Explanation:
It is given that,
The weight of the helicopter, W = 52400 N
The lift force L generated by the rotating blade makes an angle of 21.0° with respect to the vertical. The attached figure shows the free body diagram of the helicopter.
(a) Let L is the magnitude of the lift force.
The net force acting in the vertical direction,
[tex]L\ cos\theta-W=0[/tex]
[tex]L=\dfrac{W}{cos\ \theta}[/tex]
[tex]L=\dfrac{52400}{cos(21)}[/tex]
L = 56127.99 N
[tex]L\ cos[tex]L=5.61\times 10^4\ N[/tex]
(b) The net force acting in the x direction is :
[tex]L\ sin\theta-R=0[/tex]
[tex]R=L\ sin\theta[/tex]
[tex]R=5.61\times 10^4\times \ sin(21)[/tex]
[tex]R=2.01\times 10^4\ N[/tex]
Hence, this is the required solution.
We have that for the Question it can be said that the magnitude of the lift force and magnitude of the air resistance R
From the question we are told
The helicopter in the drawing is moving horizontally to the right at a constant velocity. The weight of the helicopter is W=52400 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.
a)
Generally the equation for the lift force is mathematically given as
[tex]F=\frac{w}{cos\theta}\\\\F=\frac{52400}{cos21}[/tex]
F=56128N
b)
Generally the equation for the magnitude of air resistance is mathematically given as
[tex]R=Lsin\theta\\\\R=56128N*sin 21[/tex]
R=20114N
For more information on this visit
https://brainly.com/question/23379286
