Finding M in an Equilateral Triangle of Gravitating Spheres

[zila24]

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.

 

[tiny-tim]

Welcome to PF!

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!

Show us how far you've got, and where you're stuck, and then we'll know how to help you.

 

[zila24]

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

 

[zila24]

i found the force for m y to be Gmm cos 49 / ( 7/16)d^2

 

[zila24]

but when i added them together to figure out the mass relationship, the answer was not right it should be M=m

 

[tiny-tim]

… since the mass m4 is in the center the lengths above and below it would be radical 3 over 4

Hi zila24!

Nooo … your geometry is wrong …

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

Hint: concentrate on the small right-angled triangle at the bottom!

 

[zila24]

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 =/

 

[zila24]

o wait is it cos of 60?

but how do i know the length between m4 and M

 

[tiny-tim]

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?

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.

 

[zila24]

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

 

[tiny-tim]

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?

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.