Finding M in an Equilateral Triangle of Gravitating Spheres

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SUMMARY

The discussion revolves around calculating the mass M of a sphere in an equilateral triangle configuration with two other spheres of mass m and a fourth sphere of mass m4 at the center. The net gravitational force on m4 is zero, leading to the conclusion that M must equal m. Participants utilized geometric relationships and gravitational force equations, specifically referencing the gravitational force formula GMm4/(3/16)d² and Gmm cos(30°)/(7/16)d². The correct approach involves understanding the geometry of the triangle and the distances involved.

PREREQUISITES
  • Understanding of gravitational force equations, specifically Newton's law of gravitation.
  • Familiarity with geometric principles related to equilateral triangles.
  • Knowledge of trigonometric functions, particularly sine and cosine.
  • Ability to manipulate algebraic expressions to solve for unknowns.
NEXT STEPS
  • Study the derivation of gravitational force equations in multi-body systems.
  • Learn about the center of mass calculations in triangular configurations.
  • Explore advanced trigonometric identities and their applications in physics.
  • Investigate the implications of gravitational equilibrium in various geometric arrangements.
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Students and educators in physics, particularly those focusing on gravitational forces and geometric configurations, as well as anyone interested in solving complex problems involving multiple masses and equilibrium conditions.

zila24
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In the figure below, two spheres of mass m and a third sphere mass M form an equilateral triangle, and a
fourth sphere of mass m4 is at the center of the triangle. The net gravitational force on m4 from the three
other spheres is zero; what is M in terms of m.
 
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zila24 said:
In the figure below, two spheres of mass m and a third sphere mass M form an equilateral triangle, and a
fourth sphere of mass m4 is at the center of the triangle. The net gravitational force on m4 from the three
other spheres is zero; what is M in terms of m.

Hi zila24! Welcome to PF! :smile:

Show us how far you've got, and where you're stuck, and then we'll know how to help you. :smile:
 
hi
i divided the triangle into a right triangle and found that if each side is R, for the right triangle the base would be 1/2 r and the length would be radical 3 over 2 and since the mass m4 is in the center the lengths above and below it would be radical 3 over 4
the forces of both m in the x direction would cancel each other, so we need to find out the force in the y direction, this is what I am having a problem with.

ifound the force between M and m4 (which is only in the y direction to be)- GMm4 / (3/16)d^2
 
i found the force for m y to be Gmm cos 49 / ( 7/16)d^2
 
but when i added them together to figure out the mass relationship, the answer was not right it should be M=m
 
zila24 said:
… since the mass m4 is in the center the lengths above and below it would be radical 3 over 4

Hi zila24! :smile:

Nooo … your geometry is wrong …

it's in the middle of the triangle, not the middle of that line. :frown:

Hint: concentrate on the small right-angled triangle at the bottom! :wink:
 
Oh ok so now i got the hypotenuse to be .577 and my calculation for the force to be
Gm(m4)cos 30 / (.577)^2

but it should be sin of 30 i don't get why =/
 
o wait is it cos of 60?

but how do i know the length between m4 and M
 
zila24 said:
Oh ok so now i got the hypotenuse to be .577

Yes, (√3)/6 = 0.577 is correct.

But you didn't need to calculate it, did you, since it's the same distance to all three masses? :smile:
and my calculation for the force to be
Gm(m4)cos 30 / (.577)^2

but it should be sin of 30 i don't get why =/

Because components always use the cos of the angle, and 30º isn't the angle. :smile:
 
  • #10
ok i can solve the question now thank u for your help =) ... but i just had one last question.. i just wanted to know how you got the .577
 
  • #11
zila24 said:
ok i can solve the question now thank u for your help =) ... but i just had one last question.. i just wanted to know how you got the .577

Now I'm confused … I thought you got the .577? :confused:

There's various ways of doing it.

One is to use that small right-angled triangle I mentioned, another is that the centre of mass is always at the one-third point. :smile:
 

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