Question

If $F_1$, $F_2$ makes an angle of ${30}^{\circ }$ and ${45}^{\circ }$ respectively with $F_3$ as shown in the figure, and magnitude of $F_3$ is $5\text{N}$, then the magnitude of $F_1$ and $F_2$ respectively are (Given, $\overrightarrow{F_1}+\overrightarrow{F_2}=\overrightarrow{F_3}$)

• A. $\frac{5\sqrt{2}}{(\sqrt{3}+1)}\text{N}$ and $\frac{5}{(\sqrt{3}+1)}\text{N}$
• B. $\frac{10}{(\sqrt{3}+1)}\text{N}$ and $\frac{5}{(\sqrt{3}+1)}\text{N}$
• C. $\frac{5}{(\sqrt{3}+1)}\text{N}$ and $\frac{5\sqrt{2}}{(\sqrt{3}+1)}\text{N}$
• D. $\frac{10}{(\sqrt{3}+1)}\text{N}$ and $\frac{5\sqrt{2}}{(\sqrt{3}+1)}\text{N}$

Answer 1

The correct option is D $\frac{10}{(\sqrt{3}+1)}\text{N}$ and $\frac{5\sqrt{2}}{(\sqrt{3}+1)}\text{N}$

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
$|\overrightarrow{F_{3x}}|=x\sin {30}^{\circ }-y\sin {45}^{\circ }$
or, $|\overrightarrow{F_{3x}}|=(\frac{x}{2}-\frac{y}{\sqrt{2}})$
Similarly, $|\overrightarrow{F_{3y}}|=x\cos {30}^{\circ }+y\cos {45}^{\circ }$
or, $|\overrightarrow{F_{3y}}|=(\frac{\sqrt{3}x}{2}+\frac{y}{\sqrt{2}})$


Since, $\overrightarrow{F_3}$ has no component in horizontal axis, so, $|\overrightarrow{F_{3x}}|=0$
$\Rightarrow (\frac{x}{2}-\frac{y}{\sqrt{2}})=0$
$\Rightarrow \frac{x}{2}=\frac{y}{\sqrt{2}}\Rightarrow x=\sqrt{2}y$......(i)
Also, $|\overrightarrow{F_{3y}}|=|\overrightarrow{F_3}|$
$\Rightarrow (\frac{\sqrt{3}x}{2}+\frac{y}{\sqrt{2}})=5$
$\Rightarrow \frac{\sqrt{3}\times \sqrt{2}y}{2}+\frac{y}{\sqrt{2}}=5$
$\Rightarrow y=\frac{5\sqrt{2}}{(\sqrt{3}+1)}$
and, $\Rightarrow x=\sqrt{2}y=\frac{10}{(\sqrt{3}+1)}$
Hence, the magnitude of $F_1$ and $F_2$ respectively are $\frac{10}{(\sqrt{3}+1)}\text{N}$ and $\frac{5\sqrt{2}}{(\sqrt{3}+1)}\text{N}$