(a) Forces \(F_{1} = (3N, 210°)\) and \(F_{2} = (4N, 120°)\) act on a particle

Explanation

(a)(i) \(F_{1} = \begin{pmatrix} 3 \cos 210° \\ 3 \sin 210° \end{pmatrix} ; F_{2} = \begin{pmatrix} 4 \cos 120° \\ 4 \sin 120° \end{pmatrix}\)

\(R = F_{1} + F_{2}\)

= \(\begin{pmatrix} 3 \cos 210° \\ 3 \sin 210° \end{pmatrix} + \begin{pmatrix} 4 \cos 120° \\ 4 \sin 120° \end{pmatrix}\)

= \(\begin{pmatrix} -3 \cos 30° \\ -3 \sin 210° \end{pmatrix} + \begin{pmatrix} -4 \cos 60° \\ 4 \sin 60° \end{pmatrix}\)

= \(\begin{pmatrix} -3 \times 0.866 \\ -3 \times 0.500 \end{pmatrix} + \begin{pmatrix} -4 \times 0.500 \\ 4 \times 0.866 \end{pmatrix}\)

= \(\begin{pmatroix} -2.598 \\ -1.500 \end{pmatrix} + \begin{pmatrix} -2.00 \\ 3.464 \end{pmatrix}\)

= \(\begin{pmatrix} -4.598 \\ 1.964 \end{pmatrix}\)

\(|R| = \sqrt{(-4.598)^{2} + (1.964)^{2}}\)

= \(\sqrt{21.14 +3.86} = \sqrt{25}\)

= 5N

Acceleration = \(\frac{force}{mass}\)

= \(\frac{5}{7} = 0.71 ms^{-2}\)

(ii) \(Velocity = acceleration \times time\)

= \(0.71 \times 3 = 2.13 ms^{-1}\).

(b) Let fourth force be \(F_{4} = \begin{pmatrix} a \\ b \end{pmatrix}\).

\(F_{1} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}; F_{2} = \begin{pmatrix} 0 \\ -5 \end{pmatrix} ; F_{3} = \begin{pmatrix} 6 \\ -4 \end{pmatrix}\)

Particle is in equilibrium when the sum of all forces acting on the particle = 0

\(\begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 0 \\ -5 \end{pmatrix} + \begin{pmatrix} 6 \\ -4 \end{pmatrix} + \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)

\(\implies 8 + a = 0 ; a = -8\)

\(-6 + b = 0 \implies b = 6\)

\(F_{4} = \begin{pmatrix} -8 \\ 6 \end{pmatrix}\)

\(|F_{4}| = \sqrt{(-8)^{2} + 6^{2}} = 10N\)

Let the angle be \(\theta\).

\(\tan \theta = \frac{6}{-8} = -0.75\)

\(\theta = -36.87°\)

= \(180° - 36.87° = 143.13°\)

The equilibriant is 10N and acts in the direction of N53.13°W.