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27. We choose up as the +
y
direction, so
2
ˆ
(3
.
0
0m
/
s)
j
a
=−
G
(which, without the unit
vector, we denote as
a
since this is a 1dimensional problem in which Table 21 applies).
From Eq. 512, we obtain the firefighter’s mass:
m
=
W
/
g
= 72.7 kg.
(a) We denote the force exerted by the pole on the firefighter
fp
ˆ
j
FF
=
G
and apply Eq.
51. Since
net
Fm
a
=
G
G
, we have
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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, Mass

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