Forces acting on a ring. (Help with direction)

1. Jul 14, 2007

RocketSurgery

1. Problem: The following coplanar forces pull on a ring: 200 N at 30 degrees, 500N at 80 degrees, 300N at 240 Degrees, and an unknown force. Find the magnitude and direction of the unknown force for the ring if the ring is in equilibrium.

2. Solution(So Far): Using the line 0 Degrees as the x-axis and 90 Degrees as the y-axis I got 351 N for the force (correct according to the book) however I am unable to find what angle it should be at. For the angle I got 71.7 Degrees [The book lists the answer as 71.7+180=351]. After seeing this answer I thought about it and I understand turning it 90 Degrees but why 180 degrees?

2. Jul 14, 2007

Staff: Mentor

How did you find the force to be 351 N at 71.7 degrees? Did you add the three given forces to get that answer? If so, remember the unknown force must be equal and opposite to that sum in order to produce equilibrium. To find the direction opposite to an angle, add 180 degrees. (Not 90!)
I presume this is a typo on your part?

3. Jul 14, 2007

RocketSurgery

No the book lists the answer as 351 N is that wrong? I understand why the angle needs to be opposite but is the force also wrong? If you mean the direction of the vector than that is what they asked for the angle for.

I know you need to add 180 to put it in equilibrium. I added 90 because it was 71.7 relative to the y axis instead of relative to the x axis so the angle with respect to the x axis is 161.7 so I got confused (I should have drew my diagram with all angles w/respect to the x-axis).

Last edited: Jul 14, 2007