Question

An object kept at $\text{O}$ is in equilibrium under the action of four coplanar forces shown in the figure. Find the magnitudes of $F_1$ and $F_2$.

• A. $\frac{20}{\sqrt{3}}\text{N},\frac{4}{\sqrt{3}}\text{N}$
• B. $20\sqrt{3}\text{N},4\sqrt{3}\text{N}$
• C. $\frac{4}{\sqrt{3}}\text{N},\frac{20}{\sqrt{3}}\text{N}$
• D. $4\sqrt{3}\text{N},20\sqrt{3}\text{N}$

Answer 1

The correct option is C $\frac{4}{\sqrt{3}}\text{N},\frac{20}{\sqrt{3}}\text{N}$

Resolving the components of forces in the horizontal $\left(x\right)$ and vertical directions$(y$)


Applying equilibrium condition in horizontal direction:

$\sum F_x=0$
$8+4\cos {60}^{\circ }-F_2\cos {30}^{\circ }=0$
$\Rightarrow 8+2-F_2\times \frac{\sqrt{3}}{2}=0$
$\Rightarrow F_2=\frac{20}{\sqrt{3}}\text{N}$

Applying equilibrium condition in vertical direction:
$\sum F_y=0$
$F_1+4\sin {60}^{\circ }-F_2\sin {30}^{\circ }=0$
$\Rightarrow F_1+\frac{4\sqrt{3}}{2}-\frac{F_2}{2}=0$
$\Rightarrow F_1=\frac{F_2}{2}-2\sqrt{3}=\frac{10}{\sqrt{3}}-2\sqrt{3}$
$\therefore F_1=\frac{4}{\sqrt{3}}\text{N}$