(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.
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.