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Simple Gravitation - 1 problem

  1. Apr 19, 2007 #1
    Simple Gravitation - With Answer - Help me work the middle steps

    1. The problem statement, all variables and given/known data

    Three objects -- two of mass m and one of mass M -- are located at three corners of a square of edge length l as in Figure P13.23. Find the gravitational field g at the fourth corner due to these objects. (Express your answers in terms of the edge length l, the masses m and M, and the gravitational constant G).

    2. Relevant equations

    Find the magnitude and direction of the gravitational field g.


    3. The attempt at a solution

    The direction is 45 degrees (measured counterclockwise from the positive x-axis)

    G((4m+M)/(2l^2)) was the final simplified answer I got for the magnitude, but it was wrong.
     

    Attached Files:

    Last edited: Apr 19, 2007
  2. jcsd
  3. Apr 19, 2007 #2
    Associated work:

    Base equation: Gravitational field force = GMm/(l^2)

    I see three gravitational forces at work:
    Gm/l^2 on the y axis

    Gm/l^2 on the x axis

    GM/((sqrt(2) times l)^2)

    Add all three so 2Gm/2l^2 + 2Gm/2l^2 + GM/2l^2 = (G(4m+M))/(2((l)^2))
     
  4. Apr 19, 2007 #3
    Welcome to PF.

    Careful, as in Why do all the denominators look alike when you showed the diagonal object to have longer radius?

    Never mind i see what you did.
     
  5. Apr 19, 2007 #4
    Thanks. See any easy mistakes or did I go the wrong direction?
     
  6. Apr 19, 2007 #5
    No I found no mistakes so I was sort of head scratching on this one, hoping someone else might see a problem if there is one. Of course these are vector fields, which might be the issue.
     
    Last edited: Apr 19, 2007
  7. Apr 19, 2007 #6
    I GOT THE ANSWER FROM A FRIEND:

    (G/l^2)(M/2+sqrt(2)m)

    Can anyone help me work it out now that the answer is known?
     
  8. Apr 19, 2007 #7
    It is to do with the fact that the fields are vector fields.

    You can't just arithmetically add the 3 fields up, you have to consider their direction. It's like if you push a block north with 1N and east with 1N the overall force is [tex]\sqrt{2}N[/tex] in the NE direction, not 2N. So I'd recommend drawing a diagram and doing a bit of pythagoras.
     
  9. Apr 19, 2007 #8
    No problem, resolve the diagonal force into x and y components which can be added like scalars

    cos and sin are both sqrt(2)/2 so mult your diag. expression and add to the x and y components from other forces, then calculate magnitude by taking the sqrt of sum of squares. Falls out like honey in a bucket. I suspect you'll never make this mistake again. good problem and post.
     
  10. Sep 3, 2011 #9
    Hi Guys, i guess this is the answer, i'm not sure! but? if someone has a comment reply pls.xxx
    tanθ = gy / gx
    = [ Gm / L2 + G M /2L2 sinθ ] / [ Gm /L2 + G M / 2L2 cosθ ]

    θ = tan-1 { [ Gm / L2 + G M / 2L2 sinθ ] / [ Gm /L2 + G M / 2L2 cosθ ]}
     
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