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October 2, 2014

October 2, 2014

Posted by **peter** on Friday, October 12, 2012 at 7:19am.

- physics -
**bobpursley**, Friday, October 12, 2012 at 9:07amInteresting question.

the vertical height from the edge of the brick to the center is 18 inches. The horizontal distance from the edge of the brick to the center of gravity requires some calculation.

draw the figure. from the brick edge to the center is 20 inches. sketch a horizontal line frm the edge of the brick to the vertical downward radius. The You now have a triangle, 20 hypotenuse, 18 on the downward leg, and the horizontal leg will be d=sqrt(20^2-18^2)=sqrt(400-324)=sqrt 76 check that.

Now moments around the edge sum to zero.

F*18-weight*sqrt76=0

forceat center=72lbs*sqrt76/18

at the top, the process is similar.

Force*38-72sqrt76=0 solve for Force

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