Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_25

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Determine the coordinates of the centroid of the shaded area. Problem 5/25 (a) Determination of Area Centroid Coordinates 1. Formula: The coordinates of the centroid for a composite area are given by: First, we define the boundaries. The shaded area is bounded by the x-axis, a parabola , and a circular arc . The intersection point occurs at . Since it lies on the circle: Substituting into to find : 2. Substitution: The area is split into two parts: under the parabola from to , and under the circular arc from to . 3. Calculation: (, )xˉyˉ =xˉ = A∑ Q ∑ y , = A xdA∫ yˉ = A∑ Q ∑ x A ydx∫ 2 1 2 y=kx 2 x+ 2 y= 2 r 2 P(x ,y ) 11 y = 1 r 5 3 x = 1 = r−y 2 1 2 = r − r 2 ( 5 3 ) 2 = r 25 16 2 r 5 4 P( r, r) 5 4 5 3 y=kx 2 k r = 5 3 k r ⟹ ( 5 4 ) 2 k= 16r 15 A 1 x=0 x= r 5 4 A 2 x= r 5 4 x=r A= x dx+ ∫ 0 4r/5 16r 15 2 dx ∫ 4r/5 r r−x 22 Q = y x x dx+ ∫ 0 4r/5 ( 16r 15 2 ) x dx ∫ 4r/5 r r−x 22 Q = x x dx+ ∫ 0 4r/5 2 1 ( 16r 15 2 ) 2 r−x dx ∫ 4r/5 r 2 1 ( 22 ) Total Area : Since : First Moment about y-axis : First Moment about x-axis : 4. Result: ● Final Conclusion: The coordinates of the centroid for the shaded area are and . ✨ Final Answer Summary A A = 1 = [ 16r 15 3 x 3 ] 0 4r/5 = 16r 5 ( 125 64r 3 ) r = 25 4 2 0.16r 2 A = 2 x +rarcsin = 2 1 [r−x 22 2 ( r x )] 4r/5 r −0.48−arcsin 2 r 2 ( 2 π − 2 π arcsin(0.8)=arccos(0.8)=arcsin(0.6) A = 2 arcsin(0.6)− 2 r 2 0.24r≈ 2 0.08175r 2 A=0.16r+ 2 0.08175r= 2 0.24175r 2 Q y Q = y1 x = [ 64r 15 4 ] 0