Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_6

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Determine the -coordinate of the centroid of the area under the sine curve shown. The equation of the curve is . Problem 5/6 (a) Determination of the -coordinate of the centroid ● Calculation Process 1. Present Final Formula: The -coordinate of the centroid for an area is defined by: where is a differential area element and is the -coordinate of the centroid of that element. For a vertical strip of width and height , we have: 2. Substitute Values: Substitute the expressions for and into the integrals with limits from to : 3. Partial Operations: Calculate the Total Area : Calculate the First Moment of Area : Using the trigonometric identity : y y=asin b πx y y yˉ =yˉ dA∫ y dA∫ c dAy c y dxy dA=ydx=asin dx( b πx ) y = c = 2 y sin 2 a ( b πx ) y c dA x=0x=b =yˉ = asin dx ∫ 0 b ( b πx ) sin asin dx ∫ 0 b [ 2 a ( b πx )][( b πx )] a sin dx ∫ 0 b ( b πx ) sin dx 2 a 2 ∫ 0 b 2 ( b πx ) A A= asin dx= ∫ 0 b ( b πx ) a − cos [ π b ( b πx )] 0 b A= − (cos π − π ab cos0)= − (−1− π ab 1)= π 2ab Q x sinθ= 2 2 1−cos2θ 4. Final Calculation: 5. Result: ● Final Conclusion: The -coordinate of the centroid of the area under the given sine curve is . ✨ Final Answer Summary (a) Mecademy AI Solution · ENGProblem 5/6 Q = x y dA= ∫ c sin dx= 2 a 2 ∫ 0 b 2 ( b πx ) dx 2 a 2 ∫ 0 b 2 1−cos ( b 2πx ) Q = x x− sin 4 a 2 [ 2π b ( b 2πx )] 0 b Q = x (b−0)−(0−0) = 4 a 2 () 4 ab 2 =yˉ = A Q x = π 2ab 4 ab 2 ⋅ 4 ab 2 2ab π =yˉ 8 πa =yˉ ≈ 8 πa 0.3927a y =yˉ 8 πa =yˉ