Answer to Question #288063 in Mechanics | Relativity for vera

Question

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

Answer 1

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: $\small F\times \cos\theta$.
• Then the other component (which is normal to the previous component) can be obtained with a sine: $\small F\times\sin\theta.$
• 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,

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

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

For the third situation,

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

• From the acute angle's point of view,
• The acute angle here is $\small 193-180=13^0$
• 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,

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

• 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.