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Question
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s–2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the
(a) force on the floor by the crew and passengers,
(b) action of the rotor of the helicopter on the surrounding air,
(c) force on the helicopter due to the surrounding air.
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Solution 1
(a) Mass of the helicopter, mh = 1000 kg
Mass of the crew and passengers, mp = 300 kg
Total mass of the system, m = 1300 kg
Acceleration of the helicopter, a = 15 m/s2
Using Newton’s second law of motion, the reaction force R, on the system by the floor can be calculated as:
R – mpg = ma
= mp(g + a)
= 300 (10 + 15) = 300 × 25
= 7500 N
Since the helicopter is accelerating vertically upward, the reaction force will also be directed upward. Therefore, as per Newton’s third law of motion, the force on the floor by the crew and passengers is 7500 N, directed downward.
(b) Using Newton’s second law of motion, the reaction force R’, experienced by the helicopter can be calculated as:
R' - mg = ma
=m(g+a)
=1300(10+15) = 1300 xx 25
= 32500 N
The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.
(c) The force on the helicopter due to the surrounding air is 32500 N, directed upward.
Solution 2
Here, mass of helicopter, m1= 1000 kg
Mass of the crew and passengers, m2 = 300 kg upward acceleration, a = 15 ms-2and g = 10 ms-2
(a)Force on the floor of helicopter by the crew and passengers = apparent weight of crew and passengers
= m2 (g + a) = 300 (10 + 15) N = 7500 N
(b)Action of rotor of helicopter on surrounding air is obviously vertically downwards, because helicopter rises on account of reaction to this force. Thus, force of action
F = (m1+ m2) (g + a) = (1000 + 300) (10 + 15) = 1300 x 25 = 32500 N
(c)Force on the helicopter due to surrounding air is the reaction. As action and reaction are equal and opposite, therefore, force of reaction, F = 32500 N, vertically upwards.
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