Question

Find the height of the center of mass of the equilateral triangular lamina from the side $\text{AB}$ as shown in figure. Given, side of the triangular lamina is $3\text{m}$.

• A. $\sqrt{3}\text{m}$
• B. $\frac{3}{\sqrt{2}}\text{m}$
• C. $\frac{\sqrt{3}}{2}\text{m}$
• D. $\frac{\sqrt{2}}{3}\text{m}$

Answer 1

The correct option is C $\frac{\sqrt{3}}{2}\text{m}$

Given, side of the equilateral triangular lamina is $3\text{m}$.
In this case, triangular lamina is symmetric about $y-$ axis. So, $x-$ co-ordinate of COM of the triangular lamina is $0$.


We know that, y-coordinate of $COM$ is at ${\frac{1}{3}}^{rd}$ of height from the base i.e $\frac{1}{3}\times \frac{\sqrt{3}a}{2}$

$\therefore$ Position of $COM$ of the triangular lamina is given by $(x_{COM},y_{COM})=(0,\frac{a}{2\sqrt{3}})$
So, required height $\left(H\right)$ of $COM$ above $\text{AB}$ is equal to $\frac{a}{2\sqrt{3}}$
$H=\frac{a}{2\sqrt{3}}=\frac{3}{2\sqrt{3}}\text{m}=\frac{\sqrt{3}}{2}\text{m}$.