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4 February, 18:02

Each of the four forces acting at e has a magnitude of 28 kn. Express each force as a cartesian vector and determine the resultant force.

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  1. 4 February, 18:12
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    Solution:

    KN=10^3

    we have to keep in mind nomancleature

    f=28KN

    a=4m, b=6m, c=12m

    the position in a vector and then the forces

    F_{EA}:

    R_{EA} = (b,-a,-c)

    F_{EA}=f/tfrac{r_{EA}

    }{/left | R_{EA}

    /right |}

    F_{EA} = (12,-8,-24) KN

    so, the cartesion for F_{EA} will be

    F_{EA} = (12i-8j-24k)

    F_{EB}:

    R_{EB} = (-b, a,-c)

    F_{EB}=f/tfrac{r_{EB}

    }{/left | R_{EB}

    /right |}

    F_{EB} = (-12,8,-24) KN

    so, the cartesion for F_{EB} will be

    F_{EB} = (-12i+8j-24k)

    F_{EC}:

    R_{EC} = (-b,-a,-c)

    F_{EC}=f/tfrac{r_{EC}

    }{/left | R_{EC}

    /right |}

    F_{EC} = (12,8,-24) KN

    so, the cartesion for F_{EC}

    F_{EC} = (-12i+8j-24k)

    F_{ED}:

    R_{ED} = (-b,-a,-c)

    F_{ED}=f/tfrac{r_{ED}

    }{/left | R_{ED}

    /right |}

    F_{EC} = (-12,-8,-24) KN

    so, the cartesion for F_{EC}

    F_{EC} = (-12i-8j-24k)

    Resultent force F_{r}=F_{EA}+F_{EB}+F_{EC}+F_{ED}

    = (-96K) KN
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