Question

Find the magnitude and direction of 3 coplanar forces of 1. 40 N 45 degrees east of south; 2. 10 N west; 3. 70 N 35 degrees east of north

Answer 1

Take +x = east, +y = north.

Components:
- F1 = 40 N, 45° east of south → F1x = +40 sin45° = +28.284 N, F1y = −40 cos45° = −28.284 N.
- F2 = 10 N west → F2x = −10.000 N, F2y = 0.
- F3 = 70 N, 35° east of north → F3x = +70 sin35° = +40.150 N, F3y = +70 cos35° = +57.341 N.

Sum components:
Rx = 28.284 − 10.000 + 40.150 = 58.435 N
Ry = −28.284 + 0 + 57.341 = 29.056 N

Resultant magnitude:
R = sqrt(Rx^2 + Ry^2) = sqrt(58.435^2 + 29.056^2) ≈ 65.27 N

Direction:
θ = arctan(Ry/Rx) = arctan(29.056/58.435) ≈ 26.5° north of east.

Answer: R ≈ 65.3 N at 26.5° north of east.