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35. (a) With the pivot at the hinge, Eq. 129 yields
cos
0
a
TL
F y
θ
−=
.
This leads to
T
= (
F
a
/cos
)(
y/L
) so that we can interpret
F
a
/cos
as the slope on the
tension graph (which we estimate to be 600 in SI units).
Regarding the
F
h
graph, we use
Eq. 12
−
7 to get
F
h
=
T
cos
−
F
a
= (
−
F
a
)(
y/L
)
−
F
a
after substituting our previous expression. The result implies that the slope on the
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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