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# Newton's laws of motion: Maximum load on a helicopter

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Please see the attached file and solve number 5 only. You do not have to plot the graph.

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The area of the wind stream will be A = πR2 = πD2/4 = 63.62 m2
The air speed v = 24 m/s
The density of air ρ = 1.21 kgm-3
Now as the stream is having area and speed v the rate of volume flow is given by
dQ/dt = A*dx/dt = A*v
Hence the mass of the air pushed per second is given by
dm/dt = A*v*ρ
hence the momentum imparted to the air per second is
dp/dt = (dm/dt)*v = A*v2*ρ
Thus from Newton's second law of motion force applied on air in downward direction by the fan is given by
F = dp/dt = A*v2*ρ
From the third law we know that the equal and opposite force will act on the fan by the air, which is the lifting force, hence the lifting force on the helicopter will be
F = A*v2*ρ = 63.62*242*1.21 = 44340.6 N = 44.3 kN
Hence the additional force can be lifted will be
F - W = 44.3 - 15 = 29.3 kN.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!