Question

A helicopter is moving horizontally to the right at a

Answer 1

Answer:

Explanation:

Given

Weight of helicopter $W=53800 N$

inclination of blades with vertical $\theta =21^{\circ}$

Suppose $F_l$ is the lifting Force

Now using F-B-D

$F_{net}=0$ as helicopter is moving with constant velocity

balancing forces in Vertical Force

$F_l\cos \theta -W=0$

$F_l\cos \theta =W$

$F_l=\frac{W}{\cos \theta }$

$F_l=\frac{53800}{\cos 21}$

$F_l=57,627.6\ N$

Now sin component of lift is air resistance

$F_l\sin \theta -R=0$

where R=air resistance

$R=F_l\sin \theta$

$R=57,627.6\times \sin (21)$

$R=20,651.88\ N$