Calculating the Net Force Vector for Multiple Force Vectors

[bigzee20]

Consider 3 force vectors F1, F2, and F3. The vector F1 has magnitude F1 = 36N and direction θ = 110°; the vector F2 has magnitude F2 = 22N and direction θ = - 140°; and the vector F3 has magnitude F3 = 28N and direction θ = 20°. All the direction angles θ are measured from the positive x axis: counter-clockwise for θ > 0 and clockwise for θ < 0.

What is the magnitude F or the net force vector F = F1+F2+F3? Answer in units of N.

I found the x resultants to be -12.31 + -14.14 + 26.31 = -0.14
I found the y resultants to be 33.83 + 16.85 + 9.58 = 60.26

(sqrt)-0.14^2 + 60.26^2 = 60.26

Can someone tell me were Iam going wrong?

 

[physics girl phd]

Can you show us detail about how you derived the x- and y- components for each vector?

I think you have switched x and y components on one of your vectors, and have a sign wrong. I think it's best to draw each vector separately and write out all the steps -- that way you can check if these intermediate parts are correct.

 

[bigzee20]

ok here is what i did

F1x = 36N cos110 = -12.31
F1y = 36N sin110 = 33.83
F2x = 22N cos130 = -14.14
F2y = 22N sin130 = 16.85
F3x = 28N cos20 = 26.31
F3y = 28N sin20 = 9.58

then i found the resultant for the x and y component

 

[physics girl phd]

ok here is what i did

F1x = 36N cos110 = -12.31
F1y = 36N sin110 = 33.83
F2x = 22N cos130 = -14.14
F2y = 22N sin130 = 16.85
F3x = 28N cos20 = 26.31
F3y = 28N sin20 = 9.58

then i found the resultant for the x and y component

Check your vector 2... why do you have 130 degrees?