Question

Determine the location ($\bar{x}$,$\bar{y}$,$\bar{z}$) of the centroid of the homogeneous rod. FBD Segment Area X Y X.A Y.A $\sum A =$ $\sum X.A =$ $\sum Y.A =

          Determine the location ($\bar{x}$,$\bar{y}$,$\bar{z}$) of the centroid of the homogeneous rod.
FBD
Segment	Area	X	Y	X.A	Y.A
$\sum A =$	$\sum X.A =$	$\sum Y.A =
        
Determine the location (x̅,y̅,z̅) of the centroid of the homogeneous rod.
FBD
Segment	Area	X	Y	X.A	Y.A
∑ A =	∑ X.A =	∑ Y.A =

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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please include fbd and table homogeneous rod. the location .y. of the centroid ol the FBD 00Cos30 6000 Sin30 SOTA X.A Y.A Segment Area X X.A= 2y.A= A=
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Transcript

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00:01 Hi everyone here we have to find location of x bar and y were of the centroid c and we have to calculate the product of inertia with respect to x -des and y -dice centroid x bar can be defined as summation of x -bar into a upon summation of a let us see here 9 into 18 into 150 plus 84 into 18 into 13 into 132 divided by 18 into 150 plus 18 into 132 on solving it centroid x having the value 44 .1mm centured y can be defined as submission of by bar a upon summation of a so it will be 75 into 18 into 150 plus 9 into 18 plus 132 divided by 18 into 15 plus 18 into 13 .12 that is 44 .1 millimeter.
02:01 Now moment of inertia, sorry product of inertia about x -dress and wide s can be defined as area into you have to use the concept area into x -by -war substitute the value to 18 into 150 35 .111 and 30 .89 plus 18 into 13 .132 39 .39.
02:44 Into 39 .89 into 30...
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