Question

Q:: a helicopter of mass 1000kg. rises with a vertical acceleration of 15m per second square. The crew and the passengers weigh 300kg. Give the magnitude and direction of the
a. force on floor by the crew and passengers
b. action of the rotor of helicopter on the surrounding air
c. force on the helicopter due to surrounding air.

Answer 1

a) F = mg, F=force, m=mass, g= gravitational accleration,
F = 300kg *9.8m/s² (downward force due to weight of the passengers)

c)(1000+300)kg * 9.8m/s² = F (upward force due to the lift)

Answer 2

Mass of the helicopter M = 1000 kg
Mass of passenger and crew m = 300 kg.
g=9.8m/s
Acceleration of the helicopter upwards.\[a= 15m/s ^{2}\]

(a) Now acceleration on the mass of m will be (a+g)
Thus force = m(a+g) = 300(15 + 9.8)N
Hence, weight of the passengers and crew combined becomes = 7440N.

(b) The rotor rotates and pushes air from upper side to lower side creating a pressure difference between upper and lower sides, thereby getting an upthrust according to 3rd law of motion.
Obviously the rotors will have to exert a force downwards required to keep it flying by the resulting reaction upthrust against gravitational pull + force reqd. to achieve an acceleration of \[15m/s^2\].
Now total mass of the helicopter: M + m =(1000 + 300)kg = 1300 kg
Thus force becomes (M + m)(g + a) = 1300x(15+9.8) = 32240 N.

(c) Surrounding air will give equal and opposite reaction force on the helicopter to keep it flying with the reqd. acceleration.
Force on air due to rotors = 32240 N = Force on rotor due to air (3rd law) = 32240 N.

hope this helps

Answer 3

what t-rexy has done is correct
i hve shown the free body diagram where WE DO NOT SHOW THE NET FORCE ON BODY
MANIPULATING THE FORCES ACTING ON BODY GIVES THE NET FORCE(15M/S^2*1300)