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Gravity between 3 objects

  1. Mar 20, 2015 #1
    1. The problem statement, all variables and given/known data
    PART 1: Objects with masses of 125 kg and 548 kg are separated by 0.385 m. A 63.5 kg mass is placed midway between them.
    Screenshot%2B2015-03-20%2Bat%2B6.20.08%2BPM%2B-%2BEdited.png
    Find the magnitude of the net gravitational force exerted by the two larger masses on the 63.5 kg mass. The value of the universal gravitational constant is 6.672 × 10−11 N · m2 /kg2 . Answer in units of N.

    PART 2: Leaving the distance between the 125 kg and the 548 kg masses fixed, at what distance from the 548 kg mass (other than infinitely remote ones) does the 63.5 kg mass experience a net force of zero? Answer in units of m.


    2. Relevant equations

    F = Gmm/(d^2)

    3. The attempt at a solution

    So i got part 1 no problem, the answer is .00004836 N, but part 2 is a real tough one. For my first attempt, i set up the formula ( (548)(63.5)(6.67E-11)/x^2 ) + ( (125)(63.5)(6.672E-11)/(x+.385)^2 ) = 0

    x being distance, and (x + .385) distance from smaller mass since it remains fixed

    I got 0.8165837351598118 as an answer, but that was wrong.

    Where did I go wrong? How can I solve this? Any help would be greatly appreciated

    Thanks!
     
  2. jcsd
  3. Mar 20, 2015 #2

    TSny

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    Homework Helper
    Gold Member

    I would recommend setting up the equation in symbols first. Then you can simplify by cancelling common factors on each side of the equation.

    The expression x + .385 does not represent the correct distance.

    Your equation reads ( (548)(63.5)(6.67E-11)/x^2 ) + ( (125)(63.5)(6.672E-11)/(x+.385)^2 ) = 0. Note that both terms on the left are positive. So, they can't add to zero.
     
  4. Mar 20, 2015 #3
    How would I do that? Gmm/d^2 - Gmm/d^2 = 0?
    I used - because as you said if it were addition it could not add to zero, but am I setting that up right at all? Wouldn't the gravity of the two objects on the smaller one have to end up being equal to zero?


    Here's my reasoning behind the x + .385 distance
    (its really large image, sorry, i have no idea why)
    IMG_20150320_195707066.jpg
     
  5. Mar 20, 2015 #4

    TSny

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    Homework Helper
    Gold Member

    In order for the total force on the 63.5 kg mass to be zero, does the 63.5 kg mass need to be between the other two masses or somewhere else? Think about the direction of the forces on the 63.5 kg mass.
     
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