202311070929 Exercise 21.1.2

Find (by drawing or calculation) the resultant of the following forces, in magnitude and direction.

(a)

(b)

(c)

_{Extracted from A. Godman & J. F. Talbert. (1973). Additonal Mathematics Pure and Applied in SI units.}

(a)

So that the two forces are labelled as vectors $\mathbf{OA}$ and $\mathbf{OB}$ , let their initial points for both be $O$ and the terminal points of each $A$ and $B$ .

In Cartesian coordinates,

$\begin{aligned} \mathbf{OA} & = |\mathbf{OA}|\,\hat{\mathbf{i}} \\ & = OA\,\hat{\mathbf{i}} \\ & = 12\,\mathrm{N}\,\hat{\mathbf{i}} \\ \mathbf{OB} & = |\mathbf{OB}|\,\hat{\mathbf{j}} \\ & = OB\,\hat{\mathbf{j}} \\ & = 10\,\mathrm{N}\,\hat{\mathbf{j}} \\ \end{aligned}$

With, on purpose, no diagrams being provided, let also the resultant force $\mathbf{OC}$ be directed from the same initial point $O$ to some terminal point we let be $C$ . Then,

$\begin{aligned} \mathbf{OC} & = \mathbf{OA} + \mathbf{OB} \\ & = 12\,\mathrm{N}\,\hat{\mathbf{i}} + 10\,\mathrm{N}\,\hat{\mathbf{j}} \\ OC & = |\mathbf{OC}| \\ & = \sqrt{12^2+10^2} \\ & = 15.6\,\mathrm{N}\qquad\textrm{(3 s.f.)} \\ \end{aligned}$

Trying in polar coordinates,

$\begin{aligned} \mathbf{OA} & = |\mathbf{OA}|\,\hat{\mathbf{r}} \\ & = 12\,\mathrm{N}\,\hat{\mathbf{r}} \\ \mathbf{OB} & = |\mathbf{OB}|\,\hat{\mathbf{r}} +\frac{\pi}{2}\,\hat{\boldsymbol{\theta}}\\ & = \bigg( 10\,\hat{\mathbf{r}} + \frac{\pi}{2}\,\hat{\boldsymbol{\theta}}\bigg) \,\mathrm{N} \\ \end{aligned}$

Recall the formulae for addition of vectors

$\begin{aligned} r_3 & = \sqrt{r_1^2+2r_1r_2\cos(\theta_2-\theta_1)+r_2^2} \\ \theta_3 & = \theta_1+\arctan \bigg(\frac{r_2\sin (\theta_2-\theta_1)}{r_1+r_2\cos (\theta_2-\theta_1)}\bigg) \\ \end{aligned}$

beside the rectangular form I am stupid enough to do the polar. Quit.

(b)

By tail-to-tip method, draw

$\begin{aligned} OC & = \sqrt{OB^2+BC^2} \\ & = \sqrt{OB^2+OA^2} \\ & = \sqrt{40^2+30^2} \\ & = 50\,\mathrm{N} \\ \tan\theta & = \frac{30}{40} \\ \theta & = \tan^{-1} \bigg(\frac{30}{40}\bigg) \\ & = 36.9^\circ\qquad\textrm{(3 s.f.)} \\ \end{aligned}$

Or, by parallelogram method, draw

which makes no matter how you visualise, only are the calculations the matters.

(c)

Draw a labelled diagram below:

$\begin{aligned} \mathbf{OA} & = 4\cos 30^\circ \,\hat{\mathbf{i}} + 4\sin 30^\circ \,\hat{\mathbf{j}} \\ & = 4\bigg(\frac{\sqrt{3}}{2}\bigg)\,\hat{\mathbf{i}} + 4\bigg(\frac{1}{2}\bigg) \,\hat{\mathbf{j}} \\ & = \big(2\sqrt{3}\,\hat{\mathbf{i}} +2\,\hat{\mathbf{j}}\big)\,\mathrm{N} \\ \mathbf{OB} & = \big( 6\,\hat{\mathbf{i}}\big)\,\mathrm{N} \\ \end{aligned}$

Let $\mathbf{OC}$ be the resultant vector,

$\begin{aligned} \mathbf{OC} & = \mathbf{OA} + \mathbf{OB} \\ & = \big( 2\sqrt{3}+6\big) \,\hat{\mathbf{i}} + 2\,\hat{\mathbf{j}} \\ \end{aligned}$

Write $OA$ , $OB$ , and $OC$ the magnitudes of respective vectors $\mathbf{OA}$ , $\mathbf{OB}$ , and $\mathbf{OC}$ ,

$\begin{aligned} OA & = 4 \\ OB & = 6 \\ OC & = |\mathbf{OC}| \\ & = \sqrt{(2\sqrt{3}+6)^2+2^2} \\ & = \sqrt{52+24\sqrt{3}} \\ & = 2\sqrt{13+6\sqrt{3}} \\ \end{aligned}$

With a diagram in mind,

apply cosine law,

$\begin{aligned} \cos \angle COB & = \frac{OC^2+OB^2-OA^2}{2(OC)(OB)} \\ & = \frac{(52+24\sqrt{3})+(36)-(16)}{2(52+24\sqrt{3})(6)} \\ \angle COB & = \cdots \\ \end{aligned}$

This problem is not to be attempted.