Question

Like parallel forces act at the vertices $A$, $B$ and$C$ of a triangle $ABC$ and are proportional to the lengths $BC$,$AC$ and $AB$ respectively. The center of the forces is at the

• A. Centroid
• B. Circum-center
• C. Incentre
• D. None of these

Answer 1

The correct option is C

Incentre

Step 1: Formula used:

Like parallel forces act at the vertices $A$, $B$ and$C$ of a triangle $ABC$.

The forces are proportional to the lengths $BC$,$AC$ $AB$ respectively.

The resultant vector of some vectors is represented as,
$\overrightarrow{R}$$=\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}+.......$

Step 2: Calculating the position of the center of forces

Let$\overrightarrow{{\lambda }_a},\overrightarrow{{\lambda }_b},\overrightarrow{{\lambda }_c}$ be the forces acting at the points$A,B,C$ respectively.

Let the resultant$\lambda (\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})$ of these forces act at a point$O$ inside the triangle$ABC$.

Let$AD$ be perpendicular to$BC$, $OL$perpendicular to $AB$.

The algebraic sum of the moments of the forces about$BC$ = moment of the resultant about$BC$

$\begin{array}{rcl}{\lambda }_a.AD& =& \lambda (a+b+c).OL\\ \lambda (BC.AD)& =& \lambda (a+b+c).OL\\ 2\Delta & =& 2s.OL\\ OL& =& \frac{\Delta }{s}\\ OL& =& r\end{array}$ (where $∆$denotes area of a triangle, $r=$the radius of the circle inscribed in a triangle)

The center of the force is at the incentre.

Hence option C is the correct answer.