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Math Help - Cog and Belt problem

  1. #1
    Senior Member Educated's Avatar
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    Question Cog and Belt problem

    There are 2 cogs, connected together by 1 belt. Find the length of the belt.

    The large cog has a radius of 11cm. The small cog has a radius of 4cm. The distance between the centre of the 2 cogs is 25cm.
    (Diagram not drawn to scale)



    I'm stuck because I cannot solve the angle AOB or YPZ, and I cannot solve the length of AY or BZ. Once I get those, then I can get solve the answer.

    Can anyone tell me how to get the angle or length?
    Does the angle AOB = angle YPZ?

    I got 147.48^{\circ} for angle AOB and 212.52^{\circ} for the reflex angle of YPZ because I assumed they were equal, but it doesn't seem right.
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by Educated View Post
    There are 2 cogs, connected together by 1 belt. Find the length of the belt.

    The large cog has a radius of 11cm. The small cog has a radius of 4cm. The distance between the centre of the 2 cogs is 25cm.
    (Diagram not drawn to scale)



    I'm stuck because I cannot solve the angle AOB or YPZ, and I cannot solve the length of AY or BZ. Once I get those, then I can get solve the answer.

    Can anyone tell me how to get the angle or length?
    Does the angle AOB = angle YPZ?

    I got 147.48^{\circ} for angle AOB and 212.52^{\circ} for the reflex angle of YPZ because I assumed they were equal, but it doesn't seem right.
    Draw a line parallel to BZ from P to the radius OB. Let it meet OB at Q. Note that QO = 7. Use the right-triangle QOP to get:

    1. The angle QPO and hence the angle OPZ.
    2. The angle QOP.
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  3. #3
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    Hello, Educated!

    Your diagram didn't show up, but I'll take a guess.


    There are 2 cogs, connected together by a belt.
    Find the length of the belt.

    The large cog has a radius of 11cm. The small cog has a radius of 4cm.
    The distance between the centres of the 2 cogs is 25cm.


    I got . \angle AOB = 147.48^o and . \angle YPZ = 212.52^o
    These happen to be correct.

    In the diagram, consider quadrilateral OPZB.
    . . Note that it has two right angles.


    Code:
        O *
          | @ *
        7 |       *  25
          |           *
          |               *
        Q * - - - - - - - - - * P
          |        24         |
        4 |                   | 4
          |                   |
          |                   |
        B * - - - - - - - - - * Z
                   24

    In right triangle OQP\!:\;QP^2 + 7^2 \:=\:25^2 \quad\Rightarrow\quad QP = BZ = 24

    Also: . \sin\theta \:=\:\frac{24}{25} \quad\Rightarrow\quad \theta \:\approx\:73.74^o

    and: . \angle OPZ \:=\:180^o - \theta \:=\:106.26^o


    Can you finish up?
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  4. #4
    Senior Member Educated's Avatar
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    Oh, I got it.

    The only reason I wasn't getting the right answer was because I forgot to use the pythagorus theorum on the straight length of the belt.

     Length = 22\pi * \frac{212.52}{360} + 8\pi * \frac{147.48}{360} + 24 * 2

     Length = 99.1cm
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