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Simple gravitational attraction forces problem

  1. Oct 27, 2011 #1
    Each mass in the figure below is 2.00. Find the magnitude of the net gravitational force on mass A due to the other masses in figure b (the square).

    Figure: http://i.imgur.com/fjSgU.jpg

    i used the formula F=G [(m1*m2)/r^2]. all of the forces are in the direction of D so i resolved the B and C using the above equation and SOH CAH TOA into their directions that point toward D. Then i used the above equation without SOH CAH TOA to get D and added the componenets of B and C to that of D. I got 8.88 * 10^-8 but this is wrong. where did i go wrong? thanks
  2. jcsd
  3. Oct 27, 2011 #2


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    Sow your calculation in detail, please.

  4. Oct 27, 2011 #3
    before i write it all out is this the right method?
  5. Oct 27, 2011 #4


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    The result is wrong. I can not help without seeing what you did. I am sorry.

    Good luck!

  6. Oct 27, 2011 #5
    i re-did it and got 6.67E-8 is this correct?
  7. Oct 27, 2011 #6

    D acting on A:
    F=(6.673E-11)[4/.1414^2]. i got the .1414 from pythagorean using .1 as a and b.

    B acting on A:
    F=2.669E-8 --> but this is not downward pull
    using SOH CAH TOA i did
    x=1.89E-8 --> this needs to be multiplied by 2 for both B acting on A and C acting on A (they are the same).

    so (1.89E-8)*2 + 1.3346E-8 = 5.11E-8

    i somehow get a different answer everytime i do it
  8. Oct 27, 2011 #7


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    It is correct now but you need to add the unit.

    There is less chance to make mistakes in calculations if you solve the problem symbolically and substitute the data at the end.

    So you have the resultant force along the diagonal AD. The magnitude of the force is

    F=G q^2/a^2(1/2 +2cos(45)) =6.673E-11*400(0.5+1.414)=5.11 N.

    Do not forget the unit!!

  9. Oct 27, 2011 #8
    you said it is 5.11 N. isn't it 5.11E-8 N?
  10. Oct 27, 2011 #9


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    Oppps, I left out that E-8. You are right!!

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