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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    A piece of uniform string hangs vertically so that it's free end just touches horizontal surface of a table. If now the upper end of the string is released, show that at any instant during the falling of the string, the total force on the surface is three times the weight of that part of the string lying on the surface. @Robert136 try this question|dw:1440757472828:dw|

  2. anonymous
    • one year ago
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    suppose x length has fallen v= rt (2gh) change in momentum = -dm .v dm= (nu)dx (nu)= m/l back to F=-(nu)v dx/dt = -(nu)v^2 F= (nu)2gx force by the surface= for changing momentum and fallen part F= 2(nu)gx + (nu)gx solved

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