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Newton's Law of Gravitation (again)

  1. Sep 13, 2008 #1
    1. The problem statement, all variables and given/known data

    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
    2. Relevant equations

    Possible F=m1*m2*G/R^2

    3. The attempt at a solution

    part a)
    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.
  2. jcsd
  3. Sep 14, 2008 #2


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    Staff: Mentor

    In part a, the centers of each pair of masses is separated by distance a. Remember that gravitational force is a vector, so one needs to look at horizontal (parallel with the plane of three mass) and vertical (perpendicular with plane of three mass) components of the resultant force vector.

    What is the implication of part a and part b giving the same answer?
  4. Sep 14, 2008 #3
    But wouldn't a tetrahedron be in a 3-d plane, so wouldn't I have to considered the x,y, and z components of the plane?

    Would F1 be: F1=m*m*G/(a*cos(60))^2+m*m*G/(a*sin(60))^2
    since all 3 forces have the same masses, F net would look like:

    F(net)= 3*(m*m*G/(a*cos(60))^2+m*m*G/(a*sin(60))^2)
  5. Sep 14, 2008 #4
    was my last response not understandable?
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