How Do You Calculate the Third Force in Static Equilibrium?

[starfish101]

Homework Statement

I have been given two forces and have to calculate the magnitude and direction of the third force that will result in equilibrium. The two forces I have been given are:

F1 = 2.156N at 48 degrees
F2 = 4.508N at 203 degrees

Homework Equations

I used both the cosine law (c^2 = a^2 + b^2 -2abcosC) and the sine law (sin A/a = sin C/c) to solve the problem.

The Attempt at a Solution

I started by drawing out the triangle with all the forces and angles. I then used the equation c^2 = a^2 + b^2 -2abcosC and got an answer of 2.7117N. Next, I used the sine law to find that the angle is 19.634 degrees. I think I have to calculate theta next, but I am really confused on how to do that. I am not completely sure that I have drawn my triangle correctly either. Any help would be much appreciated. Thanks in advance.

 

[Awatarn]

Solving problem is not that hard. You may use the cosine law and the sine law, but it might be lots of algebra. For me, I would like to write components of force in x and y direction. The force in each component must be canceled ( forces are balance)

So you have two unknown variables ( F3 and its angle) with two equations (forces in x and y direction), you can solve it.

Is it clear?

 

[starfish101]

I think I know what you mean. I just redid the equation by breaking the two vectors up into two right angle triangles. After I found the two x and y components, I drew a new triangle. My equilibriant force was still 2.7117N but my angle came to 183.366 degrees. The angle seems really strange to me considering the larger force is in the same quadrant. Am I still doing something incorrectly? Thanks a lot for your help so far.

 

[starfish101]

Never mind - I just figured out what I did wrong. I calculated the reactant force and not the equilibrant force. Thanks so much for all your help!