Question

Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B ? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between μs and μk.

Answer 1

Given: The mass of body Ais 5 kg, the mass of body B is 10 kg, the applied horizontal force to body Ais 200 N, the coefficient of friction between the bodies and the table is 0.15.

a)

The force of friction acting on the system is given as,

f s =μ( m A + m B )g

Where, the acceleration due to gravity is g, the mass of body A is m A , the mass of body B is m B , the coefficient of friction between the bodies and the table is μ and the force of friction is f s .

By substituting the given values in above formula, we get

f s =0.15( 10+5 )10 =1.5( 15 ) =22.5 N

The net force acting on the partition is given as,

F r =F- f s

Where, the applied force is F and the net force is F r .

By substituting the given values in the above formula, we get

F r =200−22.5 =177.5 N

Thus, the reaction of the partition will be 177.5 N and the direction of force will be the leftward.

b)

The frictional force on mass A is given as,

f A =μ m A g

By substituting the given values in the above formula, we get

f A =0.15( 5 )10 =7.5 N

The net force exerted by mass A on mass B is given as,

F r1 =F- f A

By substituting the given values in the above formula, we get

F r1 =200−7.5 =192.5 N

The acceleration of the system is given as,

a= F r m A + m B

Where, the acceleration of the system is a.

By substituting the given values in the above formula, we get

a= 177.5 5+10 =11.83 m/ s 2

The net resultant force acting on mass A is given as,

F A = m A a

By substituting the given values in the above formula, we get

F A =5×11.83 =59.15 N

The net force exerted by mass A on mass B is given as,

F N = F r1 - F A

By substituting the given values in the above formula, we get

F N =192.5−59.15 =133.35 N

Thus, the force exerted on mass A by mass B is 133.35 N and the direction of force is opposite to the direction of motion.