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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.
 Fall '11
 CalculusII
 Force

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