The friction coefficient between the two blocks shown in figure is μ but the floor is smooth. Assume the entire system is kept inside an elevator coming down with an acceleration a < g.
2m(g-a),
As it is at rest with respect to the elevator
∑Fx=0 Block M,MN=T
∑Fy=0 ⇒T=μm(g−a) .........(2)
N+ ma = Mg
⇒N=m(g-a) .........(1)
F=T+μN
=T+ μm(g-a) ..........(3)
From (2) and (3)
F= 2μm (g - a)
Case(B)
When F=4μm(g-a) For block M
For block m, ∑Fx=Ma1
∑Fx=ma1
The magnitude of acceleration will be the same as they are constrained by the string.
∴F−T−μM(g−a)=Ma1 (For block m)
T - μm(g−a)=Ma1 (For block M)
=F - 2μm(g−a)=(M+m)a1
⇒4μm(g−a)−2μm(g−a)=(M+m)a1⇒a1=2μm(g−a)M+m