IIT JEE , AIEEE Forum - Physics , Mechanics - Laws of Motion-Friction

abhishekray07

let masses of block be A & B be m and 2m where m=4kg

initially when a force of 12 N applied on a A
12- f(max)=0
f(max)=12 N .... (i)

now let the maximum force applied on B be F

let common acceleration of A and B be a
then
a= F/(m+2m) - (ii)
(considering m and 2m as a system)

now looking at the f.b.d of block A,
only frictional force(f) acts

from (i)
f=12 N
Favail=12 N (ii)
and Force required for common acceleration is ma
Freq = m*F/(3m) (iii)

for common accel

Favail>= Freq
12>=F/3
F<=36 N

therefore maximum force is 36 N

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