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A parallelogram is immersed vertically in a liquid with a corner in the surface . if a and b be the depth of the adjacent corners,find the depth of center

of pressure?



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The centre of pressure is the same as the geometric centre. And it will be (a+b)/2
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For most of the regular geometric figures, centre of mass, centre of gravity are the same as the geometric centre. And I heard about centre of pressure first time. So I assumed it to be geometric centre.
centre of pressure(c.p)=moment of resultend thrust/sum of the individual thrust
Then which point, according to you is the cp?
firstly,we consider an arbitrary pt on the figar ,then find out c.p. on this pt and then this transformed to the holl area
How good are you at integration? I don't want to do it.
The pressure will be same as the geometrical centre that is (a+b)/2
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