Question

Adobe Acrobat Reader DC iew Window Help Tools TW # 2 Solution.pdf Statics Bedford Fo TW#2.pdf 125% 3 5 Problem # 2 Locate the centroid (x, y) of the area under the curve. 4 in. 1 in. 1 in. 1 in.

Answer 1

dA = ydx to find the centroid, x^- = integral xdA/integral dA = integral x times x^2 dx/integral x^2 dx Now dA = ydx = x^2 dx Therefore x^- = integral^2_1 x^3 dx/integral^2_1 x^2 dx x co ordinate or, x^- = 1/4 x^4 Vertical-bar ^2_1/1/3 x^3 Vertical-bar ^2_1 or, x^- = 3/4 [2^4 - 1^4]/[2^3 - 1^3] = 3/4 times 15/7 = 45/28 inch = 1.61 inch y^- = integral^4_0 y dA/integral^4_0 dA = integral^4_0 y/2 ydx/integral^4_0 ydx = 1/2 integral^4_0 y^2 dx/integral^4_0 y dx (we are taking y/2 because the rectangular strip has centroid y/2 above) \r\n y^- = 1/2 integral^4_0 x^4 dx/integral^4_0 x^2 dx or, y^- = 1/2 times 1/5 x^5 Vertical-bar ^4_0/1/3 times x^3 Vertical-bar ^4_0 or, y^- = 3/5 times 2 times (4^5 - 0)/(4^3 - 0) or, y^- = 3/10 times 4^2 or, y^- = 3/10 times 4 times 4 or, y^- = 24/5 inch = 4.8 inch Therefore (x, y) of centroid = (1.61, 4.8) inch