(a) Given that (p = (4i – 3j)) and (q = (-i + 5j)), find r such that (|r| = 15) and is in the direction ((2p + 3q)).
(b)
Forces of magnitude 8N, 6N and 4N act at the point P, as shown in the above diagram. Find the : (i) magnitude ; (ii) direction of the resultant force.
Explanation
(a) (p = (4i – 3j) ; q = (-i + 5j))
(2p + 3q = begin{pmatrix} 8i – 3i \ -6j + 15j end{pmatrix})
= (begin{pmatrix} 5i \ 9j end{pmatrix})
Let (r = k begin{pmatrix} 5i \ 9j end{pmatrix} = begin{pmatrix} 5ki \ 9kj end{pmatrix})
(|r| = sqrt{(5k)^{2} + (9k)^{2}} = 15)
(sqrt{25k^{2} + 81k^{2}} = 15 implies 25k^{2} + 81k^{2} = 225)
(106k^{2} = 225 implies k^{2} = frac{225}{106} = 2.123)
(k = sqrt{2.123} = 1.457)
(therefore r = begin{pmatrix} 5i times 1.457 \ 9j times 1.457 end{pmatrix})
= (begin{pmatrix} 7.285i \ 13.113j end{pmatrix})
(b)(i)
Resolve the forces along the x and y- axis.
Let (F_{1} = begin{pmatrix} 8 cos 120° \ 8 sin 120° end{pmatrix})
(F_{2} = begin{pmatrix} 6 cos 90° \ 6 sin 90° end{pmatrix})
(F_{3} = begin{pmatrix} 4 cos 30° \ 4 sin 30° end{pmatrix})
(R = F_{1} + F_{2} + F_{3})
= (begin{pmatrix} 8 cos 120° \ 8 sin 120° end{pmatrix} + begin{pmatrix} 6 cos 90° \ 6 sin 90° end{pmatrix} + begin{pmatrix} 4 cos 30° \ 4 sin 30° end{pmatrix})
= (begin{pmatrix} -8 cos 60° \ 8 sin 60 end{pmatrix} + begin{pmatrix} 6 cos 90° \ 6 sin 90° end{pmatrix} + begin{pmatrix} 4 cos 30° \ 4 sin 30° end{pmatrix})
= (begin{pmatrix} -8 times 0.5 \ 8 times 0.866 end{pmatrix} + begin{pmatrix} 6 times 0 \ 6 times 1 end{pmatrix} + begin{pmatrix} 4 times 0.866 \ 4 times 0.5 end{pmatrix})
= (begin{pmatrix} -4 + 0 + 3.464 \ 6.928 + 6 + 2 end{pmatrix})
= (begin{pmatrix} -0.536 \ 14.928 end{pmatrix})
(|R| = sqrt{(-0.536)^{2} + (14.928)^{2}} = sqrt{0.2873 + 222.8452})
= (sqrt{223.1325} = 14.94N)
(ii)Direction (theta)
(tan theta = frac{14.928}{-0.536} = -27.851)
(theta = -87.9° = 92.1°)
The direction is N 02.1° W or a bearing of 267.9°.