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 37700 N due north, and the force from the wind is 13900 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). Hint: The net force is the vector sum of the two given forces. When doing calculations, it is often easiest to add vectors by adding their components.

Explanation / Answer

mass of plane , m = 13500 Kg

force of plane from jet, F1 = 37700 j N

force from wind. F2 = 13900 * ( -cos(75) i - sin(75) )

F2 = -3597 i - 13427 j N

Now , acceleration of the plane is a

m * a = F1 + F2

a = ( 37700 j -3597 i - 13427 j )/13500

a = -0.27 i + 1.798 j m/s^2

acceleration is sqrt(0.27^2 + 1.798^2) at arctan(1.798/0.27) degree

acceleration is 1.82 m/s^2 at 81.45 degree from west

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