Question

Three equal masses of \(\mathrm{m} \mathrm{kg}\) each are placed at the vertices of an equilateral triangle \(\mathrm{PQR}\) and a mass of \(2 \mathrm{~m} \mathrm{~kg}\) is placed at the centroid 0 of the triangle which is at a distance of \(\sqrt{2} \mathrm{~m}\) from each of vertices of triangle. The force in newton acting on the mass \(2 \mathrm{~m}\) is \(=\ldots \ldots \ldots\).. (A) 2 (B) 1 (C) \(\sqrt{2}\) (D) zero

Short answer

Since the net force acting on the 2m mass at the centroid is zero, the correct answer is: (D) zero

Step-by-step solution

Understand the gravitational force

The gravitational force between two masses M and m separated by a distance r is given by the formula: \( F = G * \frac{M * m}{r^2} \) where G is the gravitational constant with a value of approximately 6.674 x 10^-11 N m^2/kg^2.

Calculate the force between 2m and each corner mass

We will first calculate the gravitational force between the mass 2m at the centroid and the mass m at vertex P. Using the formula from step 1: \( F_{P} = G * \frac{2m * m}{(\sqrt{2}m)^2} \) Simplify the expression: \( F_{P} = \frac{2 * G * m^2}{2m^2} \) \( F_{P} = G \) Now, since all vertices of the equilateral triangle are symmetrical, the gravitational force between the mass 2m and the vertex masses at Q and R will also be equal to G: \( F_{Q} = F_{R} = G \)

Analyze the resultant force acting on the 2m mass

Let's draw the vector representation of the forces acting on the 2m mass at the centroid. The mass at P will exert a force F_P on the 2m mass in the direction of the PO vector. Similarly, the masses at Q and R will exert forces F_Q and F_R on the 2m mass in the directions of the QO and RO vectors, respectively. Now, we need to find the resultant force acting on the 2m mass. As an equilateral triangle, angles POQ, QOR, and POR are each equal to 120 degrees. Since the forces on the 2m mass are symmetrical, the components of F_P, F_Q, and F_R acting in the orthogonal directions will cancel each other out, resulting in a net force of zero acting on the 2m mass.

State the answer

Since the net force acting on the 2m mass at the centroid is zero, the correct answer is: (D) zero

Key concepts

Equilateral Triangle

An equilateral triangle is a fascinating geometric figure! Here's why: it has all three sides of equal length. This symmetry is important in understanding many physical problems because it simplifies complex calculations.
All three angles in an equilateral triangle are also equal, and they measure 60 degrees each. This uniformity not only makes them aesthetically pleasing but also mathematically convenient.
In the context of gravitational forces, the symmetry of the equilateral triangle ensures that any force applied at a vertex or the centroid experiences a uniform distribution.

• Each vertex of the triangle supports an identical mass in the given exercise.
• The centroid is the point where the medians of the triangle intersect, and it is the center of balance.
• In an equilateral triangle, the centroid divides each median into a 2:1 ratio.

This uniform distribution of equal masses results in equal forces exerted, making the analysis of the forces acting on a point within the triangle simpler and more intuitive.

Vector Addition

Understanding vector addition is essential when dealing with forces, especially when those forces are acting in different directions. In the scenario where forces exert gravitational pull, these forces are vectors, which have both magnitude and direction.
Vector addition helps determine the net force acting on an object by adding these forces together. The method of vector addition involves breaking down vectors into their components and then summing these components.
For an equilateral triangle, the gravitational forces from each vertex toward the centroid can be decomposed into x and y components.

• Vectors can be represented graphically as arrows, with direction and length showing orientation and magnitude.
• For the forces from the vertices to the centroid, the symmetry of an equilateral triangle simplifies the process, as the forces tend to cancel each other out due to symmetrical opposite directions.
• This cancellation means that the sum of these vector forces results in a net force of zero.

Thus, understanding vector addition not only aids in visualizing the problem but also in finding that resultant forces are zero, which is critical in this exercise.

Newton's Law of Gravitation

Newton's Law of Gravitation is a fundamental principle describing the attractive force between two masses. It forms the mathematical backbone for many problems involving gravitational forces.
This law states that every mass attracts every other mass through a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The formula is \( F = G \frac{M \cdot m}{r^2} \)
where

• \(F\) is the gravitational force,
• \(G\) is the gravitational constant, approximately 6.674 × 10^-11 N m²/kg²,
• \(M\) and \(m\) are the two masses, and
• \(r\) is the distance between the centers of the two masses.

In the given exercise, using this law allows calculation of gravitational forces between the mass at the centroid and those at each vertex of the triangle.
Because all masses and distances are the same, the forces calculated are equal in magnitude. By utilizing vector addition as previously outlined, the resultant force on the mass at the centroid is zero, showcasing the elegance and power of Newton’s insights.