Answer to Question #180565 in Mechanics | Relativity for Jynx

Question

Find the magnitude and direction of the resultant of the following vectors using Laws of Sines and Cosines:10 N, E50°N and 8 N, W10°S.

Answer 1

Explanations & Calculations

• Refer to the sketch attached.

• Two forces can be rearranged in a vector triangle as shown where laws of sines & cosines can be applied then.
• Due to the geometry the angle BOA = 40 degrees
• And the angle OBA can be found to be $\small (190-\theta)$
• Then by cosine rule, the magnitude of the resultant can be found

$\qquad\qquad \begin{aligned} \small \cos40&=\small \frac{OB^2+OA^2-AB^2 }{2OB\times OA}\\ \small 0.7660&=\small \frac{8^2+10^2-R^2}{2\times 8\times 10}\\ \small R&=\small \bold{6.437\,N} \end{aligned}$

• If it makes an angle $\small \theta$ with the north, then using sine rule,

$\qquad\qquad \begin{aligned} \small \frac{10}{\sin(190-\theta)}&=\small \frac{R}{\sin40}\\ \small\sin(190-\theta)&=\small 0.99858\\ \small 190-\theta &=\small 86.946\\ \small \theta&=\small \bold{103.05^0[N\,of\,E]} \end{aligned}$