Question #1b250

Jan 18, 2017

We need to use law of parallelogram of vector addition. For two vectors $\vec{P} \mathmr{and} \vec{Q}$ having an angle $\theta$ between the two the resultant vector $\vec{R}$ is given by
$| \vec{R} | = \sqrt{| \vec{P} {|}^{2} + | \vec{Q} {|}^{2} + 2 | \vec{P} | | \vec{Q} | \cos \theta}$
and angle of the resultant $\alpha = {\tan}^{-} 1 \left(\frac{| \vec{Q} | \sin \theta}{| \vec{P} | + | \vec{Q} | \cos \theta}\right)$

Inserting given values we get
$| \vec{R} | = \sqrt{{\left(6000\right)}^{2} + {\left(6000\right)}^{2} + 2 \times 6000 \times 6000 \times \cos 60}$
$\implies | \vec{R} | = \sqrt{{\left(6000\right)}^{2} + {\left(6000\right)}^{2} + 2 \times 6000 \times 6000 \times 0.5}$
$\implies | \vec{R} | = 6000 \sqrt{3}$
$\implies | \vec{R} | = 10392.3 N$ rounded to one decimal place.

and $\alpha = {\tan}^{-} 1 \left(\frac{6000 \times \sin 60}{6000 + 6000 \cos 60}\right)$
$\implies \alpha = {\tan}^{-} 1 \left(\frac{\frac{\sqrt{3}}{2}}{1 + \frac{1}{2}}\right)$
$\implies \alpha = {\tan}^{-} 1 \left(\frac{1}{\sqrt{3}}\right)$
$\implies \alpha = {30}^{\circ}$ measured as per measurement of $\angle \theta$