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.). 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 Solution to (b) change, when the bodies are in motion? Ignore the difference between μ_{s} and μ_{k}.

Asked by Abhisek | 1 year ago | 78

Mass of the body, m_{A} = 5 Kg

Mass of the body, m_{B} = 10 kg

Applied Force = 200 N

Coefficient of friction between the bodies and the table μ_{s} = 0.15

**(a)** The force of friction is given by the relation:

f_{s} = μ(m_{A} + m_{B})g

= 0.15 (5 + 10) x 10

= 1.5 x 15 = 22.5 N, towards left

Therefore, the net force on the partition is 200 – 22.5 = 177. 5 N rightward

According to Newton’s third law, action and reaction are in the opposite direction

Therefore, the reaction of the partition will be 177.5 N, in the leftward direction

**(b)** Force of friction on mass A

F_{A} = μm_{A}g

= 0.15 x 5 x 10 = 7.5 N leftward

The net force exerted by mass A on mass B = 200 – 7.5 = 192.5 N rightwards

An equal amount of reaction force will be applied on mass A by B, i.e., 192.5 N acting leftward

When the wall is removed, the two bodies move in the direction of the applied force

The net force acting on the moving system = 177. 5 N

The equation of motion for the system of acceleration a, can be written as

Net force = (m_{A} + m_{B}) a

a = \( \dfrac{ Net \;force}{(m_A + m_B)}\)

= \(
\dfrac{177.5}{ (5 + 10)}\) = 177. \(\dfrac{5}{15}\) = 11. 83 m/s^{2}

Net force causing mass A to move

F_{A} = m_{A} a = 5 x 11.83 = 59. 15 N

Net force exerted by the mass A on mass B

= 192.5 – 59.15= 133. 35 N

This force acts in the direction of motion. As per Newton’s third law of motion, an equal amount of force will be exerted by mass B on mass A, i.e., 133.3N, acting opposite to the direction of motion.

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