Question

An object is in equilibrium under four concurrent forces in the directions shown in figure. Find the magnitude of $F_1$ and $F_2$ (in N).

• A. $4\sqrt{3}$ and $20\sqrt{3}$
• B. $\frac{4}{\sqrt{3}}$ and $\frac{20}{\sqrt{3}}$
• C. $4\sqrt{3}$ and $\frac{20}{\sqrt{3}}$
• D. $\frac{4}{\sqrt{3}}$ and $20\sqrt{30}$

Answer 1

The correct option is B $\frac{4}{\sqrt{3}}$ and $\frac{20}{\sqrt{3}}$


The object is in equilibrium. Hence,
$\Sigma F_x=0$
$\therefore 8+4\cos {60}^{\circ }-F_2\cos {30}^{\circ }=0\Rightarrow 8+2-F_2\frac{\sqrt{3}}{2}=0\Rightarrow F_2=\frac{20}{\sqrt{3}}N$
and ${\Sigma }_y=0\therefore F_1+4\sin {60}^{\circ }-F_2\sin {30}^{\circ }=0\Rightarrow F_1+\frac{4\sqrt{3}}{2}-\frac{F_2}{2}=0\Rightarrow F_1=\frac{F_2}{2}-2\sqrt{3}=\frac{10}{\sqrt{3}}-2\sqrt{3}\Rightarrow F_1=\frac{4}{\sqrt{3}}N$