Answer:
68.79 N, 13.84° N of W
Explanation:
The law of cosines can be used to find the magnitude of the sum. F1 is 30° N of W, and F2 is 30° S of W, so the exterior angle of the force triangle is 30°+20° = 50°. The interior angle is the supplement of that. The angle between F1 and F2 in the force triangle representing the sum is 130°, so the sum of forces is ...
|F|^2 = |F1|^2 +|F2|^2 -2·|F1|·|F2|·cos(130°)
= 50^2 +25^2 -2·50·25·cos(130°) ≈ 4731.969
|F| ≈ √4731.969 ≈ 68.79 . . . . newtons
The angle α between F and F1 can be found from the law of sines.
sin(α)/|F2| = sin(130°)/|F|
α = arcsin(|F2|/|F|·sin(130°)) ≈ 16.16°
The diagram shows this to be the angle south of F1, so the angle of the sum vector F is 30° -16.16° N of W = 13.84° N of W.
The resultant force vector is 68.79 N at an angle of 13.84° N of W.
