Circular Motion - Help

[Pepsi]

I'm having troubles with one of my homework questions, and I was wondering if someone could help me with it.

okay here it is...

Q: Two masses, object A and object B are located 2m apart from eachother, the mass of object a is m and the mass of object b is 4m.

Showing your calculations find the point between these two objects where a third object would experiance no gravitation force.

I'm really stuck on this question, any help would be very much appreciated.

[StephenPrivitera]

"a third object would experiance no gravitation force."
what they mean is, no net gravitation force
imagine a particle C lying on the line connecting A and B - what is the force that C feels from A? what is the force that C feels from B?
What is the net force?

[Pepsi]

I dont understand that at all...

I would have thought I'd try and figure out what m was = to then, maybe do that... because from what I know I dont think I can figure out the net gravititional force from just what I was given...

A little more help? heh

[StephenPrivitera]

Are you aware that the force for gravitation is
\vec{F}=-\frac{Gm_1m_2}{r^2}\hat{r}

You have two masses m and 4m and some unknown mass m_2, so the magnitude of the gravitational attraction between m and m_2 is
F_1=\frac{Gmm_2}{r_1^2}
where r_1 is the distance between m_2 and m. The strength of the gravitational attraction between 4m and m_2 is
F_2=\frac{G(4m)m_2}{r_2^2}
where r_2 is the distance between 4m and m_2. If m_2 lies along the line between 4m and m and the distance between m and 4m is R, then
R=r_1+r_2, so the second equation becomes
F_2=\frac{G(4m)m_2}{(R-r_1)^2}
If m_2 is right between A and B, then the forces act in opposite directions and so cancel each other out. When F_1-F_2=0, m_2 will feel no force.