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Math Help - Applied pure trig question

  1. #1
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    Question Applied pure trig question

    A belt ABCDEFA of length L passes round two wheels, centres O_1 and O_2. The parts ABC and DEF of the belts are in contact with the wheels and parts AF and CD are straight. If O_1A = a and O_2F = 3a and the angle AO_1B = theta radians show that:

    4atan (theta) = 4a(theta) + L - 6a*pi

    Next, find the shortest length belt in terms of a.


    Im pretty stuck on this one. Any ideas??
    I am looking at the equation to see what its saying but can't see where tan(theta) comes from - Obviously a triangle is drawn and the Opposite and Adjacent sides are used some how.
    The arc AC on the small wheel is 2a(2*pi - (theta)) but I'm just kinda stuck
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  2. #2
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    Hello dojo
    Quote Originally Posted by dojo View Post
    A belt ABCDEFA of length L passes round two wheels, centres O_1 and O_2. The parts ABC and DEF of the belts are in contact with the wheels and parts AF and CD are straight. If O_1A = a and O_2F = 3a and the angle AO_1B = theta radians show that:

    4atan (theta) = 4a(theta) + L - 6a*pi

    Next, find the shortest length belt in terms of a.


    Im pretty stuck on this one. Any ideas??
    I am looking at the equation to see what its saying but can't see where tan(theta) comes from - Obviously a triangle is drawn and the Opposite and Adjacent sides are used some how.
    The arc AC on the small wheel is 2a(2*pi - (theta)) but I'm just kinda stuck
    What you don't say - but what I deduce from your comment about the arc AC - is that BO_1O_2E is a straight line.

    So look at the attached diagram. The key fact that you seem to be missing is that when we draw O_1G parallel to AF,\; O_1O_2G is a right-angled triangle with \angle O_1O_2G = \theta and O_2G = 2a. So O_1G = AF = 2a\tan\theta = CD.

    So now add the arc lengths ABC and D EF to AF + CD\; (=4a\tan\theta), equate the sum to L, and the result follows.

    The shortest belt possible is when O_1O_2 = a + 3a = 4a. This gives, from \triangle O_1O_2G, \;\cos\theta = \tfrac12. So with this value of \theta, solve the equation for L and you're there.

    Can you complete it now?

    Grandad
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  3. #3
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    Perfect. It's kind of abstract but I see it so clearly now. Many thanks
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