Coplanar force 4N, 8N, 6N, 4N and 5N act at a point as shown in the diagram. If the 6N force acts in the direction 090°, calculate the :
(a) magnitude of the resultant force;
(b) direction of the resultant force.
Explanation
With the 6N force acting in the direction 090°, the diagram is redrawn as :
Resultant = (begin{pmatrix} 4 cos 65° \ 4 sin 65° end{pmatrix} + begin{pmatrix} 6 cos 90° \ 6 sin 90° end{pmatrix} + begin{pmatrix} 8 cos 145° \ 8 sin 145° end{pmatrix} + begin{pmatrix} 4 cos 245° \ 4 sin 245° end{pmatrix} + begin{pmatrix} 5 cos 295° \ 5 sin 295° end{pmatrix})
= (begin{pmatrix} 4 cos 65 \ 4 sin 65 end{pmatrix} + begin{pmatrix} 6 cos 90 \ 6 sin 90 end{pmatrix} + begin{pmatrix} -8 cos 35 \ 8 sin 35 end{pmatrix} + begin{pmatrix} -4 cos 65 \ -4 sin 65 end{pmatrix} + begin{pmatrix} 5 cos 65 \ -5 sin 65 end{pmatrix})
= (begin{pmatrix} 4 times 0.4226 \ 4 times 0.9063 end{pmatrix} + begin{pmatrix} 0 \ 6 end{pmatrix} + begin{pmatrix} -8 times 0.8192 \ 8 times 0.5736 end{pmatrix} + begin{pmatrix} -4 times 0.4226 \ -4 times 0.9063 end{pmatrix} + begin{pmatrix} 5 times 0.4226 \ -5 times 0.9063 end{pmatrix})
= (begin{pmatrix} 1.6904 \ 3.6252 end{pmatrix} + begin{pmatrix} 0 \ 6 end{pmatrix} + begin{pmatrix} -6.5536 \ 4.5888 end{pmatrix} + begin{pmatrix} -1.6904 \ -3.6252 end{pmatrix} + begin{pmatrix} 2.113 \ -4.5315 end{pmatrix})
= (begin{pmatrix} -4.4406 \ 6.0573 end{pmatrix})
(|Resultant| = sqrt{(-4.4406)^{2} + (6.0573)^{2}} = sqrt{56.41})
= 7.5107N.
The magnitude of the resultant = 7.5N.
(b) Let (theta) be the direction of the resultant.
(tan theta = frac{6.0573}{-4.4406} = – 1.364)
(theta = -53.75° )
= (180° – 53.75° = 126.25°) with the positive x- axis.