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I've looked at this quite and few times and can't get it. Probably something simple...
F being force, a area, d weight density, h depth of liquid.
We have the wall of a dam, with a height of 100 ft and a width of 200 ft.What is the total force on the wall if the artificial lake behind it has risen to within 10 ft of the top? equation: F = 1/2Adh weight density of water = 62.4 lb/ft`3 (ft`3 = ft cubed) height of wall subjected to water pressure = 90 ft area of wall next to water = 18,000 ft`2
F = 1/2(18,000 ft`2)(90ft)(62.4 lb/ft`3)
The answer is F = 5.1 x 10`7 lb
ok, to the question.... I get the number answer, but what I don't get is why only lb is left over. If you were multiplying, wouldn't you add the exponents? So why is there no ft. left over? Hold on.... do you divide (18,000ft`2)(90 ft) / ft`3? Can you do that on one side of an equation?
you just need some confidence...you've figured it out. One of the best techniques to seeing if you have the 'right' answer is to look a the units. You're correct about adding the exponents ...and all on one side of the equation if you like. Ft^2 *Ft* lb/Ft^3.. Add the exponents. Ft^3*lb/Ft^3 or Ft^3*lb*FT-3.. = lb. The quick term is the units of Ft cancel out. yielding only force as your answer.
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