- #1
bigzee20
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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?
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?