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PROBLEM 9.164 CONTINUED
(
b
) Because of the symmetry of the body,
xy
I
and
yz
I
can be deduced
by considering the circular permutation of
( )
,,
xyz
and
( )
abc
.
Thus
1
20
xy
I
mab
=
W
1
20
yz
I
mbc
=
W
Alternative solution for part
a
First divide the tetrahedron into a series of thin horizontal slices of
thickness
dy
as shown.
Now
1
ay
xy
a
a
bb
=−
+ =
−
and
1
cy
zy
c
c
−
The mass
dm
of the slab is
2
11
1
22
y
dm
dV
xzdy
ac
dy
b
ρρ
ρ
==
=
−
Now
, Area
zx
zx
dI
tdI
=
where
td
y
=
and
, Area
1
24
zx
dI
x z
=
from the results of Sample Problem 9.6
Then
()
2
2
1
24
yy
dIzx
dy
a
c
−
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This note was uploaded on 07/11/2011 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.
 Spring '08
 Jenkins
 Statics

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