(a) Forces \(F_{1} = (3N, 210°)\) and \(F_{2} = (4N, 120°)\) act on a particle of mass 7kg which is at rest. Calculate the :
(i) acceleration of the particle ; (ii) velocity of the particle after 3 seconds.
(b) \(F_{1} = (2i + 3j)N, F_{2} = (-5j)N \) and \(F_{3} = (6i - 4j)N\) act on a body. Find the magnitude and direction of the fourth force that will keep the body in equilibrium.
(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.
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