1.

Two bodies A and B of masses 4 kg and 10 kg in contact with each other rest on a table against a rigid partition (see fig). The coefficient of  friction between the bodies and the table is 0.15 A force of 200 N is applied horizontally at A. What are (a) the reaction of the partition, (b) the action-reaction forces between A and B? What happens when the partition is removed? Does the answer to (b) change, when the bodies are in motion? (Ignore the difference between μs and μk)

Answer»

Let m1 and m2 be masses of the bodies A and B respectively. Then

m1 = 5 kg, m2 = 10 kg

The coefficient of friction between the bodies and the table,

μ = 0.15

Force applied on the system of the two bodies,

P = 200 N

(a) If F is the limiting friction, then

F = μ(m1 + m2) g

= 0.15 x (5 + 10) x 9.8 = 22.05 N

Therefore, net force exerted on the partition, when a force of 200 N is applied horizontally on A is given by

P' = P - F

= (200 - 22.05) N = 177.95 N (towards right)

Therefore, the reaction of the partition = 177.95 (towards left)

(b) Force of limiting friction on body A

F1 = μm1g = 0.15 x 5 x 9.8 = 7.35 N

Therefore, net force exerted by body A on B

P1 = P - F1

= (200 - 7.35) N = 192.65 N

Since action and reaction are equal and opposite, reaction of body B on A also 192.65 N

When the partition is removed: In such a situation, the two bodies move under the effect of force.

P' = 177.95 N

If a is acceleration produced in the motion of the system of two bodies. then

a = {P1}/{M1 + M2} = {177.95}/{5 + 10} = 11.86 ms-2

Therefore, force producing motion in the body A, is

m1a = 5 x 11.86 = 59.3 N

Hence, force exerted by body A and B, when partition is removed

= P' - 59.3 = 192.65 - 59.3

= 133.35 N

Also, the reaction of the body B on A is also 133.35 N.


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