Force on a mass by two fixed masses

[kubaanglin]

Homework Statement

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

Homework Equations

$$F = G\frac {m_1m_2}{r^2}$$

The Attempt at a Solution

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

 

[TSny]

You're on the right track. In your expression for the force, F_{M 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.

 

[kubaanglin]

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

 

[TSny]

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