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Gravitational Force HELP

  1. Jun 18, 2007 #1
    1. The problem statement, all variables and given/known data

    Three uniform spheres are located at the corners of an equilateral triangle. Each side of the triangle has a length of 3.80 m. Two of the spheres have a mass of 3.30 kg each. The third sphere (mass unknown) is released from rest. Considering only the gravitational forces that the spheres exert on each other, what is the magnitude of the initial acceleration of the third sphere?

    2. Relevant equations

    F=Gm1m2/d^2
    F=ma

    3. The attempt at a solution

    I tried doing it by setting the 2 above equations = to each and so I got:

    G(distance b/w 2 spheres)/d^2=a

    and I don't know if I'm almost there or not....
     
  2. jcsd
  3. Jun 18, 2007 #2

    Dick

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    You will need to consider the direction of the forces as well as their magnitude. Take the vector sum of the forces from the 3.30kg masses remembering that the forces point along the edges of the triangle. The forces will contain the unknown mass of the third mass, but that will divide back out again going from force to acceleration.
     
  4. Jun 18, 2007 #3
    That is true for one sphere. You have two spheres attracting the third sphere, and the direction of the two forces are different.
     
  5. Jun 18, 2007 #4
    so i got the a to equal 3 x 10^-11 for one sphere....is that right? i'm a little confused
     
  6. Jun 18, 2007 #5

    Dick

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    You mean the acceleration produced by ONE 3.30kg mass??? I get something more like 1.5*10^-11. How would you combine two of these with a 60 degree angle between them?
     
  7. Jun 18, 2007 #6
    Oh I have no clue! :( I don't even know how that would help me figure out the problem.
     
  8. Jun 18, 2007 #7

    Dick

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    Haven't you ever drawn a force diagram of an object with several different forces acting on it it different directions?? Like a mass on an inclined plane?? BTW how DID you get 3*10^-11?
     
  9. Jun 18, 2007 #8
    I'm in intro physics and my teacher hasn't gotten to that yet is why I'm trying to ask other people how to go about doing it because the book isn't very good.

    i plugged in G, 6.6 for d and got the distance to be 3.8.
     
  10. Jun 18, 2007 #9
    I read on another person's post about this question that taking the midpoint b/w the two objects could be used (i couldnt get past the pt the person on this post got to)-

    the link is here-https://www.physicsforums.com/showthread.php?t=112713
     
  11. Jun 18, 2007 #10

    Dick

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    You mean you plugged in 6.6 for mass. This is ok only if the two forces point in the same direction. I'm not sure this is a very good problem for you if you haven't done vector addition of forces.
     
  12. Jun 18, 2007 #11
    Ah, I see. Still confused about the problem, I've been mulling over it for an hour and a half... Any tips for me? I reallyy appreciate you taking time to read this, btw.
     
  13. Jun 18, 2007 #12

    Dick

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    Without having done vector addition this is going to be a stretch, but I'll try. You know the magnitude of the equal forces caused by both masses, call it F. They point in directions 60 degrees apart and as you've pointed out the TOTAL force (vector sum of the two) acts in a direction that bisects the 60 degree angle. What you then want to do is sum only the portion of each force that points in that direction. So you split each force up into parts that are parallel to that direction and perpendicular to that direction. These part magnitudes are respectively F*cos(30) and F*sin(30) from trigonometry. The perpendicular parts cancel, since they point in opposite directions and the parallel parts sum to 2*F*cos(60) since they point in the same direction. This can only become totally clear when you learn to sum vectors.
     
  14. Jun 18, 2007 #13

    Chronos

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    Each known mass is offset 30 degrees from the unknown mass so F x cos(30) is the contribution of each known mass to the force drawing the unknown mass toward the center of the leg connecting the known masses. Notice how the net force is zero when the unknown mass is located directly between the two known masses [cos(90) = 0]. Note that once you determine the net force, you must then calculate the initial acceleration [a simple matter] of the unknown mass - hint substitute for F and M.
     
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