Question #288063

resolve each of the following vector into components.






•F1=11×10⁴N at 33°to the positive x-axis






•F2=15 GN at 28° to the positive x-axis






•F3=11.3kN at 193° to the positive x-axis






•F4=125 ×10⁵ N at 317° to the positive x-axis







1
Expert's answer
2022-01-17T09:59:15-0500

Explanations & Calculations


  • Resolution of forces is normally done into 2 components orthogonal to each other.
  • The component resolved into the side of the given angle is written with a cosine: F×cosθ\small F\times \cos\theta.
  • Then the other component (which is normal to the previous component) can be obtained with a sine: F×sinθ.\small F\times\sin\theta.
  • When the angle between the force and the x-axis is obtuse or reflex (they form acute angles as well), you can try these 2 to resolve components,
  1. use the angle without thinking so much in the formulae of resolution.
  2. find the acute angle and resolve with that angle.


For the first situation,

Falong x-axis=(11×104N)×cos33=9.23×104NFalong y-axis=(11×104N)×sin33=5.99×104N\qquad\qquad \begin{aligned} \small F_{\text{along x-axis}}&=\small (11\times10^4\,N) \times \cos33\\ &=\small 9.23\times10^4\,N\\ \small F_{\text{along y-axis}}&=\small (11\times10^4\,N)\times\sin33\\ &=\small 5.99\times10^4\,N \end{aligned}

The procedure is the same for the second situation as the angle is acute.




For the third situation,

Falong x-axis=11.3kN×cos193=11.01kN(along negative x-axis)Falong y-axis=11.3kN×sin193=2.54kN(along negative y-axis)\qquad\qquad \begin{aligned} \small F_{\text{along x-axis}}&=\small 11.3\,kN\times\cos193\\ &=\small -11.01\,kN\cdots(\text{along negative x-axis})\\ \small F_{\text{along y-axis}}&=\small 11.3\,kN\times\sin193\\ &=\small -2.54\,kN\cdots(\text{along negative y-axis}) \end{aligned}

  • From the acute angle's point of view,
  • The acute angle here is 193180=130\small 193-180=13^0
  • Now you can visualize the alignment of the force with the x-axis, it is so close to the negative side and the components will also be along negative axes.
  • Then the resolution looks like this,

Falong (-) x-axis=11.3×cos13=11.01kNFalong (-) y-axis =11.3×sin13=2.54kN\qquad\qquad \begin{aligned} \small F_{\text{along (-) x-axis}}&=\small 11.3\times\cos13\\ &=\small 11.01\,kN\\ \small F_{\text{along (-) y-axis }}&=\small 11.3\times\sin13\\ &=\small 2.54\,kN \end{aligned}

  • In the first case, you are notified of this behaviour with a negative sign in prefix to the values.


This is the same procedure for the fourth case as well and the acute lies below the positive x-axis.

  • You can give it a try.

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