A jumbo jet of mass 21500-kg is flying in windy skies. At some point in time, wh

Question

A jumbo jet of mass 21500-kg is flying in windy skies. At some point in time, when the airplane is pointing due north, the wind is blowing from the north and east. If the force on the plane from the jet engines is 41300 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 counter-clockwise).

Explanation / Answer

Force component of plane Fxp = 0 & Fyp = 41300 N

Force component of wind Fxw = - 14900 cos 75 = -3856.4 N

Fyw = - 14900 sin 75 = -14392.3 N

The net force on the plane is Fp + Fw = (Fxp+Fxw ) + (Fyp+ Fyw)

= -3856.4 i + 26908 j

the net force is F = (3856)^2 + (26908)^2 = 27197 N

Acceleration is a = F/ m = 27183 / 21500 = 1.264m/s^2

Direction theta = tan^-1( - 26908/3856) = -81.84o

Direction with respective west is 81.84 in clock wise direction (positive).

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