Question

two blocks A & B of masses 5kg & 3kg respectively rest on a smooth horizontal surface with B over A.The co-efficient of friction b/w A & B is 0.5 .The maximum horizontal force that can be applied to A, so that there will be motion of A and B without relative slipping,is ??

Answer 1

Given Data:
Mass of block A = m₁ = 5 kg
Mass of block B = m₂ = 3 kg
The co-efficient of friction b/w A & B = μ = 0.5
We have to find the maximum horizontal force that can be applied to A, so that there will be motion of A and B without relative slipping.
Solution:
First we calculate the frictional force of block B over A.
Let it be Fₓ ,then
Fₓ = μ m₂
Fₓ = (0.5) (3)
Fₓ = 1.5 kg.wt
Now we calculate the acceleration with which the bodies move over one another.
Acceleration = a = Fₓ / m₂
a = 1.5 / 3
a = 0.5 m/s²
Let F₍max₎ be the maximum force which is applied on both A and B such that there is no relative motion between both the blocks.
Therefore,
F + Fₓ is the net force acting on the block combination.
F₍max₎ + Fₓ  = m₁₂ x a  
where m₁₂ = 5+3 = 8 kg
Hence, 
F₍max₎ = (8 x 0.5) + 1.5
F₍max₎ = 4 - 1.5
F₍max₎ = 2.5 kg.wt
which is the required answer.
Hope it will help you. Thanks.

Answer 2

Explanation:

Block B is over block A, so normal force between the blocks will be weight of block B. =30N.

so maximum friction force that can be generated is f=μ.N=0.5×30=15N

so maximum acceleration of block B is f/m= 15/3 =5m/s

so minimum acceleration of both the blocks can be 5m/s if both blocks don't want to move together.

so minimum force that required for relative motion is m.a=8×5=40N

means force should be less then 40N, converting to (in kg wt.)(divide by 10) we get maximum force is less then 4kg wt