Two bodies A and B of masses 5 kg and 10 kg in contact with each other

1.	Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wail (Fig). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are1. the reaction of the partition 2. the action-reaction forces between A and B? 3. 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. As the blocks are at rest against the rigid walls, reaction of the partition = – (force applied on A) = 200 N towards left. 2. The action-reaction force between A & B are 200 N each. 3. when the wall is removed, the pushing force gives acceleration to the system. On taking the coefficient of friction into account, 200 – µ (m₁ + m₂) g = (m₁ + m₂) a a = \(\frac{200-0.15(5+10)\times10}{5+10}\) = 11.8 ms^-2 Let the force exerted by A on B be F_BA.On considering the equilibrium of the only block A, 200 – fk₁ = m₁ a + F_BA F_BA = 200 – µ m₁ g – m₁ a = 200 – 7.5 – 59 = 133.5 N towards left.

Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wail (Fig). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are1. the reaction of the partition 2. the action-reaction forces between A and B? 3. 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. As the blocks are at rest against the rigid walls, reaction of the partition = – (force applied on A)

= 200 N towards left.

2. The action-reaction force between A & B are 200 N each.

3. when the wall is removed, the pushing force gives acceleration to the system. On taking the coefficient of friction into account,

200 – µ (m₁ + m₂) g = (m₁ + m₂) a

a = \(\frac{200-0.15(5+10)\times10}{5+10}\)

= 11.8 ms^-2

Let the force exerted by A on B be F_BA.On considering the equilibrium of the only block

A, 200 – fk₁ = m₁ a + F_BA

F_BA = 200 – µ m₁ g – m₁ a

= 200 – 7.5 – 59

= 133.5 N towards left.