Question

Question 5.34 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.


Fig. 5.21

Chapter Laws Of Motion Page 113

Answer 1

(a) Mass of body A, mA = 5 kg

Mass of body B, mB = 10 kg

Applied force, F = 200 N

Coefficient of friction, μs = 0.15

The force of friction is given by the relation:

fs = μ (mA + mB)g

= 0.15 (5 + 10) × 10

= 1.5 × 15 = 22.5 N leftward

Net force acting on the partition = 200 – 22.5 = 177.5 N rightward



Hence, 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 × 5 × 10 = 7.5 N leftward

Net force exerted by mass A on mass B = 200 – 7.5 = 192.5 N rightward



Net force acting on the moving system = 177.5


Net force causing mass A to move:

FA = mAa

= 5 × 11.83 = 59.15 N

Net force exerted by mass A on mass B = 192.5 – 59.15 = 133.35 N

Answer 2

Given,
mass of body A (Ma) = 5kg
mass of body B ( Mb) = 10 Kg
coefficient of friction between the blodies and table ( u) = 0.15
force applied in horizontal = 200 N
(a)
after applying horizontal force , limiting friction acting to the left = u ×Normal reaction on the floor
= u( Ma + Mb) g
= 0.15 × (5 + 15)× 9.8 = 22.05 N
hence, net Force acting on the partition
= Force applied in horizontal - limiting friction
= 200 N - 22.05 N
= 177.95 N { towards right }
use , Newton's 3rd Law of motion,
The reaction of the partition = - Net Force acting on the partition
= -177.95 N
hence, reaction of the partition 177.95 N towards left.

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(b) Let fa is the limiting force acting on body A and fb is the limiting force acting on body B.
fa = u × Normal reaction on the floor by body A
= 0.15 × Ma× g
= 0.15 × 5 × 9.8
= 7.35 N

Hence, net Force applied on Body B by body A = force applied in horizontal - limiting frictional force acting on body A
= 200 N - 7.35 N
= 192.65 N { towards right }

apply Newton's 3rd law ,
Normal reaction apply by body B on body A = - normal reaction applied by body A on body B
= - 192.65 N { towards left }

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after removing partition , the system of bodies start to move and then kinetic friction act into it .
now, Let (Ma +Mb) is a system which moves with acceleration a
Net acting force toward right = 200 N - 22.05 N
(Ma + Mb) × a = 177.95
a = 177.95/(5 + 10) = 11.86 m/s²

hence, the system of these two bodies moves with 11.86 m/s² .
hence, system are in motion so, answer of (b) will be change .
then, normal reaction on body B by body A = F - (fa + Ma× a )
= 200 N - ( 7.35 + 5×11.86)
= 200 N - 66.65 N
= 133.35 N { towards right }
similarly ,
Normal reaction on body A by body B =-133.35 N { towards left }