Question

a system of masses is shown in the figure with masses and coefficient of friction indicated. the frictional force between a and b​

Answer 1

Answer:

Whole system moves with acceleration a

which is given by

a=20+30+40(F−uN)=(90F−88.2)ms−2........(1)

u=coefficient of friction between A and ground 

N= Normal rxn force b/w ground and A

For no slip condition of block A

mAa<F    [F=frictionless force b/w A and B]

ma<mAg...........(2)

using eq (1) and eq (2)

F−88.2<umg⇒F−88.2<90×0.1×98

⇒F−88.2<88.2⇒F≤176.4N

hope so it helps u