An object is in equilibrium when four concurrent forces, acting in the same plane are in the directions…
An object is in equilibrium when four concurrent forces, acting in the same plane are in the directions shown in the figure.
Find the magnitudes of $F_1$ and $F_2$.
$\frac{2}{\sqrt{3}} \mathrm{~N}$ and $\frac{20}{\sqrt{3}} \mathrm{~N}$
$\frac{4}{\sqrt{3}} \mathrm{~N}$ and $\frac{20}{\sqrt{3}} \mathrm{~N}$
$\frac{\sqrt{3}}{2} \mathrm{~N}$ and $\frac{\sqrt{3}}{20} \mathrm{~N}$
$\frac{4}{\sqrt{3}} \mathrm{~N}$ and $\frac{10}{\sqrt{3}} \mathrm{~N}$
Solution
The object is in equilibrium,
Hence, $\Sigma F_x=0$ and $\Sigma F_y=0$
From the given figure,
$\begin{array}{ll} & \Sigma F_x=0 \\ \therefore & 8+4 \cos 60^{\circ}-F_2 \cos 30^{\circ}=0 \\ \Rightarrow & 8+2-F_2 \frac{\sqrt{3}}{2}=0 \\ \Rightarrow & F_2=\frac{20}{\sqrt{3}} \mathrm{~N}\end{array}$
Also $\quad \Sigma F_y=0$
$
\begin{array}{cc}
\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} \\
& =\frac{4}{\sqrt{3}} \mathrm{~N}
\end{array}
$
Hence, $F_1=\frac{4}{\sqrt{3}} \mathrm{~N}$ and $F_2=\frac{20}{\sqrt{3}} \mathrm{~N}$
