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Unformatted text preview: PROBLEM 5.90 The composite body shown is formed by removing a hemisphere of
radius r from a Cylinder of radius R and height 2R. Determine (a) the
y coordinate of the centroid when r : 3R/4, (b) the ratio n'R for which
)7 : 71.2R. SOLUTION Note, for the axes shown )7 V
7R 723R:
*21‘ lIIYA
s 4
.1
—2.'r{R4 — L]
s R4—1r4
=__ 8
Iii—1r}
3
4
11[L]
=L
1 W
3 R
“1(3)“
(0) r=§R Ez— 341R
4 11 E'
3 4
or §=—1.113R C
AF)“
(in) ;:—1.2R: —1.2R=— 8 R 3R
_ 1F]
3 R
r 4 3
or [i] 73.2[i +1.6=0
R R
Solving numerically %= 0.8844 PROBLEM 5.104 A ZOOmmdiameter cylindrical duct and a 100 x ZOOmm rectangular
duct are to be joined as indicated. Knowing that the ducts are fabricated
from the same sheet metal, which is of uniform thickness, locate the center
of gravity of the assembly. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides yith the
centroid of the corresponding area. Also note that symmetry implies Z = 0 { z—eemxm 6: \ Removers
*4» mm». at s REMM': l A, m2 I Sclm Rm 4 mm3 )7A,m3
 1 ”(0.2)(03) = 0.1885 0 0.15 0 0.028274
7!
__ , _ = — . 2 0.1
2 2 (0 2)(° 1) O 0314 ( ) = 0.06366 0.25 —0.02000 —0.007854
72'
It 2 4(0'1) — 0 04244 A
3 2(0'1) : 0.01571 3,, ‘ ‘ ' 0.30 .0.000667 0.004712
4 (0.3)(02) = 0.060 0.15 0.30 0.00900 0.001800
5 (0.3)(0.2) = 0.060 0.15 0.20 0.00900 0.001200
—1r(0.1)z = «0.1571 40“) 0.20 —0.000667 —0.003142
2 37r
(0.3)(0.1) = 0.030 0.15 0.25 0.004500 0.007500
(0.3)(0.1) = 0.030 0.15 0.25 0.004500 0.007500
0.337080 0.023667 0.066991
Have X’ZA = 22A: )‘((0.337080 mmz) = 0.023667 mm3
or X = 0.0702 In )7 = 70.2 mm 4 172A 2 WA: 17(0337080 m2) = 00669911111113 or Y = 0.19874m )7 = 198.7 mm 4 PROBLEM 5.125 Determine by direct integration the location of the centroid of onehalf of a
thin, uniform hemispherical shell of radius R. SOLUTION First note that symmetry implies f 2 0 4 The element of area (1A of the shell shown is obtained by cutting the shell
with two planes parallel to the xy plane. Now
M = (m)(Rd¢9) A 2r
yEL : ‘—
where r = Rsint9
so that dA : ”R2 sin 0‘16
YEL = igsinﬁ
It
Then A = £2:er2 sinﬂdﬂ : KR2 [—0056]?
= ”R2
and IFELdA = §(—3551n9](z122sin9d6)
n
l
= #2R3[2 _ sinlﬂ]2
2 4 0
= —£R3
2
_ , _ It
Now yA z ijLdA: y(ﬂR2) = ER3
or  —lR<
y 2 Symmetry implies ...
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 Spring '08
 Jenkins
 Statics

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