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11.60:
a) The center of mass of the beam is 1.0 m from the suspension point. Taking
torques about the suspension point,
)
m
00
.
2
)(
N
100
(
)
m
00
.
1
)(
N
0
.
140
(
)
m
00
.
4
(
=
+
w
(note that the common factor of sin

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**Unformatted text preview: **° 30 has been factored out), from which N. . 15 = w b) In this case, a common factor of sin ° 45 would be factored out, and the result would be the same....

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