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Math Help - forces on an inclined plane

  1. #1
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    Question forces on an inclined plane

    my problem has been resolved this far but i cant transpose or what ever you call it

    rearrange the formula

    ive been getting help and have ended up with this

    24.5 - x.cos(30). = u(42.44 + x.sin(30))

    where u is cofficient of friction 1/7 or 0.143

    also should the brackets be around the 30??

    30 is the angle X is at


    so 5g is resolved but cant work out X
    49cos30= 42.44
    49sin30= 24.5
    but i cannot work it out. sure you all know what its like after a day of exams.
    and a break down of how to work it would be great
    Last edited by peterfurness12; February 10th 2011 at 11:04 AM. Reason: further info needed
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  2. #2
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    e^(i*pi)'s Avatar
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    Use the unit circle to see that \sin(30) = \dfrac{1}{2} and \cos(30) = \dfrac{\sqrt{3}}{2} and you're given \mu = \dfrac{1}{7}

    \displaystyle 24.5 - \dfrac{x \sqrt{3}}{2} = \dfrac{1}{7} (28.3 + \frac{x}{2})
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  3. #3
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    ive added more info to the original thread and ive gotta be honest

    1. it has to be more straight forward than that

    2.i dont understand any of that

    have a look and see if the equation i got is even right 5g i can resolve but i cant resolve X
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  4. #4
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    solved it

    how do show its solved or is this enough
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  5. #5
    MHF Contributor Unknown008's Avatar
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    I'm not sure what even is the question... are all those forces in equilibrium and you need to generate two equations showing the relationship of the forces?

    If so, resolve along and perpendicular to the plane:

    X \cos(30) + F = 5g \sin(30)

    5g \cos(30) + X\sin(30) = R
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