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The first step is to relate the radial or normal force F (Newtons/radian) of a rope wrapped around a capstan to the tension, T (Newtons), in the rope. This is classically done with a force balance method using vectors, trigonometry, and a little calculus. The answer can be found more simply with
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Isang magandang pagbati para sa lahat at sa mga opisyal at sa mga residente ng nasasakupan, sikapin natin ang tahimik at mapayapang halalan isang parehas na kaganapan para sa kaayusan ng susunod na mga mamumuno Hindi lang sa Sweden at sa ibat ibang bansa na magaganap na halalan. Aking suporta at
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an energy argument as follows: Assume that the rope is inelastic (nonstretching) and the capstan is frictionless and compressible. Then the energy required to squeeze down the capstan radius R by an amount ∆R with one turn of rope equals the energy supplied by pulling the rope.
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Let the coefficient of friction between the rope and capstan be k = friction force/radial force. So the frictional force over a wrap angle d phi is
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lubos niyo po maasahan ang aking suport sa abot ng aking kakayahan maraming salamt at god bless all.
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The increase in tension dT over an angle d phi is the frictional force over that angle so
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The capstan equation derivation seem needlessly complex. Why not do it this way:
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80:Cancel 2 π ∆R on each side leaving F = T
110:T = exp(k phi + C) = exp( C) exp( k phi)
480:notice, but please explain why in your
345:notice, but please explain why in your
21:We do not need a separate artice: the
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460:File:Capstan-derivation-diagram.svg
70:Derivation of the Capstan Equation
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166:Another one of your uploads,
116:Tout = Tin exp(k phi) QED
101:Integrating both sides gives
95:Divide both sides by T gives
23:proof of the capstan equation
113:For phi = 0 T = Tin so
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