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Three masses each of mass M are placed at a vertices of an equilateral triangle ABC of side l the force acting on a mass 2m placed at a centriole 4 of the triangle is?
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Three masses each of mass M are placed at a vertices of an equilateral...
If the mass at vertex A gets doubled, then resultant force will be 2FA - FB = FA
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Three masses each of mass M are placed at a vertices of an equilateral...
Force Acting on Mass 2m at Centroid 4 of the Triangle

To determine the force acting on a mass 2m placed at the centroid 4 of the equilateral triangle ABC, we can use the concept of gravitational force between objects. The force acting on the mass at the centroid is the vector sum of the individual gravitational forces exerted by the other masses at the vertices.

Step 1: Identifying the Forces
Let's label the three masses at the vertices of the triangle as M1, M2, and M3. The mass at the centroid is labeled as M4. We need to find the force acting on M4 due to the gravitational pull of M1, M2, and M3.

Step 2: Calculating the Magnitude of the Forces
The magnitude of the gravitational force between two objects is given by Newton's law of universal gravitation:

F = G * (m1 * m2) / r^2

Where:
- F is the magnitude of the gravitational force
- G is the gravitational constant (approximately 6.67430 x 10^-11 N m^2/kg^2)
- m1 and m2 are the masses of the objects
- r is the distance between the centers of the objects

Step 3: Calculating the Distance
To find the distance between the centroid and the vertices, we can use the properties of an equilateral triangle. The centroid of an equilateral triangle divides each median in a 2:1 ratio, where the longer segment is twice as long as the shorter segment. Therefore, the distance between the centroid and the vertices is l/3, where l is the side length of the triangle.

Step 4: Calculating the Forces
Using the information from Steps 2 and 3, we can calculate the magnitudes of the forces acting on M4 due to M1, M2, and M3. Since the masses at the vertices are all M, the forces can be expressed as:

F1 = G * (M * M4) / (l/3)^2
F2 = G * (M * M4) / (l/3)^2
F3 = G * (M * M4) / (l/3)^2

Step 5: Vector Sum of Forces
To find the net force acting on M4, we need to sum the forces F1, F2, and F3 as vectors. Since the triangle is equilateral, the forces are evenly distributed and have the same magnitude. Therefore, the net force can be calculated as:

Fnet = F1 + F2 + F3 = 3 * (G * (M * M4) / (l/3)^2)

Step 6: Simplifying the Expression
To simplify the expression further, we can substitute (l/3)^2 with (l^2)/9:

Fnet = 3 * (G * (M * M4) / (l^2)/9)
= G * (3M * M4) / (l^2)/9
= G * (3M * M4) * (9/l^2)
= 27 * (G * M * M4) / l^2
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Three masses each of mass M are placed at a vertices of an equilateral triangle ABC of side l the force acting on a mass 2m placed at a centriole 4 of the triangle is?
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Three masses each of mass M are placed at a vertices of an equilateral triangle ABC of side l the force acting on a mass 2m placed at a centriole 4 of the triangle is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Three masses each of mass M are placed at a vertices of an equilateral triangle ABC of side l the force acting on a mass 2m placed at a centriole 4 of the triangle is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Three masses each of mass M are placed at a vertices of an equilateral triangle ABC of side l the force acting on a mass 2m placed at a centriole 4 of the triangle is?.
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