Question

A block of mass
M1 = 2.7 kg
rests on top of a second block of mass
M2 = 5.5 kg,
and the second block sits on a surface that is so slippery that the friction can be assumed to be zero (see the figure below).

(a) If the coefficient of static friction between the blocks is
μS = 0.21,
how much force can be applied to the top block without the blocks slipping apart?

Incorrect: Your answer is incorrect.
N

(b) How much force can be applied to the bottom block for the same result?

Answer 1

considering the limiting condition \[m1a=F(applied on topblock)-\mu m1 g\] =>\[F= m1(a+\mu g)\] where "a" is th maximum accelaration that can be created whitout slipping between the blocks. and the lower block moves only because of one force that is the ;friction \[a(\max)=\frac{ \mu m1 g }{ m2 }\] finaly \[F=m1 \mu g [\frac{ m1 }{ m2 }+1]\]
so on putting the values we get .085 newton aproximately