A jumbo jet of mass 13500-kg is flying in windy skies. At some point in time, wh

Question

A jumbo jet of mass 13500-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 41500 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

mass of jet = 13500 N

force in north direction = 41500 N

force at 75 degree south of west = 14900 N

x component of the force = -14900 * cos(75)

x component of the force = -3856.404 N

y component of the force = 41500 - 14900 * sin(75)

y component of the force = 27107.705 N

resultant force = sqrt(3856.404^2 + 27107.705^2)

resultant force = 27380.641 N

force = mass * acceleration

27380.641 = 13500 * acceleration

acceleration = 2.028 m/s^2

direction = tan^-1(27107.705 / 3856.404)

direction = 81.9 degree

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