7. We take the force of the left pedestal to be
F
1
at
x
= 0, where the
x
axis is along the
diving board. We take the force of the right pedestal to be
F
2
and denote its position as
x
=
d
.
W
is the weight of the diver, located at
x
=
L
. The following two equations result
from setting the sum of forces equal to zero (with upwards positive), and the sum of
torques (about
x
2
) equal to zero:
12
1
0
()
0
FFW
Fd W L d
+−=
+−
=
(a) The second equation gives
1
3.0m
(580 N)= 1160 N
1.5m
Ld
FW
d
§·
−
=−
−
¨¸
©¹
which should be rounded off to
3
1
1.2 10 N
F
×
. Thus,
3
1

1
.
2
1
0
N
.
F
=×
(b) Since
F
1
is negative, indicating that this force is downward.
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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

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