Question

A box of mass $m$ is in equilibrium under the application of three forces as shown below. If the magnitude of $\overrightarrow{f_1}$ is $10N$, what is the magnitude of $\overrightarrow{f_3}?$

• A. 5 N
• B. 15 N
• C. 20 N
• D. 30 N

Answer 1

Answer: (C)

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
$\left|F_2\right|\cos \left(60^{\circ }\right)=\left|F_1\right|\cos \left(30^{\circ }\right)$
$\left|F_2\right|\times \frac{1}{2}=10\times \frac{\sqrt{3}}{2}$
$\left(\because \left|F_1\right|=10N\text{ given }\right)$
$\left|F_2\right|=10\sqrt{3}N$
Again, resolving the force in $y$-direction, we get
$\left|F_3\right|=\left|F_1\right|\sin \left(30^{\circ }\right)+\left|F_2\right|\sin \left(60^{\circ }\right)$
$=10\times \frac{1}{2}+10\sqrt{3}\times \frac{\sqrt{3}}{2}$
$=5+15=20N$