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  1. #1
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    mechanics question

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    can some explain to me how they get tana=r/2r-x as i do not see how triangle with angle a is right angled which it has to be to use tan. just because the three forces are going through o would not make it right angled
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  2. #2
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    Re: mechanics question

    I'm going to do this problem in static equilibrium a little differently than the way the answer does. See attached diagram ...



    $\displaystyle \sum F_x = 0$ ... force right = force left

    $N = T\sin{\alpha}$


    $\displaystyle \sum F_y = 0$ ... force up = forces down

    $T\cos{\alpha} + f = W$

    since the sphere is about to slip, the force of static friction is a maximum $\implies f = \mu \cdot N = \mu \cdot T\sin{\alpha}$

    $T\cos{\alpha} + \mu \cdot T\sin{\alpha} = W$


    $\displaystyle \sum \tau = 0$ ... torque CCW = torque CW

    about the point where the sphere touches the wall ...

    $T \cdot 2r\sin{\alpha} = W \cdot r \implies W = 2T \sin{\alpha}$

    returning to the equilibrium equation in the y-direction & substituting for $W$ ...

    $T\cos{\alpha} + \mu \cdot T\sin{\alpha} = 2T \sin{\alpha}$

    divide every term by $T\cos{\alpha}$ ...

    $1 + \mu \cdot \tan{\alpha} = 2\tan{\alpha}$

    $1 = 2\tan{\alpha} - \mu \cdot \tan{\alpha}$

    $1 = \tan{\alpha}(2 - \mu)$

    $\dfrac{1}{2-\mu} = \tan{\alpha}$
    Attached Thumbnails Attached Thumbnails mechanics question-equilibrium2.jpg  
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  3. #3
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    Re: mechanics question

    Thanks . The diagram is really helpful. It will take me a day or a few to take this all in.
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