Question

An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 7.0 N ; a second force has a magnitude of 4.8 N and points in the negative y direction.

a) Find the magnitude of the third force acting on the object.

b) Find the direction of the third force acting on the object.

Answer 1

a)

Consider the forces in unit vector notation with unit vector "i" pointing in x-direction and unit vector "j" pointing in y-direction.

= First force acting on the object = 7 i + 0 j

= Second force acting on the object = 0 i - 4.8 j

= Third force acting on the object

Given that :

+ + = 0

7 i + 0 j + 0 i - 4.8 j + = 0

= - 7 i + 4.8 j

|| = magnitude of third force = sqrt((-7)² + (4.8)²) = 8.5

b)

= Direction = 180 + tan^-1(4.8/(-7)) = 145.55 deg        CCW from + x-direction