Forces \(F_{1} = (10 N, 090°), F_{2} = (20 N, 210°)\) and \(F_{3} = (4 N, 330°)\) act on a body at rest on a smooth table. Find, correct to one decimal place, the magnitude of the resultant force.
Resolve each force into components with bearings measured clockwise from north, using \(x = F\sin\theta\) (east) and \(y = F\cos\theta\) (north).
- \(F_1 = 10\,\text{N at }090^\circ:\ x = 10\sin 90^\circ = 10,\ y = 10\cos 90^\circ = 0\)
- \(F_2 = 20\,\text{N at }210^\circ:\ x = 20\sin 210^\circ = -10,\ y = 20\cos 210^\circ = -17.32\)
- \(F_3 = 4\,\text{N at }330^\circ:\ x = 4\sin 330^\circ = -2,\ y = 4\cos 330^\circ = 3.46\)
Sum the components:
\[\sum x = 10 - 10 - 2 = -2,\qquad \sum y = 0 - 17.32 + 3.46 = -13.86\]
Magnitude of the resultant:
\[R = \sqrt{(-2)^2 + (-13.86)^2} = \sqrt{4 + 192.0} = \sqrt{196.0} = 14.0\,\text{N}\]
The magnitude of the resultant force is \(14.0\,\text{N}\) (to 1 decimal place).