A particle is under the action of forces \(P = (4N, 030°)\) and \(R = (10N, 300°)\). Find the force that will keep the particle in equilibrium.

Answer Details

Resolve each force into east (x) and north (y) components using its bearing, where east \(= F\sin\theta\), north \(= F\cos\theta\).

\( \mathbf{P} = (4\sin 30^\circ,\ 4\cos 30^\circ) = (2.000,\ 3.464) \).
\( \mathbf{R} = (10\sin 300^\circ,\ 10\cos 300^\circ) = (-8.660,\ 5.000) \).

Resultant: \( \mathbf{P} + \mathbf{R} = (-6.660,\ 8.464) \).

The equilibrant \( \mathbf{E} \) is equal and opposite: \( \mathbf{E} = (6.660,\ -8.464) \).

Magnitude: \[ |\mathbf{E}| = \sqrt{6.660^2 + 8.464^2} = \sqrt{44.36 + 71.64} = \sqrt{116} \approx 10.8\text{ N}. \]

Direction (bearing): the equilibrant points east and south, so \[ \text{bearing} = 180^\circ - \tan^{-1}\!\left(\frac{6.660}{8.464}\right) = 180^\circ - 38.2^\circ \approx 142^\circ. \]

The equilibrating force is about \( 10.8\text{ N} \) on a bearing of \( 142^\circ \).