A helicopter is moving horizontally to the right at a
constantvelocity. The weight of the copter is 53800 newtons. The
lift forcegenerated by the rotating blades make an angle of 21
degrees withrespect to the vertical. a) what is the magnitude of
the liftforce?
B) Determine the magnitude of the air resistance that
opposesthe motion.

Question

contestada

Respuesta :

nuuk nuuk · Answer 1 · 2020-01-09T06:39:41+01:00

Answer:

Explanation:

Given

Weight of helicopter [tex]W=53800 N[/tex]

inclination of blades with vertical [tex]\theta =21^{\circ}[/tex]

Suppose [tex]F_l[/tex] is the lifting Force

Now using F-B-D

[tex]F_{net}=0[/tex] as helicopter is moving with constant velocity

balancing forces in Vertical Force

[tex]F_l\cos \theta -W=0[/tex]

[tex]F_l\cos \theta =W[/tex]

[tex]F_l=\frac{W}{\cos \theta }[/tex]

[tex]F_l=\frac{53800}{\cos 21}[/tex]

[tex]F_l=57,627.6\ N[/tex]

Now sin component of lift is air resistance

[tex]F_l\sin \theta -R=0[/tex]

where R=air resistance

[tex]R=F_l\sin \theta [/tex]

[tex]R=57,627.6\times \sin (21)[/tex]

[tex]R=20,651.88\ N[/tex]