Answer to Question #180565 in Mechanics | Relativity for Jynx

Question #180565

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.


1
Expert's answer
2021-04-13T06:29:14-0400

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 (190θ)\small (190-\theta)
  • Then by cosine rule, the magnitude of the resultant can be found

cos40=OB2+OA2AB22OB×OA0.7660=82+102R22×8×10R=6.437N\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 θ\small \theta with the north, then using sine rule,

10sin(190θ)=Rsin40sin(190θ)=0.99858190θ=86.946θ=103.050[NofE]\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}


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