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 $\mathbf{F}_3$ ?

• A

  5N
• B

  15 N
• Correct answer
  C

  20 N
• D

  30 N

Solution

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 $\begin{aligned} \left|\mathbf{F}_3\right| & =\left|\mathbf{F}_1\right| \sin \left(30^{\circ}\right)+\left|\mathbf{F}_2\right| \sin \left(60^{\circ}\right) \\ & =10 \times \frac{1}{2}+10 \sqrt{3} \times \frac{\sqrt{3}}{2} \\ & =5+15=20 \mathrm{~N} \end{aligned}$

Hence, the correct answer is option: 'C'.