# Gravitation problem

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. (Picture attached.) Show that the gravitational force on the third mass is directed inward along the perpendicular bisector and has a magnititude of :

$F = \frac{2GMmx}{(x^2+R^2)^\frac{3}{2}}$

2. Relevant equations

$F=G\frac{m1m2}{r^2}$

3. The attempt at a solution

First I figured out the magnitude of the force of gravitation between one of the two mass Ms and the third mass m. To find the distance between them I used the Pythagorean theorem where:

$r^2 = x^2 + R^2$

plugging this and the two masses into the gravitation formula:

$F = G\frac{Mm}{x^2+R^2}$

That is the magnitude of the force of gravitation between each M and m. To find the magnitude and direction of the total force I should find the vector sum of these two forces? This is where I end up lost. The Pythagorean theorem wont work here. Can anyone help point me in the right direction? I know that after i find the magnitude of the resultant force I can use arctan to find the direction.

Attached Images
 pointmasses.jpg (18.9 KB)

http://ift.tt/1hdAzFT