Question

A block A of mass 3kg and another block B
of mass 2 kg are connected by a light
inextensible string as shown in figure. If the
coefficient of frictionbetween the surface
of the table and A is 0.5. What maximum
mass C is to be placed on A so that the system
is to be in equlibrium?
​

Answer 1

Answer:

Explanation:

Mass of block A = m₁ = 3 kg (Given)

Mass of block B = m₂ = 2 kg (Given)

The co-efficient of friction b/w A & B = μ = 0.5

Calculating the frictional force of block B over A.

Let the frictional force be =  Fx ,then

Fₓ = μ m₂

Fₓ = (0.5) (2)

Fₓ = 1 kg.wt

Calculating the acceleration with which the bodies move over one another.

Acceleration = a = Fₓ / m₂

a = 1.5 / 2

a = 0.75 m/s²

Let F₍max₎ be the maximum force applied on both A and B so that there is no relative motion between both the blocks.

Therefore,

The net force acting on the block combination - F + Fx

F₍max₎ + Fx = m₁₂ x a  

where m₁₂ = 3+2 = 5 kg

Hence, 

F₍max₎ = (5 x 0.5) + 0.75

= 2.5 - 0.75

= 1.75

Therefore, the maximum mass should be 1.75 kg.

Answer 2

Answer:

Hope this helps u mate....,,