Question

A metal rod of length $50\text{cm}$ having mass $2\text{kg}$ is supported on two edges placed $10\text{cm}$ from each end. A $3\text{kg}$ load is suspended at $20\text{cm}$ from one end. Find the reactions at the edges (take $g=10{\text{m/s}}^2$)

• A. $30\text{N},10\text{N}$
• B. $20\text{N},20\text{N}$
• C. $30\text{N},20\text{N}$
• D. $30\text{N},30\text{N}$

Answer 1

The correct option is C $30\text{N},20\text{N}$


As the rod is uniform and homogeneous, therefore COM is at the centre of the rod.
For translational equilibrium, ${\sum }_i\overrightarrow{F_i}=0$
$\Rightarrow R_1+R_2-3\times 10-2\times 10=0$
($R_1$ and $R_2$ are acting in vertically upward direction)
$\Rightarrow R_1+R_2=50$ ... $\left(1\right)$
For rotational equilibrium, ${\sum }_i\overrightarrow{{\tau }_i}=0$
(Taking counter-clockwise moment as positive)
$\Rightarrow -R_1\left(EC\right)+30\left(CD\right)+R_2\left(FC\right)=0$

$\Rightarrow -R_1\left(0.15\right)+30\left(0.05\right)+R_2\left(0.15\right)=0$
$\Rightarrow R_1-R_2=10$ ... $\left(2\right)$
Adding $\left(1\right)$ and $\left(2\right)$, we get
$2R_1=60\Rightarrow R_1=30\text{N}$
and $R_2=R_1-10=30-10=20\text{N}$
$\therefore R_2=20\text{N}$
So, the reaction forces at the edges $E$ and $F$ are $30\text{N}$ and $20\text{N}$ respectively.