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July 30, 2014

July 30, 2014

Posted by **Corin** on Monday, November 12, 2007 at 10:51pm.

An object in equilibrium has three forces exerted on it. A 33 N force acts at 90 degrees from the x-axis and a 44 N force acts at 60 degrees. What are the magnitude and direction of the third force?

I tried drawing a triangle, but I don't know what to solve for. I put 33 N as the vertical line and 44 N 30 degrees off from this line.

I'm so confused!!!

Thanks to anyone that helps! :-)

- Physics -
**drwls**, Tuesday, November 13, 2007 at 12:14amLet Fx be the x component of the third force and Fy be the y compnent of that force.

Set the sum of the forces along the x and y axes equal to zero.

Fx + 33 + 44 sin 60 = 0

Fy + 44 cos 60 = 0

Now you can solve for the two components Fx and Fy

The magnitude of the third force is

F = sqrt (Fx^2 + Fy^2)

and the angle to the x axis is arctan Fy/Fx.

You could also do it by drawing a triangle with the force vectors as arrows placed end to end. The third force arrow would have to close the triangle. But we can't do graphics here. The answer would be the same.

- Physics -
**Corin**, Tuesday, November 13, 2007 at 12:50amThank you!

- Physics -

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