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Math Help - finding the angle

  1. #1
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    finding the angle

    I'm having trouble with this question. I was thinking of using Lami's theorem, but its not helping too much.

    A body of mass 5 kg is in equilibrium when it is acted upon by three concurrent coplanar forces P, Q and R as shown in the diagram below. P = 10 newtons and Q = 20 newtons. The angle between P and Q is 170 degrees and the angle between P and R is \theta degrees




    Find \theta.
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  2. #2
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    Quote Originally Posted by scorpion007 View Post
    I'm having trouble with this question. I was thinking of using Lami's theorem, but its not helping too much.

    A body of mass 5 kg is in equilibrium when it is acted upon by three concurrent coplanar forces P, Q and R as shown in the diagram below. P = 10 newtons and Q = 20 newtons. The angle between P and Q is 170 degrees and the angle between P and R is \theta degrees




    Find \theta.
    Resolve all forces into components in the direction of Q and normal to Q
    (with lets say positive upwards). Then as the body is in equilibrium the
    sum of each components is zero. so

    Put phi = theta-10 degrees

    p_1 = P(parrallel to Q) = -10 cos(10)
    p_2 = P(orthog to Q) = -10 sin(10)

    r_1 = R(parrallel to Q) = -R cos(phi)
    r_2 = R(orthog to Q) = R sin(phi)

    then:

    p_1 + r_1 = Q = -20
    p_2 + r_2 = 0.

    Which is a pair of simultaneous non-linear equationswhich you will have to
    solve for phi and R.

    RonL
    Last edited by CaptainBlack; October 29th 2006 at 11:26 PM.
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  3. #3
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    hmm, what i have so far is these equations:

    10.15 - R\cos(\phi) = 0
    -1.74 - R\sin(\phi) = 0

    I'm not too sure how to eliminate one of the pronumerals from here...
    Also, im not 100% i got up to here correctly. Anyone care to check? lol sorry.
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  4. #4
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    Quote Originally Posted by scorpion007 View Post
    hmm, what i have so far is these equations:

    10.15 - R\cos(\phi) = 0
    -1.74 - R\sin(\phi) = 0

    I'm not too sure how to eliminate one of the pronumerals from here...
    Also, im not 100% i got up to here correctly. Anyone care to check? lol sorry.
    Two comments. Following CaptainBlack's notation the second equation should be:
    -1.74 + R\sin(\phi) = 0

    The second comment is that would keep a few extra digits on your numbers to stave off rounding errors, but that you otherwise calculated them correctly.

    Solve the equations thusly:
    R\cos(\phi) = 10.15

    R\sin(\phi) = 1.74

    Now divide the second by the first:

    \frac{R sin( \phi )}{R cos( \phi )} = \frac{1.75}{10.15}

    tan( \phi ) = 0.17241

    Which you can solve for \phi.

    You can now choose any way you like to get R, but the method I recommend is:

     \left ( R sin( \phi ) \right )^2 + \left ( R cos( \phi ) \right )^2 = 1.74^2 + 10.15^2

    R^2 sin^2( \phi ) + R^2 cos^2( \phi ) = 106.0501

    R^2 (sin^2( \phi ) + cos^2( \phi ) ) = 106.0501

    R^2 = 106.0501

    Now you can solve for R.

    (The reason I choose this method over the others is that this equation uses fewer values that have been calculated (as opposed to given), which reduces rounding error.)

    -Dan
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  5. #5
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    ahh, thank you very much!

    i was concerned with some values for a sec, but i realised i made some stupid arithmetic mistakes.

    The answers are indeed correct

    EDIT: Although, i do have a minor question, CaptainBlack, why did you have p_1 and p_2 negative? shouldn't the sum of the (parallel to Q) components of R and P = 20?
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  6. #6
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    Quote Originally Posted by scorpion007 View Post
    ahh, thank you very much!

    i was concerned with some values for a sec, but i realised i made some stupid arithmetic mistakes.

    The answers are indeed correct

    EDIT: Although, i do have a minor question, CaptainBlack, why did you have p_1 and p_2 negative? shouldn't the sum of the (parallel to Q) components of R and P = 20?
    The way the axes are chosen Q's in these axes is (+20,0), the other forces
    components parallel to Q are in the opposite direction so the - signs.

    RonL
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  7. #7
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    Hmm, but if i translate your first equation, wouldnt this be it?

    -29.848 - R\cos(\phi) = 0 ?

    which is different to what I got.
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  8. #8
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    Quote Originally Posted by scorpion007 View Post
    Hmm, but if i translate your first equation, wouldnt this be it?

    -29.848 - R\cos(\phi) = 0 ?

    which is different to what I got.
    Mistake on my part should equal -20

    That is the sum of the components of the forces with the correct signs should be 0 for equilibrium to prevail.

    RonL
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  9. #9
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    that's how i understand it too. Which is what i did. Thanks for the clarification anyway.
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