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thepatient
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http://i272.photobucket.com/albums/jj198/1000_2008/sgsdfgsfsd/Capture.jpg" [Broken]
I just had a question regarding the determination of force P on the body of water CDE. For weight P, they used 1/2(18ft)(1ft)(18ft)(62.4lb/ft^3). I understand that this is because we are using volume*specific weight to find net weight P on body CDE. What I don't understand is why did they use 18ft as the base of the triangle? Is this because the pressure varies with the depth in the dam? Will this always be the case where there is a 1:1 ratio between height and width when considering the force per length on a distributed load due to pressure? Thanks. :]
I just had a question regarding the determination of force P on the body of water CDE. For weight P, they used 1/2(18ft)(1ft)(18ft)(62.4lb/ft^3). I understand that this is because we are using volume*specific weight to find net weight P on body CDE. What I don't understand is why did they use 18ft as the base of the triangle? Is this because the pressure varies with the depth in the dam? Will this always be the case where there is a 1:1 ratio between height and width when considering the force per length on a distributed load due to pressure? Thanks. :]
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