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# The vertical components dont quite cancel something

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Unformatted text preview: since the bit is not moving horizontally. • The vertical components don’t quite cancel; something remains to pull the bit back to equilibrium. So the slope of the string tells us the direction of the tension: rise/run = F1y/F or F2y/F. That is: force points down, but slope is up. ￿ ￿ ∂y ￿ F1 y ￿ =− F ∂x x and 34 Fig. 15.13 ￿ ∂y ￿ F2 y ￿ = ￿ ∂ x x+∆x F Fig. 15.13 Let’s rewrite that to make it a bit clearer. Deﬁne" (S is for Slope :) So the net vertical force on this bit of string is: Then for that last line: Plug this into F = ma: m Rearrange: ￿￿...
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## This note was uploaded on 03/11/2013 for the course PHYS 101 taught by Professor Calculusii during the Fall '11 term at University of Alberta.

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