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Math Help - (HELP)Vector of moment

  1. #1
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    (HELP)Vector of moment

    a force magnitude 2 units acts at the point a with coordinates (1,1,3). the force is applied in the direction of the vector 3i-j+2k
    (I) Find the vector of the force
    (II) Find the position vector, OA
    (III) Find the moment of the force about O
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  2. #2
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    Re: (HELP)Vector of moment

    Quote Originally Posted by waynewkng View Post
    a force magnitude 2 units acts at the point a with coordinates (1,1,3). the force is applied in the direction of the vector 3i-j+2k
    (I) Find the vector of the force
    (II) Find the position vector, OA
    (III) Find the moment of the force about O
    this is all pretty straightforward if you understand anything at all about vectors and moments.

    Do you have any idea how to get started? Do you have any work you can show?

    I) you're given a magnitude and a vector in the direction of the force, what do you have to do to that vector to make it a directional vector? So what then is the resulting force vector?

    II) you're given the coordinates of P, just turn this into a vector

    III) you need to know the vector formula for a moment about a point given the force and the position vector. Look it up in your book.
    Thanks from topsquark
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  3. #3
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    Re: (HELP)Vector of moment

    Sorry, i haven't idea~~
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  4. #4
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    Re: (HELP)Vector of moment

    You really need to review your textbook. This problem is about as straightforward as they come. I'll work this for you as a Christmas present.

    I) The magnitude of the force is 2 (Newtons I assume) and it is in the direction of 3i - j + 2k.

    To get the directional or unit vector in this direction we must divide it by it's length. The length of a vector is the square root of the sum of the squares of it's components.

    \left|3i - j + 2k\right|=\sqrt{3^2+1^2+2^2}=\sqrt{9+1+4}=\sqrt{14  }

    \mbox{so the unit vector in this direction is given by $u=\frac{1}{\sqrt{14}}(3i-j+2k)$}

    Now the force vector is the magnitude of the force times this directional vector.

    \vec{F}=\frac{2}{\sqrt{14}}(3i-j+2k)N

    II) Find the position vector OA. A is given by the point {1,1,3} (meters?) so the position vector is just the vector


    \vec{OA}=(A_x-0)i+(A_y-0)j+(A_z-0)k=A_x i+A_y j+A_z k= (i+j+3k)m


    III)
    \mbox{The moment of force }\vec{F}\mbox{ acting at A about 0 is found as $\vec{F}\times \vec{OA}$}

    \vec{F}\times \vec{OA}=\frac{2}{\sqrt{14}}(3i-j+2k)\times (i+j+3k)=\frac{2}{\sqrt{14}}(-5i-7j+4k)N\cdot m
    Last edited by romsek; December 26th 2013 at 03:38 AM.
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