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Shady01

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My teacher gave our class this in 2U, I get the vibe its a 3U ques.

An open rectangualr tank "a" units deep and "b" units wide holds water and is tilted so that the base BC makes an angle Q with the horizontal. When BC is returned to the horizontal, show that the depth of water is a^2x cotQ/2b
 

rama_v

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Shady01 said:
My teacher gave our class this in 2U, I get the vibe its a 3U ques.

An open rectangualr tank "a" units deep and "b" units wide holds water and is tilted so that the base BC makes an angle Q with the horizontal. When BC is returned to the horizontal, show that the depth of water is a^2x cotQ/2b
The question if from the green 3 unit fitzpatrick book. From what I remember it can be done by 2 unit techniques but its not especially easy.
 

airie

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... I don't quite get what the question asks...Where was the water level when the tank was tilted? And did you mean that the whole length of the tank is always in contact with the ground? (as in, the tank was not standing on a corner or something :p)
 

SoulSearcher

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After digging out my notes of the past term, here's my answer. (assuming its the question in the 3u fitzpatrick book)

You have the base of the tank making an angle of Q with the horizontal. The water is parallel to the horizontal, making the angle the water line makes with the base BC also equal to Q (refer to diagram)

Let the side of the triangle on the base BC be equal to x

therefore
a/x = tan Q
x/a = 1/tan Q (flipping the equation)
therefore
x = a/tan Q
x = a cot Q
therefore area of a triangle is
1/2 * a * x * sin 90 = a/2 * a cot Q * 1
= (a2cot Q)/2

now since the volume of the water is equal to the height multiplied by the base when the tank lays flat,
(a2cot Q)/2 = height * base
(a2cot Q)/2 = height * b, as b is the base length therefore
a2cot Q/2b = height of the water when the tank lays flat.
 

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