3)
=F (minimum)
where mu is the coefficient of static friction between M and m2
Substitute the values to get the answer

The second part of the question

The procedure is the same and I am not going into elaborate discussion:

For the block m1;

f_s = mu*m1g
= m1a_x------3
so therefore again using the equation 1 and 3 we get the answer as :
[M+m1+m2+m3] mu*g =F

Both the above answers are dimensionally correct

For the third part u can get the answer by evaluating the above expressions.

---Pinnaka