Question

A box of mass $m$ is in equilibrium under the application of three forces as shown below. If the magnitude of $\vec{f_{1}}$ is $10\, N$, what is the magnitude of $\vec{f_{3}} ?$

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