Meriam, Kraige & Bolton — Distributed Forces: Centroids and Centers of Gravity: Problem 5_19

⚡ Mecademy AIENG정역학 · ch5  Problem Statement Locate the centroid of the area shown in the figure by direct integration. (Caution: Carefully observe the proper sign of the radical involved.) Problem 5/19 (a) Determination of the -coordinate of the centroid, 1. Formula: We define the coordinate system with the origin at the bottom-left corner of the square. The circular arc has a radius and is centered at the point . The equation of this circle is: Solving for , we get . Since the arc lies to the left of its center , we take the negative sign: The shaded area is bounded by the -axis (), the line , and this arc. Using horizontal differential elements : 2. Substitution: First, calculate the total area : Now, substitute the expression for into the integral for : 3. Calculation: Expand the integrand: xxˉ a(a,0) (x−a)+ 2 y= 2 a 2 xx=a± a−y 22 x=a x=a− a−y 22 Ayx=0y=a dA=xdy A= xd y = ∫ 0 a (a− ∫ 0 a )d ya−y 22 =xˉ x dA= A 1 ∫ c (xd y )= A 1 ∫ 0 a 2 x x d y 2A 1 ∫ 0 a 2 A A=[ay] − 0 a d y = ∫ 0 a a−y 22 a− 2 = 4 πa 2 a1− 2 ( 4 π ) xAxˉ A=xˉ (a− 2 1 ∫ 0 a )d ya−y 22 2 A=xˉ ( a − 2 1 ∫ 0 a 2 2a + a−y 22 a− 2 y)dy 2 Integrate term by term: Evaluating from to : Finally, calculate : 4. Result: ● Final Conclusion: The -coordinate of the centroid is , which is approximately . (b) Determination of the -coordinate of the centroid, 1. Formula: Using the same differential element , the centroid of the element is at its midpoint height . 2. Substitution: Substitute and set up the integral: 3. C