Question

Determine the centroid of the bracket shown knowing the radius of the hole, r, is 0.5 in, and the height, h, is 2.2 in Glve the answer to the nearest .001 inch, An integral table ( x values are all in radians) COS +2) dx = sin( + 2x 9.81859 in? -2 cos) + 2) dar sin + cosa O in -2 +2) da 2 sin + 2 x cosa 10.9747 in' 2 + 2 dir sin 24 + 2 sin 12.44799 in 13 sin(2)+11 22 COS 21. 6234 in ? sin 3 - - 2

Answer 1

•Cos(x) +2. r=0.5 in. 2.2 in ke 2 in 2 in od T. YdA E AY sy JdA EA А, 9, Ý = Ay, A2₂ A,+A₂ A3 (oux2.2) ( 22 ) A4, = = 9.68 inch? Adx A2 = 1922 ly A= yda Ay = y.dxxy AzY₂ = da -2 y = f(x) = Cos(x) + 2 [ Cocos (2) + 2)²] . du A292 N POCO X2

AzY2 - 1 / ( Cosca) + 2) Cos(x) +2) da from table, entegration 2 [ (cos(x) +2) Jaux = 12.44799 in? - 2 3 .. A₂Y2 12.44799 inch AzY3 (7x0.52) (2.2) AY 1.7278 inch² 2 A = (4x2.2) 8.8 inch 2 2 Az = - 2 1 (cos(x) + 2) dx 9.81859 inch (from table) Tx0.5² - 0.7854 inch ² Az = Ay+ A₂Y₂ AY TE Ait A₂ - A3 9.68 + 12.44799 - 1-7278 8.8 + 9.81859 - 0.7894 Ý = 1.144 inch.

X - Aix, + A₂ - AX3 A+ A₂-Az Ax = (2.2x4) (0) y y -0. A282= sta t x da Area da = y da. xtda 2 2. 2 Ax2 = x. (y dx). - 2 2 AX2 2 = x (Cosca) +2) dx - 2 (from integration table x (Colx+2) dx (since it is symmetrical about y-axis). i.Az x2 = 0. A₃9₂ (IT x 0.5²) + (0) 20.

A, x, + A1 - A₂ X 3 X = A + A₂-A3 oto-O Ait A₂-A3 Since the cross section is Symmetrical about y-axis, x=0) Centroid is located on y-axis and y 1.44 inches. ž=0 - location of centriod Y = - 1.44 inches,
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