Question

two blocks A and B each of mass m are placed on a smooth horizontal surface . two horizontal forces F and 2F are applied on both the blocks A and B respectively. Block A does not slide over block B. Then the normal reaction force acting between the blocks is? a)F B)F/2 C)F/(sqrt of 3) d) 3 F answer is 3F. Can anyone guide me how to get there?

Answer 1

After the pieces come in contact the net power would be toward the10 N constrain. Along these lines, the net power will be (10 – 5 ) N = 5 N. 
With some geometry, you will find that the ordinary response on the square on the left hand side by the piece on the righthand side makes a point of 60 .After both the pieces come in contact we can regard the pieces as one framework so the acceleration of both the pieces moves toward becoming F/m = 10 – 5/1 + 1 = 5/2 As the net power on the piece on the left hand side is 5 N toward the power consequently, we get the condition as takes after 
N cos 60 – 5 = m a 
N cos 60 – 5 = 5/2 
N cos 60 = 135/2N/2 = 15/2 
N = 15 
Also for the other block,, the condition framed will be 
10 – N cos 60 = 5/2 
Settle and we get 
N = 15