Question

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

At a point $P$ on the line CG such that $PG =\frac{1}{\sqrt{2}} m$

Answer 2

$F _{ A } = F _{ B }=\frac{ G m _{1} m _{2}}{ r ^{2}}=\frac{ G 8 \times 2}{1^{2}}= G (16)$ $F _{ AB } =\sqrt{ F _{ A }^{2}+ F _{ B }^{2}+2 F _{ A } F _{ B } \cos \theta}$ $=\sqrt{ F _{ A }^{2}+ F _{ B }^{2}+2 F _{ A } F _{ B } \cos 120}$ $=\sqrt{ F _{ A }^{2}+ F _{ B }^{2}+2 F _{ A } F _{ B }\left(-\frac{1}{2}\right)}$ $F _{ AB } = F _{ A }= G (16)$ For resultant force on $2\, kg$ to be zero $\overrightarrow{ F }_{ CG }=-\overrightarrow{ F }_{ AB }$ $\Rightarrow \frac{ G 2 \times 4}{ X ^{2}}= G (16)$ $X ^{2}=\frac{2 \times 4}{16}=\frac{1}{2}$ $X =\frac{1}{\sqrt{2}}$