Question

Find the magnitude and direction of 3 coplanar forces of:

i) 40 N 45 degrees east of south
ii) 10 N west
iii) 70 N 35 degrees east of north

Answer 1

Take east as +x and north as +y.

1) 40 N, 45° east of south:
- Fx1 = 40 sin45° = 40(0.7071) = 28.284 N (east)
- Fy1 = −40 cos45° = −40(0.7071) = −28.284 N (north negative → south)

2) 10 N west:
- Fx2 = −10 N, Fy2 = 0

3) 70 N, 35° east of north:
- Fx3 = 70 sin35° = 70(0.5736) = 40.150 N (east)
- Fy3 = 70 cos35° = 70(0.8192) = 57.341 N (north)

Sum components:
- Rx = 28.284 − 10 + 40.150 = 58.435 N (east)
- Ry = −28.284 + 0 + 57.341 = 29.056 N (north)

Resultant magnitude:
R = sqrt(Rx^2 + Ry^2) ≈ sqrt(58.435^2 + 29.056^2) ≈ 65.26 N

Direction (vector in NE quadrant):
- angle east of north = arctan(Rx/Ry) ≈ arctan(58.435/29.056) ≈ 63.6° east of north
- equivalently, 26.4° north of east.

Answer: Resultant ≈ 65.3 N, directed 26.4° north of east (or 63.6° east of north).