Chapter Six: Centroid and Centers of Gravity ....................... ............................ Introduction
119
Dr.MohammedZuhair&Dr. MayadahW.
Example No. 2: Locate the centroid of the shaded area in Figure.
Solution:
Shape
𝑨
𝒙
𝒚
𝑨∙𝒙
𝑨 ∙ 𝒚
Rectangular
90 ×60 =5400
1/2 × 90
=45
1
2×60 =30
243000
162000
quarter
circle
−𝜋
4×302
= −706.86
4𝑟/3𝜋
=12.73
60 −4𝑟
3𝜋
=47.27
-8998.3
-33413.3
Triangle
−1
2×45 ×60
= −1350
45 +2
3×45
=75
1
3×60 =20
-101250
-27000
Sum
3343.14 𝑚𝑚2
132751.7 𝑚3
101586.7 𝑚3
