Question

A force $\overrightarrow{F_1}=A\overrightarrow{j}$ is applied to a point whose radius vector is $\overrightarrow{r_1}=a\overrightarrow{i}$, while a force $\overrightarrow{F_2}=B\overrightarrow{i}$ is applied to the point whose radius vector is $\overrightarrow{r_2}=b\overrightarrow{j}$. Both radius vectors are determined relative to the origin of coordinates $O$. It is given $\overrightarrow{i}$ and $\overrightarrow{j}$ are the unit vectors of the x and y axes and $a$, $b$, $A$, $B$ are constants. Find the arm $l$ of the resultant force relative to the point $O$.

• A. $l=\frac{aA-bB}{\sqrt{A^2+B^2}}$
• B. $l=\frac{ab+AB}{\sqrt{A^2+B^2}}$
• C. $l=\frac{ab-AB}{\sqrt{A^2-B^2}}$
• D. None of these

Answer 1

The correct option is A $l=\frac{aA-bB}{\sqrt{A^2+B^2}}$

Relative to point $O$, the net moment of force,
$\overrightarrow{N}=\overrightarrow{r_1}\times \overrightarrow{F_1}+\overrightarrow{r_2}\times \overrightarrow{F_2}=(a\overrightarrow{i}\times A\overrightarrow{j})+(b\overrightarrow{j}\times B\overrightarrow{i})$
$=aA\overrightarrow{k}+bB(-\overrightarrow{k})=(aA-bB)\overrightarrow{k}$
Resultant of the external force
$\overrightarrow{F}=\overrightarrow{F_1}+\overrightarrow{F_2}=A\overrightarrow{j}+B\overrightarrow{i}$

Now we want to find out the arm length of the resultant force, which will give same torque.
As angle between N and F is ${90}^0$ we can write

$|N|=|F|l||$

$l=\frac{|N|}{|F|}=\frac{aA-bB}{\sqrt{A^2+B^2}}$