Question

Two forces each numerically equal to $25\text{N}$ are acting as shown in the following figure. What is the resultant vector?

• A. $|\overrightarrow{R}|=25\text{N}$ at an angle of ${60}^{\circ }$ clockwise from $\overrightarrow{A}$
• B. $|\overrightarrow{R}|=25\text{N}$ at an angle of ${120}^{\circ }$ clockwise from $\overrightarrow{A}$
• C. $|\overrightarrow{R}|=25\text{N}$ at an angle of ${30}^{\circ }$ clockwise from $\overrightarrow{A}$
• D. $|\overrightarrow{R}|=30\text{N}$ at an angle of ${75}^{\circ }$ clockwise from $\overrightarrow{A}$

Answer 1

The correct option is A $|\overrightarrow{R}|=25\text{N}$ at an angle of ${60}^{\circ }$ clockwise from $\overrightarrow{A}$

Given, vector $\overrightarrow{A}$ and $\overrightarrow{B}$, each has $|\overrightarrow{A}|=|\overrightarrow{B}|=25\text{N}$ and from the figure it can be seen that angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is ${120}^{\circ }$ resultanatn vector $\overrightarrow{R}$,

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$
$|\overrightarrow{R}|=\sqrt{|\overrightarrow{A}{|}^2+|\overrightarrow{B}{|}^2+2\times |\overrightarrow{A}|\times |\overrightarrow{B}|\times \cos \theta }$
$|\overrightarrow{R}|=\sqrt{{25}^2+{25}^2+2\times 25\times 25\times \cos \left({120}^{\circ }\right)}$
$|\overrightarrow{R}|=\sqrt{{25}^2+{25}^2+2\times 25\times 25\times \left(\frac{-1}{2}\right)}$
$|\overrightarrow{R}|=\sqrt{{25}^2}$
$|\overrightarrow{R}|=25\text{N}$
And,
$\tan \alpha =\frac{25\times \sin {120}^{\circ }}{25+25\times \cos {120}^{\circ }}$

$\tan \alpha =\sqrt{3}$
$\alpha ={\tan }^{-1}\left(\sqrt{3}\right)={60}^{\circ }$