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Gravitational Attraction-

  1. Apr 17, 2008 #1
    Gravitational Attraction-Please Help!

    1. The problem statement, all variables and given/known data
    Two 100kg lead sphers are suspended from 100m long massless cables. The tops of the cables have been anchored 1 m apart. What is the distance between the center of the spheres?

    2. Relevant equations
    Fg= Gm1m2/r^2


    3. The attempt at a solution

    Would I say find the force between them when they are just anchored, using 1m as my value for r? Then I thought I might use a force body diagram to find the angle that the cables are hanging and then the distance between the spheres. But the problem is, isn't the force going to increase as they get closer, so wouldn't they just keep moving toward each other?
     
  2. jcsd
  3. Apr 17, 2008 #2

    Dick

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    Don't worry about it. Do what you said you were going to do. The movement toward each other is 'tiny', the correction to that is 'tiny'^2, or maybe even 'tiny'^3. Ignore it. It's completely insignificant.
     
  4. Apr 22, 2008 #3
    Ok, I summed up the forces on one sphere

    Fx= GM^2/(1m)^2 In this case it is 6.67 *10^-7 N
    Fy= T-mg T= 980 N

    But how do I get an angle from that?
     
  5. Apr 22, 2008 #4
    Any thoughts on this question anybody??
     
  6. Apr 22, 2008 #5

    Dick

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    Fy is just mg. Now call theta the angle the cable makes with the vertical. Split T into x and y components. The total force on the mass is Fx+Fy+T=0.
     
  7. Apr 22, 2008 #6
    The change in the distance because the cables will both point to the center of the earth, wich is equal to:
    [tex] \frac {d l}{R_{earth}} [/tex]
    wich is 0.016 mm is much bigger than the change because of the mutual attraction of the lead balls.
     
  8. Apr 22, 2008 #7
    Ty=980 and then does tension on x equal the force of gravity. I guess I don't understand what you mean by Fx + Fy + T = 0
     
  9. Apr 22, 2008 #8

    Dick

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    I mean that the sum of all of the forces on a stationary object is 0. Ty does equal 980N. What does Tx equal? Ty also equals T*cos(theta) and Tx equals T*sin(theta). Which you would know if you'd split T into x and y components. Does that suggest a way to find theta?
     
  10. Apr 22, 2008 #9
    I can't think of anything other than gravity between the spheresthat Tx would equal. When I draw my force body diagram, I have gravity from the other sphere along x and mg along y. But that doesn't seem to make the forces sum to zero. Once I figure Tx out though, I can draw a triangle and use trigonometry to find that angle.
     
  11. Apr 22, 2008 #10

    Dick

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    Tx does equal the gravitational attraction between the spheres! If you can use trig from there then you are good to go. They do sum to zero Ty=-Fy and Tx=-Fx.
     
  12. Apr 22, 2008 #11
    arctan(6.67E-10/980) = 6.8 * 10-10 degrees that's very tiny!
     
    Last edited: Apr 22, 2008
  13. Apr 22, 2008 #12

    Dick

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    You meant 6.67*10^(-7), right? And your final answer is correct but in radians, not degrees. But yes. Still very tiny. Gravitational force is extremely weak.
     
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