Three horizontal forces are pulling on a ring, at rest. F1 is 100 N at a 45.0° angle, and F2 is 135 N at a 160° direction. What is the y-component of F3?

Answer 1

Since the ring is at rest, the vector sum of F1 + F2 + F3 = 0

That means that the sums of their x-components and their y-components are 0.

The y-component of any force is the force's magnitude multiplied by the sin of its angle.

Therefore the answer to the question is

-(F1 sin(45°) + F2 sin(160°)) = -116.9N