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Chapter 6: Problem 679

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

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

Step by step solution

01

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.
02

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: \( F_{Q} = F_{R} = G \)
03

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.
04

State the answer

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

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Most popular questions from this chapter

The density of a newly discovered planet is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of earth. If the radius of the earth is \(\mathrm{R}\), the radius of planet would be (A) \(2 \mathrm{R}\) (B) \(4 \mathrm{R}\) (C) \(1 / 4 \mathrm{R}\) (D) \(\mathrm{R} / 2\)

Match the following Table-1 \(\quad\) Table-2 (A) kinetic energy (P) \([(-\mathrm{GMm}) /(2 \mathrm{r})]\) (B) Potential energy (Q) \(\sqrt{(\mathrm{GM} / \mathrm{r})}\) (C) Total energy (R) - [(GMm) / r] (D) orbital velocity (S) \([(\mathrm{GMm}) /(2 \mathrm{r})]\) Copyright (O StemEZ.com. All rights reserved.

Which one of following statements regarding artificial satellite of earth is incorrect (A) The orbital velocity depends on the mass of the satellite (B) A minimum velocity of \(8 \mathrm{kms}^{-1}\) is required by a satellite to orbit quite close to the earth. (C) The period of revolution is large if the radius of its orbit is large (D) The height of geostationary satellite is about \(36000 \mathrm{~km}\) from earth

The escape velocity of an object from the earth depends upon the mass of earth (M), its mean density ( \(p\) ), its radius (R) and gravitational constant (G), thus the formula for escape velocity is (A) \(U=\mathrm{R} \sqrt{[}(8 \pi / 3) \mathrm{Gp}]\) (C) \(\mathrm{U}=\sqrt{(2 \mathrm{GMR})}\) (D) \(U=\sqrt{\left[(2 \mathrm{GMR}) / \mathrm{R}^{2}\right]}\)

Gravitational potential at any point inside a spherical shall is uniform and is given by \(-(\mathrm{GM} / \mathrm{R})\) where \(\mathrm{M}\) is the mass of shell and \(\mathrm{R}\) its radius. At the center solid sphere, potential is \([-\\{(3 \mathrm{GM}) /(2 \mathrm{R})\\}]\)(1) There is a concentric hole of radius \(\mathrm{R}\) in a solid sphere of radius \(2 \mathrm{R}\) mass of the remaining portion is \(\mathrm{M}\) what is the gravitational at center? (A) \(-[(3 \mathrm{GM}) / 7 \mathrm{R}]\) (B) \(-[(5 \mathrm{GM}) / 7 \mathrm{R}]\) (C) \(-[(7 \mathrm{GM}) / 14]\) (D) \(-[(9 \mathrm{GM}) /(14 \mathrm{R})]\)
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