Finding resultant force from two forces in quadrants II and III

[spokorne]

I've been stuck on this problem for too long and need help. Please explain the answer if possible.

F1 has a magnitude of 8.80 n and is directed at an angle of $$\alpha$$= 65.0 above the negative x-axis in the second quadrant. F2 has a magnitude of 6.40 N and is directed at an angle of $$\beta$$ = 53.3 below the negative x-axis in the third quadrant.


What is the x component Fx of the resultant force?

What is the y component Fy of the resultant force?

What is the magnitude F of the resultant force?

What is the angle $$\gamma$$ that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis.

 

[LowlyPion]

Welcome to PF.

What have you attempted?

 

[spokorne]

i tried making them into triangles and then using the lengths as the force. then i figured out the upward force and minused the downard force to find the total vertical force but it isn't right

 

[LowlyPion]

Maybe if you calculated the x,y components of the vectors it would be a little easier?

 

[spokorne]

how do you go about doing that?

 

[LowlyPion]

The dog on a leash is an example at the bottom of this link:
http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/vectors/u3l1e.html

 

[spokorne]

That's what I did previously and then then i found the difference in the two and it was incorrect

 

[LowlyPion]

I can't see your effort, so you should check that you have the correct sign of the x components when you add them and the y components.

After you have added them then you use what Pythagoras taught.