Finding the horizontal force on the gate if the density of water is w. helpp!? - parabolic gate moon
I asked them twice to check but no answers on the back of the book. I must really know how my exams this please help me and Really Really appreciate your time. well heres the quesiton again.
A grid control in the form of a parabolic segment of the base 12 and a height of 4 is immersed in water, so that its base is 2 units below the water surface. Find the horizontal force on the door when the density of water W \\ \\ \\ \\ \\ \\ \\ \\ U0026lt; is the end of the matter.
For further details: The door of the parabolic equation 9y = x ^ is 2nd and there are points (-6.4) and (6.4) in the door theres a flat in the text of these formulas. P = F / D = F = WH WH A How can I set it up? and what would be the limit for me?
Wednesday, February 17, 2010
Parabolic Gate Moon Finding The Horizontal Force On The Gate If The Density Of Water Is W. Helpp!?
4 comments:
So find the intersections with the surface of the water:
"height 4 is immersed in water, so that its base is 2 units below the surface of the water."
So y = 1 / 9 x ^ 2 teams of 2, y = 1 / 9 x ^ 2-2
And intersections with the surface y = 0:
0 = 1 / 9 x ^ 2-2
x ^ 2 = 18
x = + / - 3 √ 2
F = Pressure x Area
But the pressure is also a function of depth and (or D, oh, whatever):
Pressure = w * g * and ( "W" = Rho, as a rule)
Thus, the increase in the strength and depth, because a gang of dy infinitely deep x-(-x) = 2x as broad as follows:
dF = DP * DA
= (W * g * y) * (2x dy)
It is easier to configure and to x.
If we know, therefore, y, x is equal to:
y = 1 / 9 x ^ 2-2
x = 3 √ (y 2)
So
df = (W * g * y) * (6 √ (y 2) dy)
df = (6WG) and sqrt (y 2) dy
Then, the integration between y =- 2 and y = 0:
F = (6WG) ∫ √ y (y 2) dy from y =- 2-0
A substitution z = y 2, so that dy = dz
F = (6WG) ∫ ((Z-2) √ z) between z = 0 and 2 dz
= (6WG) ∫ z ^ (3 / 2)-2z ^ (1 / 2) dz)
= (6WG) [-2 / 5 z ^ (5 / 2) + 2 (2 / 3) Z ^ (3 / 2)]
= (GT) [-12 / 5 z ^ (5 / 2) + 8z ^ (3 / 2)]
= (GT) [-12 / 5 2 ^ (5 / 2) + 8 * 2 ^ (3 / 2)]
= (GT2 √ 2) [8 to 12 / 5 * 2]
= (GT2 √ 2) [8-4.8]
= (WG) 6.4 √ 2
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sahsjing, I must say that their approach is the same as me, because I sent the first approach:)
Did not you find just the boarding area multiplied by the density?
The area is int (4 - x ^ 2 / 12) DX, -6 to 6
math55,
My approach has some errors. The thing is, not according to the photo. I guess the door is in the shape of a parabola, opens it and most of the airlines on the surface. Is that correct? But I do not understand why "there are points (-6.4) and (6.4) in the apartment door? These two points must be above the waterline, is not it?
If you are not in the image, because the only answer? I am 100% sure I will find out.
If p = WH, then
F = ∫ POA
W = ∫ (2-y) 2xdy [Y: 0 ... 2] (DA = 2xdy)
W = ∫ (2-y) 6 √ d, and [Y: 0 ... 2] (x = 3 √ y)
W = 9.0501
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bictor717,
The density is a function of height. So his ideaWork. By the way, what do you think of the parabola opens up or down?
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SMCI,
His goal is the same as mine. You have missed a sign. Check it. They should have the same answer, which I made available here.
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SMCI,
I am very happy that we finally reached the same answer I have been here. :)
In fact, this is not really necessary to change the equation that the more complicated integral.
Everything you need to do is more water pressure on the submerged part of the door to take. I'm going to my calculations, and put numbers in the past. Things are much clearer in this way, and if I heard somewhere it's easy for you to fix it.
The gate has the form y = a * x ^ 2 Here a = 1 / 9
The base (origin) is at a depth of b. Here b = 2
the density of water is k. k is not fixed, but the dimensions in feet, EUR 62.4 / m ^ 3 meters of 1000 kg / m ^ 3
Water pressure is proportional to the density, which fnction a straight line with a value of k * b at y = 0 With depth (see below) increases at a rate of k. These two points, the slope and intercept give the function of pressure. The print functionn, and that is:
p = b * k - k * y = k (b - y)
The force on a differential area is the product of pressure and environment:
df = p dA = k (b - y) dy dx
All that remains is to establish boundaries and to integrate them. varies between 0 and abyx varies the width of the door-sqrt (n / a) + sqrt (n / a). While the limits are a function of X on Y, which are provided with "integrated x" in the first place. The first integration yields: k (b - y) x dy
The application of the limit, we obtain 2 k (b - y) sqrt (y / y) dy
The integration of the Y (and simplicity), there 4ky (b / 3 - y / 5) sqrt (y / y)
The application of the boundary of 0-B
F = 4K (b / 3 - B / 5) sqrt (b / a) ^ 2 = 8kb sqrt (b / a) / 15
Putting in the numbers: F = 8 k 2 2 sqrt (18) / 15 = 9.051 k
The final result depends on theSelection of devices to choose, K.
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