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A box of mass $m$ is in equilibrium under the application of three forces as shown below. If the magnitude of $\mathbf{F}_1$ is $10 \mathrm{~N}$, what is the magnitude of $\mathbf{F}_3$ ?
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Verified Answer
The correct answer is:
20 N
All forces are resolved in two perpendicular axes $(X$ and $Y)$ as shown in the figure.
Since, block of mass $m$ is in equilibrium.
Hence, resolving the forces in $x$-direction, we get
$\begin{aligned}
& \left|\mathbf{F}_2\right| \cos \left(60^{\circ}\right)=\left|\mathbf{F}_1\right| \cos \left(30^{\circ}\right) \\
& \left|\mathbf{F}_2\right| \times \frac{1}{2}=10 \times \frac{\sqrt{3}}{2} \quad\left(\because\left|\mathbf{F}_1\right|=10 \mathrm{~N} \text { given }\right) \\
& \left|\mathbf{F}_2\right|=10 \sqrt{3} \mathrm{~N}
\end{aligned}$
Again, resolving the force in $y$-direction, we get
$\begin{aligned}
\left|\mathbf{F}_3\right| & =\left|\mathbf{F}_1\right| \sin \left(30^{\circ}\right)+\left|\mathbf{F}_2\right| \sin \left(60^{\circ}\right) \\
& =10 \times \frac{1}{2}+10 \sqrt{3} \times \frac{\sqrt{3}}{2} \\
& =5+15=20 \mathrm{~N}
\end{aligned}$
Since, block of mass $m$ is in equilibrium.
Hence, resolving the forces in $x$-direction, we get
$\begin{aligned}
& \left|\mathbf{F}_2\right| \cos \left(60^{\circ}\right)=\left|\mathbf{F}_1\right| \cos \left(30^{\circ}\right) \\
& \left|\mathbf{F}_2\right| \times \frac{1}{2}=10 \times \frac{\sqrt{3}}{2} \quad\left(\because\left|\mathbf{F}_1\right|=10 \mathrm{~N} \text { given }\right) \\
& \left|\mathbf{F}_2\right|=10 \sqrt{3} \mathrm{~N}
\end{aligned}$
Again, resolving the force in $y$-direction, we get
$\begin{aligned}
\left|\mathbf{F}_3\right| & =\left|\mathbf{F}_1\right| \sin \left(30^{\circ}\right)+\left|\mathbf{F}_2\right| \sin \left(60^{\circ}\right) \\
& =10 \times \frac{1}{2}+10 \sqrt{3} \times \frac{\sqrt{3}}{2} \\
& =5+15=20 \mathrm{~N}
\end{aligned}$
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