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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 escape velocity from the earth is about \(11 \mathrm{kms}^{-1}\). The escape velocity from a planet having twice the radius and the same mean density as the earth is \(=\ldots \ldots \ldots \ldots \mathrm{kms}^{-1}\). (A) 22 (B) 11 (C) \(5.5\) (D) \(15.5\)

If the density of small planet is that of the same as that of the earth while the radius of the planet is \(0.2\) times that of the earth, the gravitational acceleration on the surface of the planet is (A) \(0.2 \mathrm{~g}\) (B) \(0.4 \mathrm{~g}\) (C) \(2 \mathrm{~g}\) (D) \(4 \mathrm{~g}\)

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