Three point masses each of mass m are placed at the three corners of an equilateral triangle of side x. Find the resultant force acting on any one particle at the corner.

Question

Prepp Team · Accepted Answer

This problem asks us to find the net gravitational force acting on one of the three identical point masses placed at the corners of an equilateral triangle. Gravitational force is a vector quantity, so we need to consider both the magnitude and direction of the forces.

Understanding the Setup: Masses on an Equilateral Triangle

We have three point masses, each of mass $m$. They are located at the corners of an equilateral triangle with side length $x$.

Let the corners of the triangle be A, B, and C.
Mass $m$ is placed at A, B, and C.
The distance between any two corners (A to B, B to C, C to A) is $x$.

We want to find the total force acting on any one mass, say the mass at corner A.

Identifying the Forces Acting on One Mass

The mass at A experiences gravitational forces from the other two masses:

Force due to the mass at B, pulling A towards B. Let's call this $\vec{F}_{AB}$.
Force due to the mass at C, pulling A towards C. Let's call this $\vec{F}_{AC}$.

According to Newton's Law of Universal Gravitation, the magnitude of the force between two point masses $m_1$ and $m_2$ separated by a distance $r$ is given by $F = \frac{G m_1 m_2}{r^2}$, where $G$ is the gravitational constant.

Calculating the Magnitude of Individual Forces

For the force between the mass at A and the mass at B:

$m_1 = m$, $m_2 = m$
Distance $r = x$
Magnitude of $F_{AB} = \frac{G m \cdot m}{x^2} = \frac{G m^2}{x^2}$. This force is directed from A towards B.

For the force between the mass at A and the mass at C:

$m_1 = m$, $m_2 = m$
Distance $r = x$
Magnitude of $F_{AC} = \frac{G m \cdot m}{x^2} = \frac{G m^2}{x^2}$. This force is directed from A towards C.

So, the magnitudes of the two forces acting on the mass at A are equal: $|\vec{F}_{AB}| = |\vec{F}_{AC}| = \frac{G m^2}{x^2}$. Let's denote this magnitude as $F_0 = \frac{G m^2}{x^2}$.

Vector Addition to Find the Resultant Force

The angle between the lines AB and AC in an equilateral triangle is 60 degrees. The forces $\vec{F}_{AB}$ and $\vec{F}_{AC}$ act along these lines, directed inwards towards B and C, respectively. Therefore, the angle between the force vectors $\vec{F}_{AB}$ and $\vec{F}_{AC}$ is 60 degrees.

To find the resultant force $\vec{F}_{resultant}$ acting on the mass at A, we need to add the vectors $\vec{F}_{AB}$ and $\vec{F}_{AC}$ vectorially: $\vec{F}_{resultant} = \vec{F}_{AB} + \vec{F}_{AC}$.

We can use the parallelogram law of vector addition. If two vectors of magnitudes $A$ and $B$ act at an angle $\theta$, the magnitude of their resultant $R$ is given by $R = \sqrt{A^2 + B^2 + 2AB \cos\theta}$.

In our case:

$A = |\vec{F}_{AB}| = F_0 = \frac{G m^2}{x^2}$
$B = |\vec{F}_{AC}| = F_0 = \frac{G m^2}{x^2}$
$\theta = 60^\circ$

The magnitude of the resultant force is:

$|\vec{F}_{resultant}| = \sqrt{(F_0)^2 + (F_0)^2 + 2(F_0)(F_0) \cos(60^\circ)}$

Since $\cos(60^\circ) = \frac{1}{2}$:

$|\vec{F}_{resultant}| = \sqrt{F_0^2 + F_0^2 + 2F_0^2 \left(\frac{1}{2}\right)}$

$|\vec{F}_{resultant}| = \sqrt{F_0^2 + F_0^2 + F_0^2}$

$|\vec{F}_{resultant}| = \sqrt{3 F_0^2}$

$|\vec{F}_{resultant}| = F_0 \sqrt{3}$

Substitute the value of $F_0$:

$|\vec{F}_{resultant}| = \left(\frac{G m^2}{x^2}\right) \sqrt{3}$

$|\vec{F}_{resultant}| = \frac{\sqrt 3 {\rm{\;}}G{m^2}}{{{x^2}}}$

The resultant force acting on any one particle at the corner is $\frac{{\sqrt 3 {\rm{\;}}G{m^2}}}{{{x^2}}}$.

Summary of Calculation Steps

Identify the forces acting on the target mass due to other masses.
Calculate the magnitude of each individual gravitational force using Newton's Law.
Determine the angle between these force vectors.
Use vector addition (like the parallelogram law) to find the magnitude of the resultant force.

Quantity	Value	Notes
Mass of each particle	$m$	Given
Side length of triangle	$x$	Given
Magnitude of force between any two particles	$F_0 = \frac{G m^2}{x^2}$	Newton's Law
Angle between force vectors on one particle	$60^\circ$	Angle of equilateral triangle
Resultant force magnitude	$F_0 \sqrt{3}$	Vector addition ($\sqrt{F_0^2 + F_0^2 + 2F_0^2 \cos 60^\circ}$)
Final resultant force	$\frac{{\sqrt 3 {\rm{\;}}G{m^2}}}{{{x^2}}}$	Substituting $F_0$

Revision Table: Gravitational Force on Triangle Corner

Concept	Description	Formula/Key Idea
Newton's Law of Gravitation	Force between two point masses is proportional to the product of their masses and inversely proportional to the square of the distance between them.	$F = \frac{G m_1 m_2}{r^2}$
Vector Addition	Forces are vectors and must be added considering their directions.	Resultant of $\vec{A}$ and $\vec{B}$ at angle $\theta$ is $\sqrt{A^2 + B^2 + 2AB \cos\theta}$.
Equilateral Triangle Properties	All sides are equal, all interior angles are 60 degrees.	Angles are crucial for vector direction.

Additional Information: Symmetrical Mass Distributions

For symmetrical arrangements of masses, the resultant force on a mass at the center can often be zero due to cancellation of forces. However, for masses placed at the corners of a polygon (like a triangle, square, etc.), the forces on a mass at a corner from other corners typically don't cancel out completely unless the mass is isolated or the forces happen to balance due to specific arrangement (e.g., a central mass attracting corner masses, where the net force on the central mass would be zero if equally distant and same magnitude corner masses exist). In this case, the two forces on the corner mass are at a 60-degree angle, requiring vector addition.

The direction of the resultant force on the mass at A would bisect the angle between $\vec{F}_{AB}$ and $\vec{F}_{AC}$, pointing along the angle bisector of $\angle BAC$, towards the center of the opposite side BC.

Prepp Team · Answer

Zero

Prepp Team · Answer

$\frac{{3{\rm{\;}}G{m^2}}}{{{x^2}}}$

Prepp Team · Answer

$\frac{{G{m^2}}}{{{x^2}}}$