three identical particles each of mass m placed at the verticles of an equilateral triangle of slide a .find the force exerted by this system on a particle p of mass m placed at center.

To determine the force exerted by three identical particles, each of mass \( m \), positioned at the vertices of an equilateral triangle with side length \( a \), on a fourth particle \( P \) of mass \( m \) placed at the center of the triangle, we can follow a systematic approach using the principles of gravitational force and symmetry.

Understanding the Setup

First, let's visualize the scenario. We have three particles located at the vertices of an equilateral triangle. The center of this triangle is where particle \( P \) is placed. Due to the symmetry of the arrangement, the forces exerted by each of the three particles on particle \( P \) will have specific characteristics that we can analyze.

Calculating the Distance from Vertices to Center

The distance from each vertex of the triangle to the center can be calculated using the formula for the centroid of an equilateral triangle. The centroid divides each median in a 2:1 ratio. The length of a median \( m \) in an equilateral triangle can be calculated as:

• Median \( m = \frac{\sqrt{3}}{2} a \)

Thus, the distance \( r \) from each vertex to the center is:

• Distance \( r = \frac{1}{3} m = \frac{1}{3} \cdot \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{6} a \)

Calculating the Gravitational Force

The gravitational force \( F \) between two masses is given by Newton's law of gravitation:

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

In our case, the force exerted by one of the particles at a vertex on particle \( P \) is:

F_{vertex} = G \frac{m^2}{r^2}

Substituting \( r = \frac{\sqrt{3}}{6} a \):

F_{vertex} = G \frac{m^2}{\left(\frac{\sqrt{3}}{6} a\right)^2} = G \frac{m^2}{\frac{3}{36} a^2} = \frac{12 G m^2}{a^2}

Considering the Forces from All Three Particles

Since the triangle is symmetric, the forces exerted by each of the three particles on \( P \) will have the same magnitude but will be directed along the lines connecting the vertices to the center. The angle between any two forces is \( 120^\circ \). To find the resultant force, we can use vector addition.

Resolving Forces

Each force can be resolved into its x and y components. The x-components of the forces from two particles will cancel each other out due to symmetry, while the y-components will add up. The y-component of each force can be calculated as:

F_{y} = F_{vertex} \cdot \sin(60^\circ) = \frac{12 G m^2}{a^2} \cdot \frac{\sqrt{3}}{2} = \frac{6\sqrt{3} G m^2}{a^2}

Since there are three particles, the total y-component of the force acting on \( P \) is:

F_{total} = 3 \cdot F_{y} = 3 \cdot \frac{6\sqrt{3} G m^2}{a^2} = \frac{18\sqrt{3} G m^2}{a^2}

Final Result

The total force exerted by the three identical particles on the particle \( P \) at the center of the triangle is:

F_{total} = \frac{18\sqrt{3} G m^2}{a^2}

This force is directed vertically upwards along the y-axis, away from the plane of the triangle, due to the symmetry of the system. Thus, we have successfully calculated the gravitational force acting on the central particle by the three particles at the vertices of the triangle.