Question -
Answer -
(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
Theforce of friction is given by the relation:
fs = μ (mA + mB)g
=0.15 (5 + 10) × 10
=1.5 × 15 = 22.5 N leftward
Netforce acting on the partition = 200 – 22.5 = 177.5 N rightward
Asper Newton’s third law of motion, the reaction force of the partition will bein the direction opposite to the net applied force.
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
Netforce exerted by mass A on mass B = 200 – 7.5 = 192.5 N rightward
Asper Newton’s third law of motion, an equal amount of reaction force will beexerted by mass B on mass A, i.e., 192.5 N acting leftward.
Whenthe wall is removed, the two bodies will move in the direction of the appliedforce.
Netforce acting on the moving system = 177.5 N
The equation of motion for the system ofacceleration a,can be written as:
Netforce causing mass A to move:
FA = mAa
=5 × 11.83 = 59.15 N
Netforce exerted by mass A on mass B = 192.5 – 59.15 = 133.35 N
Thisforce will act 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., 133.3 N,acting opposite to the direction of motion.