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Newton's Theory of Gravity Problem help

  1. Jul 6, 2008 #1
    1. The problem statement, all variables and given/known data
    "Two 100 kg lead spheres are suspended from 100-m-long massless cables. The tops of the cables have been carefully anchored exactly 1 m apart. What is the distance between the centers of the spheres?


    2. Relevant equations
    F = (Gm1m2)/r^2 ????


    3. The attempt at a solution
    I'm not sure how this problem is related to Newton's theory and I am confused at this problem as I think the answer should simply be 1 meter because that is exactly how far the cables are anchored apart. Using the equation for F that I have provided I get 6.67x10^-7 N. This is the only equation we have really learned in class and I am unsure how this applies to the problem or how I use this to calculate the distance between the centers of the spheres. I feel dumb because this problem sounds so easy but I feel like I'm missing something. The book says it's a medium difficulty problem and that it requires material from earlier chapters but I'm not sure what. I would greatly appreciate any help on this to get me going. Thanks!
     
  2. jcsd
  3. Jul 6, 2008 #2

    tiny-tim

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    Welcome to PF!

    Hi aarunt1! Welcome to PF! :smile:

    The tops are a metre apart, but the cables can swing, so the bottoms can be nearer if the lead spheres attract each other.

    Use trigonometry, and FBDs! :smile:
     
  4. Jul 6, 2008 #3
    thanks for the welcome. I figured that the spheres would probably attract each other somewhat and I calculated the gravitational attraction (I believe) in my attempt at the solution please correct me if I am wrong tho. I am unsure how to calculate how far 6.67x10^-7 N would push each sphere in. Is there an equation I can use for this? Thanks!
     
  5. Jul 6, 2008 #4

    tiny-tim

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    The cables are at an angle, so consider the tension in one of the cables … that's what the FBD is for! :wink:
     
  6. Jul 6, 2008 #5
    I'm all for the FBD idea but I don't know how to get the angles. If they were provided in the equation that would be cake but I can't remember there being a way to calculate them. Sorry for being so noob I haven't used trig in a long time and it's over a year since I've taken a physics class.
     
  7. Jul 7, 2008 #6

    tiny-tim

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    Hi aarunt1! :smile:

    No problem … the sine of the angle (which is equal to the tangent for very small angles) is the horizontal displacement divided by the length of the cable. :smile:
     
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