Question

The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W=49000 N. The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical. (a) What is the magnitude of the lift force? (b) Determine the magnitude of the air resistance R that opposes the motion.

Answer 1

Answer:

(a) The magnitude of the lift force is 52144.71 N, approximately.

(b) The magnitude of the air resistance force opposing the movement is 17834.54 N, approximately.

Explanation:

Since the helicopter is moving horizontally at a constant velocity, we can assume that the net force acting on it is zero, then

(a) in the vertical direction we have

$L\cos(20\deg)-W=0\\L=\frac{W}{\cos(20\deg)}=\frac{49000 N}{\cos(20\deg)}\approx \mathbf{52144.71 N}$.

(b) Now horizontally,

$L\sin(20\deg)-R=0\\R=L\sin(20\deg)=52144.71 N\times \sin(20\deg) \approx \mathbf{17834.54 N}.$