## Zale101 one year ago Find the magnitude of the resultant force and the angle that makes with the positive x-axis

1. Zale101

|dw:1440990161648:dw|

2. Zale101

$$F1=75 N; θ=0$$ $$F2=50 N; θ=30$$ $$F3=40 N; θ=150$$ F1= $$f_1x=75cos(0)=75 N$$ $$f_1y=75sin(0)=0 N$$ F2= $$f_2x=50cos(30)=25\sqrt{3} N$$ $$f_2y=50sin(30)=25 N$$ F3= $$f_3x=40cos(150)=−20\sqrt{2}N$$ $$f_3y=40sin(150)=20 N$$

3. Zale101

Is this good so far?

4. jim_thompson5910

I don't agree with f3x

5. jim_thompson5910

everything else looks good

6. ganeshie8

Looks good! next simply add them component by component

7. ganeshie8

Ahh I didn't check the arithmetic..

8. ganeshie8

this should fix it : $$F1=75 N; θ=0$$ $$F2=50 N; θ=30$$ $$F3=40 N; θ=150$$ F1= $$f_1x=75cos(0)=75 N$$ $$f_1y=75sin(0)=0 N$$ F2= $$f_2x=50cos(30)=25\sqrt{3} N$$ $$f_2y=50sin(30)=25 N$$ F3= $$f_3x=40cos(150)=−20\sqrt{\color{red}{3}}N$$ $$f_3y=40sin(150)=20 N$$

9. jim_thompson5910

much better

10. Zale101

Okay. Thanks! @jim_thompson5910 and @ganeshie8 !!! I'll deal with the arithmetic, it's not the problem for me now. But the next step is what i'm having problems with.

11. jim_thompson5910

you have the vector components, add up the corresponding components

12. jim_thompson5910

u = <a,b> v = <c,d> w = <e,f> u+v+w = <a+c+e, b+d+f>

13. ganeshie8

|dw:1440991132300:dw|

14. jim_thompson5910

|dw:1440991352564:dw|

15. Zale101

Okay, thanks. makes sense !

16. jim_thompson5910

you'll end up with some vector z = <p,q> p = sum of the Fx q = sum of the Fy magnitude of z = |z| = sqrt(p^2 + q^2)

17. Zale101

18. jim_thompson5910

theta = arctan(q/p)

19. Zale101

Oh yes. I forgot about that xD

20. jim_thompson5910

theta = arctan(q/p) if theta is in Q1 or Q4 theta = arctan(q/p)+180 if theta is in Q2 or Q3

21. jim_thompson5910

to figure out which quadrant theta is in, simply look at the vector <p,q>

22. Zale101

Absolutely. Alright thanks for all the help.

23. jim_thompson5910

sure thing