applied mechanics question?

http://img85.imageshack.us/img85/9896/appliedmecha...

I have this quesiton for one of my tutorials in university its in my applied mechanics course. Despite being rushed through a very quick explanation i still dont get it. I know your meant to get out 2 equations or something but i cant seem to do it. Could someone have an attempt. btw those are the answers down the bottom (in kN)

3 Answers

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  • smci
    Lv 7
    1 decade ago
    Favorite Answer

    Just take horizontal and vertical components, that's all there is to it, or in fact any 2D equilibrium-of-forces question

    (ok maybe sometimes we prefer to take radial and tangential components, but they're always somehow orthogonal):

    W=14kN in -y-direction

    N acts @ 20deg left of vertical +y-direction

    Thus:

    [Equilibrium of forces in y-dirn]

    W=14k = N cos 20

    N = 14k/cos 20 ... gives you N

    [Equilibrium of forces in x-dirn]

    T = N sin 20

    = 14k tan 20 ... gives you T

    Solutions:

    T=5.1kN in -x-direction

    N=14.9kN @ 20deg left of vertical +y-direction

  • 1 decade ago

    The sum of the forces in the x direction =0.

    The sum of the forces in the y direction =0.

    let i and j be the unit vectors for x and y

    W = -14kN j

    T = Ti

    N = -cos(70)Ni + sin(70)Nj

    so:

    sin(70)N -14kN = 0

    N = 14898

    -cos(70)N + T = 0

    T = 5095

  • 1 decade ago

    I drew a diagram and wrote the calculations in the picture, hope this helps:

    http://img108.imageshack.us/img108/8328/test1xk0.j...

    Forgive my units.

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