Problem concernng Newton's Law of Gravitation

[Benzoate]

Homework Statement

Four particles, each of mass m , are situated at the vertices of a regular tetrahedron of the side a. Find the gravitational force exerted on anyone of the particles by the other three.

Homework Equations

F=m*M*G/R^2

The Attempt at a Solution

Since the mass of the four particles are equivalent, F=m^2*G/a^2. Each of the particles lie at each of the four vertices of the base of the tetrahedron. I assumed the base of the tetrahedron is a square since the polyhedron is a regular tetrahedron. particle one exerts a force on each of the remaining three particles that lie on the vertices. So I will write out 3 forces

F(1,2)=m^2*G/a^2

F(1,3)=m^2*G/2a^2

F(1,4)= m^2*G/a^2

F(total)=F(1,2)+F(1,3)+F(1,4)= 5/2*(m^2/a^2).

The book says that the total Force is sqrt(6)*m^2*G/a^2. What did I do wrong? I think the calculation for my net Force was slightly closed to the books answer, I happened to be off by .05 decimal places.

 

[Benzoate]

Is my problem understandable to anyone?

 

[granpa]

http://en.wikipedia.org/wiki/Tetrahedron

A tetrahedron (plural: tetrahedra) is a polyhedron composed of four triangular faces, three of which meet at each vertex.

 

[Benzoate]

http://en.wikipedia.org/wiki/Tetrahedron

A tetrahedron (plural: tetrahedra) is a polyhedron composed of four triangular faces, three of which meet at each vertex.

What is wrong with my Force equations?

 

[Dick]

What is wrong with my Force equations?

The total force is the sum of three vectors. These three vectors are the same length but don't point in the same direction.

 

[granpa]

I assumed the base of the tetrahedron is a square

the base is a triangle