Question

Two bodies of masses $8kg$ are placed at the vertices $A$ and $B$ of an equilateral triangle $ABC$. A third body of mas $2kg$ is placed at the centroid $G$ of the triangle. If $AG=BG=CG=1m$, where should a fourth body of mass $4kg$ be placed so that the resultant force on the $2kg$ body is zero?

• A. At C
• B. At a point $P$ on the line CG such that $PG=\frac{1}{\sqrt{2}}m$
• C. At a point P on the line CG such that PG = 0.5 m
• D. At a point P on the line CG such that P G = 2 m

Answer 1

Answer: (B)

$F_A=F_B=\frac{G{m}_1{m}_2}{r^2}=\frac{G8\times 2}{1^2}=G(16)$
$F_{AB}=\sqrt{F_A^2+F_B^2+2F_A{F}_B\cos \theta }$
$=\sqrt{F_A^2+F_B^2+2F_A{F}_B\cos 120}$
$=\sqrt{F_A^2+F_B^2+2F_A{F}_B\left(-\frac{1}{2}\right)}$
$F_{AB}=F_A=G(16)$
For resultant force on $2kg$ to be zero
${\overrightarrow{F}}_{CG}=-{\overrightarrow{F}}_{AB}$
$\Rightarrow \frac{G2\times 4}{X^2}=G(16)$
$X^2=\frac{2\times 4}{16}=\frac{1}{2}$
$X=\frac{1}{\sqrt{2}}$