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A net gravitational problem

  1. May 7, 2008 #1
    Hello everyone,

    Could someone please help me with this problem?

    1. The problem statement, all variables and given/known data
    A 145kg object and a 445kg object are separated by .420m.
    a. Find the net gravitational force exerted by these objects on a 41.0kg object placed midway between them.
    b. At what position can the 41.0kg object be placed so as to experience a net force of zero?


    2. Relevant equations
    F=Gm1m2/r^2


    3. The attempt at a solution

    (6.67X10^-11Nm^2/kg)(145kg)(41.0kg)/.420^2=-2.2X10^-6

    (6.67X10^-11Nm^2/kg)(445kg)(4)/.420^2=6.7X10^-7

    -2.2X10^-6+6.67X10^-7=2.87X10^-6

    Thank you very much
     
  2. jcsd
  3. May 7, 2008 #2

    Redbelly98

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    If the 41.0 kg object is "placed midway between them", how far is it from the first two masses?

    Other than that, looks like you have (a) pretty much.
     
  4. May 8, 2008 #3
    Thank you very much

    Could you please tell me if a. looks correct?

    For part b., would you just take the first one plus the middle one plus the third one and set it equal to the total force and solve for the middle one?

    Thank you
     
  5. May 9, 2008 #4

    Redbelly98

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    a is not quite correct. You have the right idea except that:
    1. As I said before, if the 41.0 kg object is "placed midway between them", how far is it from the other two masses?
    2. Why do you get a smaller force due to the larger 445 kg mass? Larger masses should exert larger forces. Something is wrong here.

    For b, you'll need to write out an expression for the force on the 41 kg mass.
     
  6. May 11, 2008 #5
    Thank you very much

    For part a., would I just take 4.7 X 10^-6 and divide it by 2?

    Thank you
     
  7. May 11, 2008 #6

    Redbelly98

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    I don't see where "4.7 X 10^-6" comes in, so I'll have to say no, that is not what to do.

    For the 3rd time:
    If the 41.0 kg object is "placed midway between them", how far is it from the other two masses?
     
  8. May 11, 2008 #7
    It's in the middle. Is that what you mean? So, wouldn't you just take

    6.67 X 10^-11(145)(41/2)/.42^2?

    6.67 X 10^-11(445kg)(41.0kg)/2/(.42)^2?

    Thank you
     
    Last edited: May 11, 2008
  9. May 11, 2008 #8

    Redbelly98

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    It's almost (but not quite) what I mean. What distance is "in the middle" from the other masses? I'm look for the actual distance, in meters, from "in the middle" to one end or the other.

    We can get to that later, but first it would be helpful to answer the distance question I asked above.
     
  10. May 11, 2008 #9
    Well, in the problem it states that the two objects are separated by .420 m. Is that what you mean? Thus, the 41.0 object is at .420 m, right?

    Thank you
     
  11. May 11, 2008 #10

    Redbelly98

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    Have you drawn a diagram of the 3 masses? You are just not picturing the arrangement correctly.

    Sorry, I can't offer any more advice or hints to you.
     
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