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 μs and μk.

Asked by Abhisek | 1 year ago |  78

##### Solution :-

Mass of the body, mA = 5 Kg

Mass of the body, mB = 10 kg

Applied Force = 200 N

Coefficient of friction between the bodies and the table μs = 0.15

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

fs = μ(mA + mB)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

FA = μmAg

= 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 = (mA + mB) a

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

$$\dfrac{177.5}{ (5 + 10)}$$ = 177. $$\dfrac{5}{15}$$ = 11. 83 m/s2

Net force causing mass A to move

FA = mA 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.

Answered by Pragya Singh | 1 year ago

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