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

The Correct Option is B

To solve this problem, we need to determine the position of a fourth body of mass \(4\, \text{kg}\) such that the resultant gravitational force on the \(2\, \text{kg}\) body placed at the centroid \(G\) of the equilateral triangle \(ABC\) is zero. 

First, consider the gravitational force exerted by the masses at points \(A\), \(B\), and \(C\) on the \(2\, \text{kg}\) mass at \(G\). The masses at \(A\) and \(B\) are both \(8\, \text{kg}\) and are located \(1\, \text{m}\) away from \(G\).

The gravitational force \(F\) between two masses \(m_1\) and \(m_2\) at a distance \(r\) is given by:

\(F = \frac{G \cdot m_1 \cdot m_2}{r^2}\)

Where \(G\) is the gravitational constant. However, for this analysis, we are only interested in the relative magnitudes of the forces as \(G\) will cancel out, ultimately.

Step 1: Calculate the force exerted by masses at \(A\) and \(B\)

The force due to the \(8\, \text{kg}\) masses at \(A\) and \(B\) on the \(2\, \text{kg}\) mass at \(G\) will point directly towards \(A\) and \(B\), respectively. Since \(A\) and \(B\) are symmetrically placed relative to \(G\), their horizontal components will cancel each other out.

Let's denote \(F_{AG}\) and \(F_{BG}\) as the forces on \(G\) due to masses at \(A\) and \(B\):

\(F_{AG} = F_{BG} = \frac{8 \cdot 2}{1^2} = 16 \, G \cdot \text{N} \cdot m^2/kg^2\)

Both forces are directed symmetrically and thus their vertical components add up, contributing no net force horizontally which aligns in the direction of \(CG\).

Step 2: Determine the position of the \(4\, \text{kg}\) mass

The force due to the \(4\, \text{kg}\) mass, placed at a point \(P\) on line \(CG\), should cancel out the resultant of \(F_{AG}\) and \(F_{BG}\) on the \(2\, \text{kg}\) body.

Since \(F_{AG}\) and \(F_{BG}\) only have a net vertical component, place \(P\) such that the force \(F_{PG}\) counteracts this resultant force.

The force due to \(4\, \text{kg}\) mass at \(P\) is:

\(F_{PG} = \frac{4 \cdot 2}{PG^2}\)

Equating the forces in the direction along \(CG\):

\(16 = 8 \cdot \frac{1}{PG^2}\) 
\(PG = \frac{1}{\sqrt{2}} \, \text{m}\)

Therefore, the correct position of the \(4\, \text{kg}\) mass is 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 $\overrightarrow{ F }_{ CG }=-\overrightarrow{ F }_{ AB }$ $\Rightarrow \frac{ G 2 \times 4}{ X ^{2}}= G (16)$ $X ^{2}=\frac{2 \times 4}{16}=\frac{1}{2}$ $X =\frac{1}{\sqrt{2}}$