Net gravitational force on an object

[Jaccobtw]

Homework Statement

A test object of mass m is placed at the origin of a two dimensional coordinate system. An object 1, of the same mass, is at (d, 0), and an object 2, of mass 2m, is at (-d, l). What is the magnitude of the vector sum of the gravitational forces exerted on the test object by the other two objects?

Relevant Equations

F(gravity) = G * (M * m)/ (r^2)
G = 6.6738 * 10^-11 N * m^2

I suppose we can just find the net x components and y components and then go from there.
Σ Fx = F(mass 1) - Fx(mass 2)
G* (m^2)./d^2) - something

I'm not sure how to express the component forces of the 2nd mass

 

[HallsofIvy]

Object 1 is at (d, 0). The distance from object 0 to object 1 is r=d. The direction vector from object 1 to object 0 is <-d, 0> the unit vector in that direction is <-1, 0>. The magnitude of the gravitational force object 1 exert on object 0 is $\frac{Gm^2}{d^2}$ and the vector force is $\frac{Gm^2}{d^2}\left<-1, 0\right>$. Since you are asking about object 2 I presume you got that.

Object 2 is at (-d, l). The distance from object 0 to object 2 is $r=\sqrt{d^2+ l^2}$. The direction vector from object 2 to object 0 is <d, -l>. The magnitude of the gravitational force object 1 exerts on object 0 is $\frac{2Gm^2}{d^2+ l^2}$ and the vector force is $\frac{2Gm^2}{d^2+ l^2}\left<d, -l\right>$.

The resultant is the sum of those, $\left<\frac{-Gm^2}{d^2}, 0\right>+ \left<\frac{2dGm^2}{d^2+ l^2}, \frac{-2lGm^2}{d^2+ l^2}\right>$.

 

[kuruman]

The resultant is the sum of those, $\left<\frac{-Gm^2}{d^2}, 0\right>+ \left<\frac{2dGm^2}{d^2+ l^2}, \frac{-2lGm^2}{d^2+ l^2}\right>$.

Of course you meant to write
"The resultant is the sum of those, $\left<\frac{-Gm^2}{d^2}, 0\right>+ \left<\frac{2dGm^2}{(d^2+ l^2)^{3/2}}, \frac{-2lGm^2}{(d^2+ l^2)^{3/2}}\right>$."