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Law of gravitation

  1. Dec 30, 2007 #1
    [SOLVED] Law of gravitation

    In Figure 13-34, a square of edge length 15.0 cm is formed by four spheres of masses m1 = 5.00 g, m2 = 4.00 g, m3 = 1.50 g, and m4 = 5.00 g. In unit-vector notation, what is the net gravitational force from them on a central sphere with mass m5 = 2.10 g?

    I know that since the mass of 1 and 4 are equal they cancel out and i know that the radius is .075*Square root(2). which gives me .106m. I use the equation Gm2m5/r25^2+Gm3m5/r35^2 and i get the answer 6.85e-14 which is not the right answer. I know that the masses are .004, .0015, and .0021 since they are in grams. Can some one help me
     
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  3. Dec 30, 2007 #2

    Doc Al

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    (1) You didn't provide the figure, but I assume that m1 and m4 are at adjacent corners not opposite corners. So they don't cancel out.
    (2) The forces are vectors and must be added as such. Direction counts!
     
  4. Dec 30, 2007 #3
    sorry Doc Al M1 and M$ are on opposite corners here is the picture given
     

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  5. Dec 30, 2007 #4

    Doc Al

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    OK. Realize that the forces due to m2 and m3 point in opposite directions.
     
  6. Dec 30, 2007 #5
    that should mean i should subtract their forces but i tried that and that did not work i got a value of 3.11e-14 which is also incorrect
     
  7. Dec 30, 2007 #6

    Doc Al

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    I get the same answer. But that's the magnitude of the force. They asked for the force in unit vector notation, which means they want x and y components.
     
  8. Dec 30, 2007 #7
    But how would i do that I know how to go from unit vector notation but i do not understand how to do the opposite
     
  9. Dec 30, 2007 #8

    Doc Al

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    Just find the x and y components of the net force. What direction is that force? What angle does it make with the x-axis?
     
  10. Dec 30, 2007 #9
    It is moving towards mass 2 and the angle should be 45
     
  11. Dec 30, 2007 #10

    Doc Al

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    Right. So what are the x and y components of a vector at an angle of 45 degrees to the horizontal?
     
  12. Dec 30, 2007 #11
    so i set up the sin 45=o/h and solve so i get hsin 45=0 and i get 2.64e-14 but that is not the correct answer. i do the same thing with cos and get 1.63e-14. where should i go from here
     
  13. Dec 30, 2007 #12

    Doc Al

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    If a force F makes an angle [itex]\theta[/itex] with the x-axis, its components are:

    [tex]F_x = F \cos \theta[/tex]

    [tex]F_y = F \sin \theta[/tex]
     
  14. Dec 30, 2007 #13
    Isn't that what i did, i just wrote it in different terms
     
  15. Dec 30, 2007 #14

    Doc Al

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    How did you get different answers? ([itex]\sin 45 = \cos 45[/itex].)
     
  16. Dec 30, 2007 #15
    ooo i am in radian mode
     
  17. Dec 30, 2007 #16
    Thank you Doc Al for your help
     
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