# Force on a mass by two fixed masses

1. Oct 24, 2016

### kubaanglin

1. The problem statement, all variables and given/known data
Two identical point masses, each of mass $M$, always remain separated by a distance of $2R$. A third mass $m$ is then placed a distance $x$ along the perpendicular bisector of the original two masses. Show that the gravitational force on the third mass is directed inward along the perpendicular bisector and has a magnitude of
$$F = \frac {2GMmx}{(x^2+R^2)^\frac{3}{2}}$$
2. Relevant equations
$$F = G\frac {m_1m_2}{r^2}$$
3. The attempt at a solution

Am I on the right track? If so, what do I do from here? If not, where did I go wrong?

2. Oct 24, 2016

### TSny

You're on the right track. In your expression for the force, FM on m, you wrote the denominator as (R2 + x2)1/2. The power of 1/2 is not correct.

You can simplify cosθ. There is no need to write it as cos(tan-1(R/x)). You should be able to "read off" cosθ from one of your triangles in your picture.

3. Oct 24, 2016

### kubaanglin

Why is $(R^2+x^2)^{1/2}$ not correct? Isn't that the distance between $M$ and $m$?

4. Oct 24, 2016

### TSny

Yes, it's the distance between M and m. But what is the denominator in Newton's law of gravity?