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Addition of forces acting on object

  1. Feb 16, 2009 #1
    This should be simple, i've already solved several problems just like this one but it won't come out right however i try... apologies in advance for non-native english.

    1. The problem statement, all variables and given/known data
    In a 3-dimensional coordinate system, a hook is placed in A=(0,0,3). Two lines connect the hook down to a hatch that resides completely within the XY-plane. These lines keep the hatch horizontal. The lines attach to the corners of the hatch in B=(4,0,0) and C=(4,5,0). The magnitudes of the forces in the lines are S[AB]=P and S[AC]=sqrt(2)*P. (Did not know how to index S properly). What is the magnitude of the total force acting on the hook?

    3. The attempt at a solution
    First, i figure out the complete description of the line forces one by one by multiplying the magnitude of the force with the force vector divided by the length of the force vector, F=P*(AB)/|AB|;
    F[AB]= P*(B-A)/|B-A| = P*(4,0,-3)/sqrt(4²+3²) = P*(4,0,-3)/5
    F[AC]= sqrt(2)*P*(C-A)/|C-A| = sqrt(2)*P*(4,5,-3)/|4²+5²+3²| = sqrt(2)*P*(4,5,-3)/(5*sqrt(2)) = P*(4,5,-3)/5

    Then, i sum the line forces= P*(4,0,-3)/5 + P*(4,5,-3)/5 = P*(8,5,-3)/5.

    The magnitude of the total force acting upon the hook would then be 1/5*P. But it isn't. The answer says sqrt(5)*P. What did i miss?

    Sorry if it's not optimally legible, i tried to tidy it up as best i could... thankful for any help!
  2. jcsd
  3. Feb 16, 2009 #2
    I see some problems in your unit vectors. Your unit vectors should be

    e1=(-4,0,3)/5 and e2=(-4,-5,3)/(5*sqrt(2))

    Always draw a diagram! It will help ;).
  4. Feb 16, 2009 #3
    Hmm... the way i've done it thus far, and is described in my mechanics book, is to form the vector from A to B by subtracting A from B... so AB in the problem would be B-A=(4,0,0)-(0,0,3)=(4,0,-3), right? Or have i left something out?

    Also, how would this change the magnitude of anything? I don't mean to argue, i just want to understand...
  5. Feb 16, 2009 #4
    It depends on which way the forces are acting. Think about it this way, the hook is at x=0, B is at x=4, so to since the hook is pulling UP on the hatch, you know that the x component of the force must be negative. The vector is from B to A!!
  6. Feb 16, 2009 #5
    http://img222.imageshack.us/img222/1154/vectorsms2.jpg [Broken]
    Last edited by a moderator: May 4, 2017
  7. Feb 16, 2009 #6
    Oh... you're right, of course! That's helpful. But that only reverses the direction of my final unit vector, it still has the same magnitude...

    The problem is marked as "more difficult" in the book, and it doesn't seem all that hard so i guess there's something that i don't immediately consider.
  8. Feb 16, 2009 #7


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    Homework Helper

    Aren't we trying to find the force on the hook?
    AB = [4/5P, 0, -3/5P],
    AC = [4P/5,5P/5,-3P/5]

    Hope you guys don't mind me playing with this, too. I've never seen one like it.
    I like the vector approach you took - much better than my trigonometry!

    EDIT - simplified root(2)/root(50) to 1/5
    Last edited: Feb 16, 2009
  9. Feb 16, 2009 #8
    oops, ya know what, you were right with the unit vectors :$, sorry :(, but yea, no matter which way you do it you'll get the same answer! (opposite and equal forces) Im so sorry mate.


    When adding your vectors you had -3 + -3 = -3, should be -6.

    sum the line forces= P*(4,0,-3)/5 + P*(4,5,-3)/5 = P*(8,5,-6)/5.If you do that and find the magnitude of the force you will find is [tex]\sqrt{5}[/tex]P


    You did the problem correctly, just a simple calculation error :).
  10. Feb 16, 2009 #9


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    Homework Helper

    Do you have a short way to find the magnitude? I combined the first two with the Pythagorean theorem, then combined that result with the 3rd the same way. It does come out to root(5).

    EDIT: Oh, I see it now - thanks!
    I think I took a course on this in 1970 - Linear Algebra.
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