Question

A rod of length $20\text{cm}$ is kept along $x$ axis. Two forces $5\text{N}$ and $10\text{N}$ are applied at distances $10\text{cm}$ and $15\text{cm}$ from point $A$ as shown in figure. Find the point of application of net force from point $A$.

• A. $20\text{cm}$
• B. $12.5\text{cm}$
• C. $15\text{cm}$
• D. $17.5\text{cm}$

Answer 1

The correct option is A $20\text{cm}$

Given, length of rod $\left(l\right)$ = $20\text{cm}$
Position vector of point of application of $F_1=5\text{N}(-\hat{j})$
$\overline{r_1}=10\text{cm}\hat{i}$
Position vector of point of application of $F_2=10\text{N}\left(\hat{j}\right)$
$\overline{r_2}=15\text{cm}\hat{i}$
Torque due to $F_1$
$\overrightarrow{{\tau }_1}=\overrightarrow{r_1}\times \overrightarrow{F_1}=50\times {10}^{-2}\text{Nm}(-\hat{k})$
Torque due to $F_2$
$\overrightarrow{{\tau }_2}=r_2\times \overrightarrow{F_2}=150\times {10}^{-2}\text{Nm}\left(\hat{k}\right)$

Net force acting on rod $\left(F_{net}\right)=\overrightarrow{F_1}+\overrightarrow{F_2}=5\text{N}\left(\hat{j}\right)$
Since, $\overrightarrow{r_1}\times \overrightarrow{F_1}+\overrightarrow{r_2}\times \overrightarrow{F_2}=\overrightarrow{r}\times \overrightarrow{F_{net}}$ we get
$150\times {10}^{-2}-50\times {10}^{-2}=5r\Rightarrow r=20\text{cm}$