Gravitational Forces on three masses at the corners of an equilateral triangle

In summary, the conversation discusses solving a math problem using vector rule and finding the answer to be \sqrt{3}\frac{Gm^2}{a^2} (-\hat{j}). The question is raised as to why (-\hat{j}) is added and why it is negative. The answer is that the positions of the masses in terms of the coordinate system are needed. The conversation also addresses a misunderstanding about what it means to know the positions of the masses in terms of the coordinate system.
  • #1
hasibx
2
0
Homework Statement
Three identical masses m are kept at the vertices of equilateral triangle of side 'a'. Find the force on A due to B and C
Relevant Equations
F =\frac{Gm_{1}m_{2}}{r^2}
I solved the math using vector rule
R= \sqrt{F^2 +F^2 +2F^2cos\frac{\pi}{3}} =\sqrt{3}\frac{Gm^2}{a^2}
But the answer is showing: \sqrt{3}\frac{Gm^2}{a^2} (-\hat{j})

My question is:

Why is (-\hat{j}) added here? Why is it negative?
 
Physics news on Phys.org
  • #2
We would need to know the positions of the masses in terms of the coordinate system.
 
  • #3
haruspex said:
We would need to know the positions of the masses in terms of the coordinate system.
 

Attachments

  • Screenshot at Jul 15 16-02-01.png
    Screenshot at Jul 15 16-02-01.png
    33.6 KB · Views: 63
  • #4
That's not what I asked for, but the given answer seems to be assuming that ##\hat j## is straight up the page in that diagram. If you were not told that then I do not see how you could be expected to get that answer.
 
  • #5
@hasibx you do not seem to understand what it means to
haruspex said:
know the positions of the masses in terms of the coordinate system.
WHERE do the points sit relative to the x-y coordinates?
 

FAQ: Gravitational Forces on three masses at the corners of an equilateral triangle

How do you calculate the gravitational force between two masses?

The gravitational force between two masses can be calculated using Newton's law of universal gravitation: \( F = G \frac{m_1 m_2}{r^2} \), where \( F \) is the gravitational force, \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)), \( m_1 \) and \( m_2 \) are the masses, and \( r \) is the distance between the centers of the two masses.

How do you determine the net gravitational force on one mass in an equilateral triangle configuration?

To determine the net gravitational force on one mass, calculate the gravitational forces exerted by the other two masses separately using Newton's law of universal gravitation. Then, break these forces into their components and sum the components in the x and y directions. Finally, use vector addition to find the net force.

What is the significance of the symmetry in an equilateral triangle for gravitational forces?

The symmetry of an equilateral triangle simplifies the problem because the distances between all pairs of masses are equal, and the angles between the forces are 60 degrees. This symmetry helps in breaking down the forces into components and ensures that the magnitudes of the forces due to each pair of masses are identical.

How do you resolve the gravitational forces into components?

To resolve the gravitational forces into components, use trigonometric functions. For a force \( F \) making an angle \( \theta \) with the horizontal, the horizontal (x) component is \( F \cos(\theta) \) and the vertical (y) component is \( F \sin(\theta) \). In the case of an equilateral triangle, the angles involved are typically 60 degrees or 120 degrees, which simplifies the trigonometric calculations.

What is the net gravitational force if all three masses are equal?

If all three masses are equal, the gravitational forces between any two masses will be the same in magnitude. Due to symmetry, the net gravitational force on each mass will point towards the centroid of the triangle and have the same magnitude. The exact calculation involves adding the vector components of the forces, which will result in a net force that can be found using the formula for the resultant of two vectors at 60 degrees to each other.

Back
Top