Posted by Corin on .
This is probably a really easy question, but I just don't know how to set it up.
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! :-)
Let 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.
The answer is 74N. You can get this answer by breaking down 44N. 33N at 0 degrees is already broken down; the x component is zero and the y component is 33N so for 44N draw it out. Once you draw it out find the sides which should equal out to 22.2N for the x component and 38N forthe y component. Once you have these sides add up ONLY the x components which should give you 22.2. Now add up the why components which should give you 71N (33+38). Now you know that equilibrium should equal zero, so x components and y components should be zero.In order for them to zero the third force added must be the negative of the numbers added up (-71 and -22.2). Now draw out a triangle with the sides 22.2 and 71, in the negative direction (south and west) then solve for the hypotenuse and then find the angle, and you will get 74N and angle 72 degrees or 18 degrees!