Question

A 21500 kg jet airplane is flying through some high winds. At some point in time, the airplane is pointing due north, while the wind is blowing from the north and east. If the force on the plane from the jet engines is 35100 N due north, and the force from the wind is 14900 N in a direction 75.0° south of west, what will be the magnitude and direction of the plane's acceleration at that moment? Enter the direction of the acceleration as an angle measured from due west (positive for clockwise, negative for counterclockwise).

Answer 1

Answer:$\theta =79.47 North\ of\ west$

Explanation:

Given

Mass of airplane =21500 kg

Force due to jet engines=35100 N

Force from wind is 14900 N at angle of 75 south of west

Resolving forces

$F_x=14900cos75=-3856.403 N$

$F_y=35100-14900sin75$

$F_y=35100-14392.29=20707.70 N$

Therefore acceleration in x direction

$a_x=\frac{F_x}{m}=\frac{-3856.403}{21500}=-0.179 m/s^2$

$a_y=\frac{F_y}{m}=\frac{20707.70}{21500}=0.963 m/s^2$

Direction of acceleration

$tan\theta =\frac{a_y}{a_x}=\frac{0.963}{0.179}=-5.38$

$\theta =79.47\ North\ of\ west$