# Question #82392

Aug 3, 2017

$| \vec{R} | = 143.7 N$
Angle between $100 N$ force and $\vec{R}$=${\tan}^{- 1} \left(\frac{57}{282}\right)$

#### Explanation:

$\theta =$ angle between vectors

$\alpha$ = angle between resultant and ${F}_{1}$

${F}_{1} = 100 N$

${F}_{2} = 50 N$

Angle with vertical = ${15}^{o}$

$\therefore \theta = {35}^{o}$

Applying parallelogram law of vector addition,

$| \vec{R} | = \sqrt{{\left({F}_{1}\right)}^{2} + {\left({F}_{2}\right)}^{2} + 2 \left({F}_{1}\right) \left({F}_{2}\right) \left(\cos \theta\right)}$

$\Rightarrow | \vec{R} | = \sqrt{{\left(100\right)}^{2} + {\left(50\right)}^{2} + 2 \left(100\right) \left(50\right) \left(\cos {37}^{o}\right)}$

$\Rightarrow | \vec{R} | = 30 \sqrt{23} N$

$\Rightarrow | \vec{R} | = 143.7 N$

$\tan \alpha = \frac{{F}_{2} \sin \theta}{{F}_{1} + {F}_{2} \cos \theta}$

$\Rightarrow \tan \alpha = \frac{50 \sin 35}{100 + 50 \cos 35}$

$\Rightarrow \tan \alpha = \frac{28.5}{141}$

$\Rightarrow \tan \alpha = \frac{57}{282}$

$\Rightarrow \alpha = {\tan}^{- 1} \left(\frac{57}{282}\right)$