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we can increase eficiency of rotating shaft.

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  • jigar patel
    Swamishriji Dear Sir, The figure is in cross section. As shown in figure, There are two cases. In each case there are two pulleys of same diameter. Each
    Message 1 of 10 , Feb 13, 2006
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                                                       Swamishriji
      Dear Sir,
                  
                    The figure is in cross section. As shown in figure, There are two cases. In each case there are two pulleys of same diameter. Each pulley is of exactly circular shape and diameter is 4000mm.
                     In first case 50 K.G weight is fixed with each pulley as shown in figure. The center of each pulley is fixed. Between these two pulleys there is a stationary plate. Plate will remain stationary, while rotating the pulleys because centers of pulleys are fixed. 100 K.G weight is put on this plate. The force of this weight is applied on these two pulleys in vertical downward direction. Now we try to rotate slowly first pulley in clockwise direction and second pulley in anti clockwise direction. Lubrication is provided between the contact surfaces of stationary plate and pulleys.
                   In second case two pulleys of same diameter. 50 K.G weight is also fixed with each pulley, but there is no stationary plate and 100 K.G weight. Now we try to rotate slowly first pulley in clockwise direction and second pulley in anti clockwise direction.
                 As free body diagram in first case we can rotate pulleys using less moment of force. We can lift mass using less energy.   File is attached with this mail. Please open is to see diagram. 
       
      1. In case 1) the moment necessary to rotate each pulley is

                  M=(-G/2+G1*cos(alpha))*R
      where:
              G is weight of plate (=100 kg)
              G1 is given weight (=50 kg)
              alpha is angle of rotation of each pulley
              R is radius of pulley
      2. In case 2) weigt of plate G=0.
      What is your opinion?
      Yours Truly,
      Jigar Patel
      5, Chankyanagar soc,
      Nr Nandanvan Soc,
      Nr Highway,
      Mehsana-384002,
      Gujarat
      India
      Ph:+91-9342539996,+91-2762-254342

      Send instant messages to your online friends http://uk.messenger.yahoo.com

    • hustierhof
      Hello Jigar , sorry I do not see any benefit or positiv momentum in the experiment you described. For me, the force to lift the 100kg weight for a given amount
      Message 2 of 10 , Feb 14, 2006
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        Hello Jigar ,

        sorry I do not see any benefit or positiv momentum in the experiment
        you described. For me, the force to lift the 100kg weight for a
        given amount of height is always the same.

        Hubert
        --- In AMBIENTENERGY@yahoogroups.com, jigar patel
        <jig_patel_1982@...> wrote:
        >
        > Swamishriji
        Dear Sir,
        >
        > The figure is in cross section. As shown in
        figure, There are two cases. In each case there are two pulleys of
        same diameter. Each pulley is of exactly circular shape and diameter
        is 4000mm.
        > In first case 50 K.G weight is fixed with
        each pulley as shown in figure. The center of each pulley is fixed.
        Between these two pulleys there is a stationary plate. Plate will
        remain stationary, while rotating the pulleys because centers of
        pulleys are fixed. 100 K.G weight is put on this plate. The force of
        this weight is applied on these two pulleys in vertical downward
        direction. Now we try to rotate slowly first pulley in clockwise
        direction and second pulley in anti clockwise direction. Lubrication
        is provided between the contact surfaces of stationary plate and
        pulleys.
        > In second case two pulleys of same diameter. 50 K.G
        weight is also fixed with each pulley, but there is no stationary
        plate and 100 K.G weight. Now we try to rotate slowly first pulley
        in clockwise direction and second pulley in anti clockwise
        direction.
        > As free body diagram in first case we can rotate
        pulleys using less moment of force. We can lift mass using less
        energy. File is attached with this mail. Please open is to see
        diagram.
        >
        > 1. In case 1) the moment necessary to rotate each pulley is
        >
        > M=(-G/2+G1*cos(alpha))*R
        > where:
        > G is weight of plate (=100 kg)
        > G1 is given weight (=50 kg)
        > alpha is angle of rotation of each pulley
        > R is radius of pulley
        > 2. In case 2) weigt of plate G=0.
        >
        > What is your opinion?
        > Yours Truly,
        > Jigar Patel
        > 5, Chankyanagar soc,
        > Nr Nandanvan Soc,
        > Nr Highway,
        > Mehsana-384002,
        > Gujarat
        > India
        > Ph:+91-9342539996,+91-2762-254342
        >
        >
        >
        > Send instant messages to your online friends
        http://uk.messenger.yahoo.com
        >
      • hustierhof
        photo photo2 photo3
        Message 3 of 10 , Feb 15, 2006
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        • hustierhof
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          Message 4 of 10 , Feb 15, 2006
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          • hustierhof
            DSC-00465.jpg DSC-00466.jpg DSC-00467.jpg
            Message 5 of 10 , Feb 15, 2006
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            • hustierhof
              photo photo2 photo3
              Message 6 of 10 , Feb 15, 2006
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              • hustierhof
                DSC-00465.jpg DSC-00466.jpg DSC-00467.jpg
                Message 7 of 10 , Feb 15, 2006
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                • hustierhof
                  photo photo2 photo3
                  Message 8 of 10 , Feb 16, 2006
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                  • hustierhof
                    DSC-00465.jpg DSC-00466.jpg DSC-00467.jpg
                    Message 9 of 10 , Feb 16, 2006
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                    • hustierhof
                      DSC-00465.jpg DSC-00466.jpg DSC-00467.jpg
                      Message 10 of 10 , Feb 16, 2006
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