Find the resultant of the two perpendicular forces
|F| = sqrt (76.5^2+46.3^2)
tan A = 76.5/46.3
where A is angle north of west
our force is equal and opposite
same |F|
same angle but south of east
Three forces act on a moving object. One force has a magnitude of 76.5 N and is directed due north. Another has a magnitude of 46.3 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.
1 answer