Question

Three particles, each of mass $M$ , are placed at the three corners of an equilateral triangle of side $l$. What is the force due to this system of particles on another particle of mass $m$ placed at the midpoint of any side?
$\begin{align}
  & A.\dfrac{3GMm}{4{{l}^{2}}} \\
 & B.\dfrac{4GMm}{3{{l}^{2}}} \\
 & C.\dfrac{GMm}{4{{l}^{2}}} \\
 & D.\dfrac{4GMm}{{{l}^{2}}} \\
\end{align}$

Answer 1

Hint: Calculate the force exerted by each of the particles of mass $M$ placed at the corners of the equilateral triangle to the particle of mass $m$ placed at the midpoint of any side of the triangle.

Formula Used:
$F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$

Complete step by step answer:
We know that the force exerted between two particles, placed at a distance apart from each other is called gravitational force.
Gravitational Force is directly proportional to the product of mass of two particles or objects and inversely proportional to the square of their distance. It is a vector quantity that has both magnitude and direction.
Mathematically gravitational force is given by:
$F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$

Where:
$F=$ Gravitational Force between two objects.
${{m}_{1}}=$ mass of one object
${{m}_{2}}=$ mass of another object
$r=$distance between the object
In our question three particles of mass $M$ are placed at the corners of the equilateral triangle and one particle of mass $m$ place at the midpoint at any side of the triangle.

Therefore force exerted particle placed at corner B to the particle of mass $m$ placed at midpoint D is given by
${{F}_{BD}}=\dfrac{GMm}{{{\left( \dfrac{l}{2} \right)}^{2}}}(-i)$

Because D is the midpoint of side BC therefore distance becomes $\dfrac{l}{2}$, and here $-i$represents the direction of force which is in the negative x axis.
Force exerted particle placed at corner C to the particle of mass $m$ placed at midpoint D is given by:
${{F}_{CD}}=\dfrac{GMm}{{{\left( \dfrac{l}{2} \right)}^{2}}}(i)$
Here $i$ represents the direction of force which is in the positive x axis.
Also, the force exerted by particle placed at corner A of mass $M$ to the particle placed at midpoint D:

$\begin{align}
  & {{F}_{AD}}=\dfrac{GMm}{{{\left( \dfrac{\sqrt{3}l}{2} \right)}^{2}}}(-j) \\
 & {{F}_{AD}}=\dfrac{4GMm}{3{{l}^{2}}}(-j) \\
\end{align}$

Here the distance between corner A and midpoint D which is on side BC is given by $\dfrac{\sqrt{3}l}{2}$ because triangle is equilateral and equilateral triangle has altitude $\dfrac{\sqrt{3}l}{2}$. Which can be found by Pythagoras Theorem.
And here $-j$represents the direction of force which is in negative y direction.
Therefore the net force is the vector sum of all the forces acting on particle placed at D is given by:
$\begin{align}
  & {{F}_{net}}={{F}_{AD}}+{{F}_{BD}}+{{F}_{CD}} \\
 & {{F}_{net}}=\dfrac{4GMm}{3{{l}^{2}}}(-j)+\dfrac{GMm}{{{\left( \dfrac{l}{2} \right)}^{2}}}(-i)+\dfrac{GMm}{{{\left( \dfrac{l}{2} \right)}^{2}}}(i) \\
 & {{F}_{net}}=\dfrac{4GMm}{3{{l}^{2}}}(-j) \\
\end{align}$
Taking the magnitude of this net force.
${{F}_{net}}=\dfrac{4GMm}{3{{l}^{2}}}$
Hence, the correct answer is option B.

Note: While doing such types of questions students must remember to take care on the direction of forces. To deal with it simply use vector notation to find your answer more accurately. For vector notation use i, j, k position vectors.