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22. As shown in the freebody diagram, the forces on the climber
consist of
T
G
from the rope, normal force
N
F
G
on her feet, upward static
frictional force
s
f
G
and downward gravitational force
mg
G
. Since the
climber is in static equilibrium, the net force acting on her is zero.
Applying Newton’s second law to the vertical and horizontal
directions, we have
net,
net,
0s
i
n
0c
o
s
.
xN
ys
FF
T
FT
f
m
g
φ
==
−
+
−
¦
¦
In addition, the net torque about O (contact point between her feet
and the wall) must also vanish:
net
0
sin
sin(180
)
O
mgL
TL
τθ
θ
−
°
−
−
¦
From the torque equation, we obtain
sin /sin(180
).
Tm
g
θθ
=°
−
−
Substituting the
expression into the force equations, and noting that
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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, Force, Friction, Static Equilibrium

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