Question

If $F_1$, $F_2$ makes an angle of ${60}^{\circ }$ and ${37}^{\circ }$ respectively with $F_3$ as shown in the figure, and magnitude of $F_3$ is $2\text{N}$, then the magnitude of $F_1$ and $F_2$ respectively are (Given, $\overrightarrow{F_1}+\overrightarrow{F_2}=\overrightarrow{F_3}$)

• A. $\frac{4\sqrt{3}}{(\sqrt{3}+4)}\text{N}$ and $\frac{10}{(\sqrt{3}+4)}\text{N}$
• B. $\frac{2\sqrt{3}}{(\sqrt{3}+4)}\text{N}$ and $\frac{10}{(\sqrt{3}+4)}\text{N}$
• C. $\frac{4\sqrt{3}}{(\sqrt{3}+4)}\text{N}$ and $\frac{5}{(\sqrt{3}+4)}\text{N}$
• D. $\frac{4}{(\sqrt{3}+4)}\text{N}$ and $\frac{5}{(\sqrt{3}+4)}\text{N}$

Answer 1

The correct option is A $\frac{4\sqrt{3}}{(\sqrt{3}+4)}\text{N}$ and $\frac{10}{(\sqrt{3}+4)}\text{N}$

Let the magnitude be $|\overrightarrow{F_1}|=x\text{and}|\overrightarrow{F_2}|=y$
Also, Given, $\overrightarrow{F_1}+\overrightarrow{F_2}=\overrightarrow{F_3}$
So, the given vectors can be shown as


Now, resolving the vectors along the axes we get
$|\overrightarrow{F_{3x}}|=x\cos {60}^{\circ }+y\cos {37}^{\circ }$
or, $|\overrightarrow{F_{3x}}|=(\frac{x}{2}+\frac{4y}{5})$
Similarly, $|\overrightarrow{F_{3y}}|=x\sin {60}^{\circ }-y\sin {37}^{\circ }$
or, $|\overrightarrow{F_{3y}}|=(\frac{\sqrt{3}x}{2}-\frac{3y}{5})$


Since, $\overrightarrow{F_3}$ has no component in vertical axis, so, $|\overrightarrow{F_{3y}}|=0$
$\Rightarrow (\frac{\sqrt{3}x}{2}-\frac{3y}{5})=0$
$\Rightarrow \frac{\sqrt{3}x}{2}=\frac{3y}{5}\Rightarrow x=\frac{2\sqrt{3}y}{5}$......(i)
Also, $|\overrightarrow{F_{3x}}|=|\overrightarrow{F_3}|$
$\Rightarrow (\frac{x}{2}+\frac{4y}{5})=2$
$\Rightarrow \frac{2\sqrt{3}y}{10}+\frac{4y}{5}=2$
$\Rightarrow y=\frac{10}{(\sqrt{3}+4)}$
and, $\Rightarrow x=\frac{2\sqrt{3}y}{5}=\frac{4\sqrt{3}}{(\sqrt{3}+4)}$
Hence, the magnitude of $F_1$ and $F_2$ respectively are $\frac{4\sqrt{3}}{(\sqrt{3}+4)}\text{N}$ and $\frac{10}{(\sqrt{3}+4)}\text{N}$