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PROBLEM 5.1
Locate the centroid of the plane area shown.
SOLUTION
2, inA , in.x , in.y 3, inxA 3, inyA
1 8 6 48× = 4− 9 192− 432
2 16 12 192× = 8 6 1536 1152
Σ 240 1344 1584
Then 3
2
1344 in
240 in
xAX
A
Σ= =
Σ or 5.60 in.X =
and 3
2
1584 in
240 in
yAY
A
Σ= =
Σ or 6.60 in.Y =
PROBLEM 5.2
Locate the centroid of the plane area shown.
SOLUTION
2, mmA , mmx , mmy 3, mmxA 3, mmyA
1 1
60 75 22502× × = 40 25 90 000 56 250
2 105 75 7875× = 112.5 37.5 885 900 295 300
Σ 10 125 975 900 351 600
Then 3
2
975 900 mm
10 125 mm
xAX
A
Σ= =
Σ or 96.4 mmX =
and 3
2
351 600 mm
10 125 mm
yAY
A
Σ= =
Σ or 34.7 mmY =
PROBLEM 5.3
Locate the centroid of the plane area shown.
SOLUTION
For the area as a whole, it can be concluded by observation that
( )224 in.
3Y = or 16.00 in.Y =
2, inA , in.x 3, inxA
1 1
24 10 1202× × = ( )2
10 6.6673
= 800
2 1
24 16 1922× × = ( )1
10 16 15.3333
+ = 2944
Σ 312 3744
Then 3
2
3744 in
312 in
xAX
A
Σ= =
Σ or 12.00 in.X =
PROBLEM 5.4
Locate the centroid of the plane area shown.
SOLUTION
2, mmA , mmx , mmy 3, mmxA 3, mmyA
1 21 22 462× = 1.5 11 693 5082
2 ( )( )16 9 27
2− = − 6− 2 162 54−
3 ( )( )16 12 36
2− = − 8 2 288− 72−
Σ 399 567 4956
Then 3
2
567 mm
399 mm
xAX
A
Σ= =
Σ or 1.421 mmX =
and 3
2
4956 mm
399 mm
yAY
A
Σ= =
Σ or 12.42 mmY =
PROBLEM 5.5
Locate the centroid of the plane area shown.
SOLUTION
2, mmA , mmx , mmy 3, mmxA 3, mmyA
1 120 200 24 000× = 60 120 1 440 000 2 880 000
2 ( )260
5654.92
π− = −
94.5 120 534 600− 678 600−
Σ 18 345 905 400 2 201 400
Then 3
2
905 400 mm
18 345 mm
xAX
A
Σ= =
Σ or 49.4 mmX =
and 3
2
2 201 400 mm
18 345 mm
yAY
A
Σ= =
Σ or 93.8 mmY =
PROBLEM 5.6
Locate the centroid of the plane area shown.
SOLUTION
2, inA , in.x , in.y 3, inxA 3, inyA
1 ( )29
63.6174
π=
( )( )4 9
3.89173π
−= − 3.8917 243− 243
2 ( )( )115 9 67.5
2= 5 3 337.5 202.5
Σ 131.1 94.5 445.5
Then 3
2
94.5 in
131.1 in
xAX
A
Σ= =
Σ or 0.721 in.X =
and 3
2
445.5 in
131.1 in
yAY
A
Σ= =
Σ or 3.40 in.Y =
PROBLEM 5.7
Locate the centroid of the plane area shown.
SOLUTION
First note that symmetry implies X Y=
2, mmA , mmx 3, mmxA
1 40 40 1600× = 20 32 000
2
2(40)1257
4
π− = − 16.98 21 330−
Σ 343 10 667
Then 3
2
10 667 mm
343 mm
xAX
A
Σ= =
Σ or 31.1 mmX =
and 31.1 mmY X= =
PROBLEM 5.8
Locate the centroid of the plane area shown.
SOLUTION
First note that symmetry implies 0X =
2, inA , in.y 3, inyA
1 ( )24
25.132
π− = −
1.6977 42.67−
2 ( )26
56.552
π=
2.546 144
Σ 31.42 101.33
Then 3
2
101.33 in
31.42 in
yAY
A
Σ= =
Σ or 3.23 in.Y =
PROBLEM 5.9
For the area of Problem 5.8, determine the ratio 2 1/r r so that 13 /4.y r=
SOLUTION
A y yA
1 2
12r
π−
14
3
r
π 31
2
3r−
2 2
22r
π
24
3
r
π 32
2
3r
Σ ( )2 22 1
2r r
π− ( )3 3
2 1
2
3r r−
Then Y A y AΣ = Σ
or ( ) ( )2 2 3 31 2 1 2 1
3 2
4 2 3r r r r r
π× − = −
2 3
2 2
1 1
91 1
16
r r
r r
π − = −
Let 2
1
rp
r=
[ ] 29( 1)( 1) ( 1)( 1)
16p p p p p
π+ − = − + +
or 216 (16 9 ) (16 9 ) 0p pπ π+ − + − =
PROBLEM 5.9 CONTINUED
Then 2(16 9 ) (16 9 ) 4(16)(16 9 )
2(16)p
π π π− − ± − − −=
or 0.5726 1.3397p p= − =
Taking the positive root 2
1
1.340r
r=
