Question

If three particles, each of mass $M$, are placed at the three corners of an equilateral triangle of side $a$, the forces exerted by this system on another particle of mass $M$ placed
$\left(i\right)$ at the midpoint of a side and
$\left(ii\right)$ at the centre of the triangle are respectively,

• A. $0,\frac{4G{M}^2}{3a^2}$
• B. $\frac{4G{M}^2}{3a^2},0$
• C. $\frac{3G{M}^2}{a^2},\frac{G{M}^2}{a^2}$
• D. $0,0$

Answer 1

The correct option is B $\frac{4G{M}^2}{3a^2},0$

Consider a equilateral triangle $\text{ABC}$ of side $a$ as shown in figure.

Case $\left(i\right):$ Let the particle be placed at the middle of side $\text{AB}$ as shown in figure.


Gravitational force on the particle placed at the midpoint of side $\text{AB}$ is

$\overrightarrow{F}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3$

But from the given figure it is clear that ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ are equal in magnitude but opposite in direction and hence will cancel each other.

So, net force on the particle,

$F=\left|{\overrightarrow{F}}_3\right|=\frac{G{M}^2}{(CD{)}^2}....\left(1\right)$

From figure, $\text{CD}=a\sin {60}^{\circ }=\frac{\sqrt{3}a}{2}$

From equation $\left(1\right)$,

$F=\frac{G{M}^2}{{\left(\frac{\sqrt{3}a}{2}\right)}^2}$

$\therefore F=\frac{4G{M}^2}{3a^2}$

Now,
Case $\left(ii\right)$ When the particle is placed at the centre of triangle:

Since the particle is placed along the median of the triangle, it is equidistant from all the three vertices.

Hence, the magnitude of gravitational forces acting due to each of the three particles is the same.


Also, it can be seen in the diagram that vectors ${\overrightarrow{F}}_1$ , ${\overrightarrow{F}}_2$ & ${\overrightarrow{F}}_3$ are making ${120}^{\circ }$ with each other.

Hence, by Lami’s theorem net force acting on the mass placed at the centre of the triangle will be,

${\overrightarrow{F}}_{net}={\overrightarrow{F}}_1+{\overrightarrow{F}}_2+{\overrightarrow{F}}_3=0$

Hence, option (b) is the correct answer.