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Gravitational force among 4 spheres

  1. Mar 20, 2008 #1
    1. The problem statement, all variables and given/known data
    Four 7.5kg spheres are located at the corners of a square of side 0.70 m.
    Calculate the magnitude of the gravitational force exerted on one sphere which is in left and down corner of the square by the other three.


    2. Relevant equations

    F_tot = F_2 + sqrt(2) * F1

    3. The attempt at a solution

    came up with that equation, but it says i am wrong.
    can anyone help/
     
  2. jcsd
  3. Mar 20, 2008 #2
    It'll be better if you explain your equation (or your solution).
     
  4. Mar 20, 2008 #3
    well since they are in a square and have equal masses F_1 = F_3 in magnitude. and for our triangle we have 1/sqrt2 are the sides and 1 is the hypotenuse. i got that. but i don't think its right
     
  5. Mar 20, 2008 #4
    am i way offf? can anybody help
     
  6. Mar 20, 2008 #5

    Doc Al

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    Staff: Mentor

    If I understand what you're saying here, this is correct. But it's not the final answer.

    What are F_2 and F_1?
     
  7. Mar 21, 2008 #6
    F1 and F3 are the 90 degree forces and F2 is the 45 degree force. what do i still need to do?
     
  8. Mar 21, 2008 #7
    Still your solution solution looks bit wrong. Show the substitution process (the values for F1 ..) and and calculation, so that I can help you better.
     
  9. Mar 21, 2008 #8

    Doc Al

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    Staff: Mentor

    That's just the schematic of the answer. Now you have to figure out F1 & F2 and plug in the numbers. I assume they want an actual answer for the force in Newtons.
     
  10. Mar 22, 2008 #9
    F1 = ((6.67 * 10 ^-11)* 7.5 *7.5 )/ .70^2 = 7.66 * 10^-9

    length of hypotnuse
    sqrt(.70^2 +.70^2); = .99

    F2 = ((6.67 * 10 ^-11)* 7.5 *7.5 )/ .99^2 = 3.83*10^-9

    F_tot = 3.83*10^-9 + sqrt(2) * 7.66*10^-9 =

    1.46 * 10^-8

    PROBLEM SOLVED. I MUST HAVE JUST PLUGGED IN SOME WRONG NUMBERS. THANKS ANYWAY
     
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