Two blocks are in contact on a frictionless table. A horizontal force is applied

Question

Two blocks are in contact on a frictionless table. A horizontal force is applied to the larger block m1. (a) If m1 = 2.3 kg, m2 = 1.2 kg, and F = 3.2 N, find the force of contact between the two blocks. (b) Show that if the same force F is applied to m2 rather than m1, the force of contact between the blocks is 2.1 N, which is not the same as the value derived in (a). Explain

so I've found the acceleration a = F/m1+m2    and if b) is 2.1 N then I know a) is

m2 * F / (m1 + m2) which would be 1.1 N    

I'm having trouble wrapping my head around this, I thought newton's 3rd law says the forces should be equal, can anyone help me understand this?

Explanation / Answer

the blocks are different on which you are applying the force,so the reaction will be different.
In the first case you applying on m1 which is 2.3kg,the net force acting on it will give the acceleration of the body i.e (F-N)=m1*and N=m2*a
in second case you are applying on m2
so the equations will be (F-N)=m2*a and N=m1*a
YOU WILL GET A BETTER PICTURE IF YOU DRAW FREE BODY DIAGRAMS AND SOLVE IT
*IN mechanics,it will be easier if you solve the problems by drawing free body diagrams

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