Question

Question 5.27 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.

Chapter Laws Of Motion Page 112

Answer 1

mass of helicopter( M) = 1000 kg
mass of the crew passengers(m) = 300 kg
acceleration of helicopter = 15 m/s^2 (vertically upward)
(a) see the figure ,
R1 is the normal reaction applied by floor on the crew and the passengers.
because , R1 - mg = ma
R1 = m ( g + a)
= 300( 10 + 5)
= 300* 15
= 7500 N { in upwards direction }

(b) action of the rotor of the helicopter on the surrounding air
= (m + M) g + (m + M) a
= (m + M)(g + a)
= (1000+300)*(10+5)
= 1300*25 N
= 32500 N { in downward direction }


(c) according to Newton's third law ,
action and reaction are of equal magnitude and directed in opposite direction so,
force on the helicopter due to the surrounding air = -32500 N
e.g negative sign indicates that force act in upward direction.

Answer 2

Answer:

Here, mass of helicopter, m1= 1000 kg

Mass of the crew and passengers, m2 = 300 kg upward acceleration, a = 15 ms-2 and 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.

Hopes this helps you!!!