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Gravitational force of three objects on a fourth

  1. Aug 12, 2012 #1
    1. The problem statement, all variables and given/known data
    Four spheres of equal 9.5 kg are located at the corners of a square of length 0.6 m per side. Find the magnitude and direction of the gravitational field on one by the other three spheres


    2. Relevant equations
    gravity = GM/r^2 with G = gravitation constant, M = mass, r = radius

    The field strength of the diagonal mass = GM/(r*.71)^2
    The sum of the field magnitudes from the two neighbored masses is GM(.71)/r^2
    The total magnitude = GM/((2r)^2) + GM(.71)/r^2

    3. The attempt at a solution

    So to clarify the ".71" is the sin of 45°, the radius of the neighboring spheres is .6 m and using phythagorean the diagonal radius is .85 m.

    Solving..... the answer in the book says....3.2 x 10^-8 N but....

    (6.67x 10^-11)(9.5)/(2 * 0.6)^2 + (.71)(6.67x10^-11)(9.5)/(.85)^2=

    4.4x10^-10 + 6.2x10^-10 = 1.06 x 10^-9

    Where do I need to be set straight? Wrong equation(s) or substitutions?
     
    Last edited: Aug 13, 2012
  2. jcsd
  3. Aug 13, 2012 #2

    gabbagabbahey

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    The gravitational field is a vector field. This is not how you find the magnitude of a sum of vectors (unless they all point in the same direction).

    [tex]||\mathbf{A}+\mathbf{B}|| \neq ||\mathbf{A}||+||\mathbf{B}||[/tex]
     
  4. Aug 13, 2012 #3
    So do I just use
    F = (G*m1*m2)/r^2 for each sphere at 90° and

    F = sin 45°(G*m1*m2)/r^2 for the diagonal sphere and add them all together?

    doing that I get 3.9 x 10^-8
     
  5. Aug 13, 2012 #4

    CWatters

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    Forget the one on the diagonal for the moment... Go look up how you add two vectors that are at right angles.
     
  6. Aug 13, 2012 #5
    You would use pythagorean method...

    A-----B
    | /
    | /
    C-----D
    so lets say we call "A" sphere that is the 4th sphere... calculating the force of gravity between A and B would give a result of

    1.67*10^-8

    using g = G (m1m2/r^2)

    and the same for A - C

    using pythagorean theorm... c^2 = a^2 + b^2 gives us a "g" between A-D of 2.4x 10^-8

    am I on the right track now? if so then do I add these together or am I waaaay off.

    btw ABCD is supposed to be a square but I can't seem to get the dash to move over.
     
  7. Aug 14, 2012 #6

    CWatters

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    Yes that right.

    and since it's a square the resulting vector of the first two points at the third sphere making it easier to add the effect of that one.
     
  8. Aug 14, 2012 #7
    Now I'm confused since the answer in the book is 3.2x10^-8

    adding 2.4 + 1.67 +1.67 =5.74 x10^-8
     
  9. Aug 15, 2012 #8

    CWatters

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    Total force = SQRT{(Force between A&B)2 + (Force between A&C)2} + (Force between A&D)

    and it points from A to D.

    You appear to have calculated SQRT{(Force between A&B)2 + (Force between A&C)2} correctly. I make that part = 2.36 * 10-8

    It's the (Force between A&D) you appear to have wrong. The book answer appears to be correct.

    Aside: Strictly speaking the equation I provide should be vector addition but you can simplify it to ordinary addition because we know that the result of adding two of the forces results in an vector that points in the same direction as the third.
     
    Last edited: Aug 15, 2012
  10. Aug 15, 2012 #9

    CWatters

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    Might help. Work out what force AD is. Then add it to the 2.36 * 10^-8 you calculated. Remember D is further away from A.
     

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    Last edited: Aug 15, 2012
  11. Aug 15, 2012 #10

    gabbagabbahey

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    You should really just choose an appropriate coordinate system and break down each of the 3 forces into their components in that coordinate system. As long as the basis vectors in the coordinate system you choose are orthogonal to each other (like Cartesian unit vectors), you can then just add the components together.

    In 2D Cartesian coordinates, for example the sum two vectors ([itex]\mathbf{A} = A_x \mathbf{e}_{x} + A_y \mathbf{e}_{y}[/itex] and [itex]\mathbf{B} = B_x \mathbf{e}_{x} + B_y \mathbf{e}_{y}[/itex]) is computed as follows:

    [tex]\mathbf{A} + \mathbf{B} = (A_x +B_x)\mathbf{e}_{x} + ( A_y + B_y ) \mathbf{e}_{y}[/tex]

    And so the magnitude of the resulting vector is just [itex]||\mathbf{A} + \mathbf{B} || = \sqrt{(A_x +B_x)^2 + (A_y +B_y)^2}[/itex].

    This is something that you will find in any introductory linear algebra textbook, and is usually taught in highschool physics. If you are having this much difficulty adding three vectors together and finding the resultant vector's magnitude and direction, you really need to go back to your textbook (or look up vector addition online) and re-read the section on adding vectors and study the examples in the textbook.
     
  12. Aug 15, 2012 #11
    Thank you CWatters! And yes Gab... I am taking a college level intro physics course with an "okay" book. It's been 15 years since I've worked with Physics like this and I'm gonna have to review more before diving in. I would have been okay with three objects but four is a bit of a stretch.
     
  13. Aug 15, 2012 #12

    CWatters

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    Its even longer since I did it.

    Gab's answer is better than mine because next problem will have unequal masses and the cheat/trick I used won't work. As he said best work out the components of all the vectors in a suitable co-ordinate system and add them up.
     
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