Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_20

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Determine the - and -coordinates of the centroid of the shaded area. Problem 5/20 (a) Determination of the constants and and the area 1. Present Final Formula: The shaded area is bounded by two curves that intersect at and . Upper curve: Lower curve: 2. Substitute Values: At the point : Thus, the equations are: 3. Partial Operations: The area is calculated by integration: 4. Final Calculation: ● Final Conclusion: The area of the shaded region is . (b) Calculation of the -coordinate of the centroid () xy k 1 k 2 A (0,0)(a,a)y = 2 k 2 xy = 1 k x 1 3 (a,a) a=k ⟹ 2 ak = 2 a a=k a⟹ 1 3 k = 1 a 2 1 y = 2 ,y =ax 1 a 2 x 3 A A= ( y − ∫ 0 a 2 y )dx= 1 − dx ∫ 0 a (ax a 2 x 3 ) A= ⋅ x− [a 3 2 3/2 4a 2 x 4 ] 0 a A= a− ( 3 2 2 4 a 2 ) A= a = 12 8−3 2 a 12 5 2 A= a 12 52 xxˉ 1. Present Final Formula: 2. Substitute Values: 3. Partial Operations: Evaluate the integral: 4. Final Calculation: ● Final Conclusion: The -coordinate of the centroid is . (c) Calculation of the -coordinate of the centroid () 1. Present Final Formula: Using vertical differential elements, the centroid of each element is at . 2. Substitute Values: 3. Partial Operations: Evaluate the integral: =xˉ x( y − A 1 ∫ 0 a 2 y)dx 1 a = xˉ( 12 5 2 ) x − dx= ∫ 0 a (ax a 2 x 3 ) x− dx ∫ 0 a (a 3/2 a 2 x 4 ) x− dx= ∫ 0 a (a 3/2 a 2 x 4 ) ⋅ x− [a 5 2 5/2 5a 2 x 5 ] 0 a = a− = ( 5 2 3 5 a 3 ) a 5 1 3 =xˉ = a 12 5 2 a 5 13 ⋅ 5 1 a= 5 12 a 25 12 x=xˉ a= 25 12 0.48a y yˉ y = c (y + 2 1 1 y )