Problem Set 3Problem 3.1 Center of Gravity for a Bent RodDetermine the distance xCand yCto the center of gravity of the bentrod.Figure P9.1: Problem 9.1Problem 3.2 Centroid of a Composite SectionDetermine the location yCof the centroid of the beam’s cross sectionalarea. Neglect the size of the corner welds at A and B for the calculation.Figure P9.2: Problem 9.2Problem 3.3 Centroid of a Tapered Cross SectionLocate the centroid yCof the concrete beam with the tapered cross sectionshown.Figure P9.3: Problem 9.3Problem 3.4Find the x-coordinate of the centroid of the indicated region where A =2 m and k = π/8 m−1.xyOy = A cos(kx)Figure 3.1: Problem 3.1Problem 3.5Determine the area and the centroid of the area.Solution% y(x) = x^(3/2)% where 0<x<a and 0<y<aa = 1; % m% differential element% dA = y dx = x^(3/2) dx% centroid of the differential element is at% x and y/2 = (1/2) x^(3/2)% A = int(dA) where 0<x<aA = int( x^(3/2),x,0,a);% A = 0.400 (m^2)% My = int(x dA) where 0<x<aMy = int(x*x^(3/2),x,0,a);xC = My/A;% xC = 0.714 (m)% Mx = int(y/2 dA) where 0<x<aMx = int((1/2)*x^(3/2)*x^(3/2),x,0,a);yC = Mx/A;% yC = 0.312 (m)Problem 3.6Determine the area and the centroid of the area.Problem 3.7Determine the location of the centroid C of the area where a = 6 in,b = 6 in, c = 3 in, and d = 6 in.yOxdbac(1)(2)(3)Figure 3.7: Problem 3.7Problem 3.8Locate the center of mass of the homogeneous block assemblySolution% rectangular prism 1V1 = 150*150*550; % mm^3x1 = 75; % mmy1 = 275; % mmz1 = 75; % mm% rectangular prism 2V2 = 150*150*200; % mm^3x2 = 225; % mmy2 = 450; % mmz2 = 75; % mm% triangular prism 3V3 = 150*150*100/2; % mm^3x3 = 200; % mmy3 = 50; % mmz3 = 50; % mm% total areaV = V1 + V2 + V3;xC = (x1*V1+x2*V2+x3*V3)/V;yC = (y1*V1+y2*V2+y3*V3)/V;zC = (z1*V1+z2*V2+z3*V3)/V;% V = 1.8e+07 (mm^3)% xC = 120.312 (mm)% yC = 304.688 (mm)% zC = 73.438 (mm)Problem 3.9Locate the centroid of the paraboloid defined by the equation x/a =(y/b)2where a = b = 4 m. The material is homogeneous.yb 2=xayxabFigure 3.6: Problem 3.6Solutiona = 4; % mb = 4; % my = b*sqrt(x/a);% dV = pi*y^2 dyV = int(pi*y^2, x, 0, a);% V = 32 pi (m^3)% xC = (1/V) int(x dV)xC = int(x*pi*y^2, x, 0, a)/V;% xC = 2.667 (m)Problem 3.10A circular V-belt has an inner radius r = 600 mm and a cross-sectionalarea with the dimensions a = 25 mm, b = 50 mm, and c = 75 mm. Determinethe volume of material required to make the belt.bacar(1)(2)(3)Figure 3.10: Problem