a). Four particles, each of mass m , are situated at the vertices of a regular tetrahedron of side a. Find the Gravational force exerted on any one of the particles by the other three
b) Three uniform rigid spheres of mass M and radius a are placed on a horizontal table and are pressed together so that their centres are at the vertices of an equilateral triangle. A fourth uniform rigid sphere of mass M and radius a is placed on top of the other three so that all four spheres are in contact with each other. Find the gravitional force exerted on the upper phere by the three lowers one
The Attempt at a Solution
What is really going to be difficult for me to calculate is the distance between the mass that lies on top of the vertices and the of one of the three masses that lies on the three vertices of the equilateral triangle; the base of the tetrahedron is an equilateral triangle since the polyhedron is a regular tetrahedron.
attempting calculation at height: since triangle is equilateral, angles will be 60 degrees each. I draw a line segment that bisects one of the four equilateral triangles: therefore a breaks into two segments of a/2. I also consider that a*cos(30)= a*root(3)/2 . using the pythagorean theorem my Radius should be: R=Root((a*root(3)/2)^2 +(a/2)^2)=Root(3*a^2+a^2/4)) ; that isn't right .What other method should I apply to obtain the radius.
part b) Almost the same as above: h=a*root(3)/2 ; therefore R=root((3/4)*a+(3/4)*a)=.5*root(6*a) ; still wrong
how is part b different from part a , other than the fact that the masses of part b have a spherical shape.