Question

If two forces $F_1$ and $F_2$ act on a body in such a way that the resultant force $F_3$ has magnitude of $150\text{N}$ at ${15}^{\circ }$ East of North. If the magnitude of $F_1$ is $100\text{N}$ at ${10}^{\circ }$ West of North, what is the magnitude and direction of $F_2$?
[Take $\sin {15}^{\circ }=0.258,\cos {15}^{\circ }=0.966,$$\sin {10}^{\circ }=0.174,\cos {10}^{\circ }=0.985$$\text{and}{\tan }^{-1}\left(0.827\right)={39.6}^{\circ }$]

• A. $|\overrightarrow{F_2}|=72.8$ and $\theta ={39.6}^{\circ }$ East of North
• B. $|\overrightarrow{F_2}|=72.8$ and $\theta ={39.6}^{\circ }$ North of East
• C. $|\overrightarrow{F_2}|=68.3$ and $\theta ={39.6}^{\circ }$ North of East
• D. $|\overrightarrow{F_2}|=68.3$ and $\theta ={39.6}^{\circ }$ East of North

Answer 1

The correct option is B $|\overrightarrow{F_2}|=72.8$ and $\theta ={39.6}^{\circ }$ North of East

The said condition can be shown as
Given, $|\overrightarrow{F_3}|=150\text{N}$, $|\overrightarrow{F_1}|=100\text{N}$


we can also write as,
$\overrightarrow{F_2}=\overrightarrow{F_3}-\overrightarrow{F_1}$
$\overrightarrow{F_3}=\hat{i}\left[150\sin \right({15}^{\circ }\left)\right]+\hat{j}\left[150\cos \right({15}^{\circ }\left)\right]$
$\overrightarrow{F_3}=\hat{i}\left(38.7\right)+\hat{j}\left(144.9\right)$
$\overrightarrow{F_1}=\hat{i}[-100\sin ({10}^{\circ }\left)\right]+\hat{j}\left[100\cos \right({10}^{\circ }\left)\right]$
$\overrightarrow{F_1}=\hat{i}(-17.4)+\hat{j}\left(98.5\right)$
we can write as
$\overrightarrow{F_2}=\hat{i}(38.7-(-17.4)+\hat{j}(144.9-98.5)$
$\overrightarrow{F_2}=\hat{i}\left(56.1\right)+\hat{j}\left(46.4\right)$
$|\overrightarrow{F_2}|=\sqrt{{56.1}^2+{46.4}^2}$
$|\overrightarrow{F_2}|=72.8\text{N}$
the angle $\theta ={\tan }^{-1}\left(\frac{46.4}{56.1}\right)={39.6}^{\circ }$ North of East