Fn = 62.6
Fw = 45.9
|F| = sqrt (62.6^2+45.9^2) = 77.6
angle north of west = tan^-1 (62.6/45.9)
= 53.8 degrees
so to cancel that
77.6 at 53.8 south of east
Three forces act on a moving object. One force has a magnitude of 62.6 N and is directed due north. Another has a magnitude of 45.9 N and is directed due west. What must be (a) the magnitude and (b) the direction of the third force, such that the object continues to move with a constant velocity? Express your answer as a positive angle south of east.
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