Finding the horizontal force on the gate if the density of water is w. helpp!?
i posted this twice already but neither of the answers checked with the back of the book. i really need to understand how to do this for my test so please help me and i REALLY REALLY appreciate your time. okay heres the quesiton again.
A control gate in the form of a parabolic segment of base 12 and height 4 is submerged in water so that its base is 2 units below the surface of the water. Find the horizontal force on the gate if the density of the water is w. < thats the end of the question.
more details: the parabolic gate has the equation 9y=x^2. and there are points (-6,4) and (6,4) on the parabolic gate in the text theres these formulas. p=F/D=wh F=w h A how would i set this up? and what would be my limits ?
- smciLv 71 decade agoFavorite Answer
So first find the intersection points with the water surface:
"height 4 is submerged in water so that its base is 2 units below the surface of the water."
So y=1/9 x^2 shifted down by 2 becomes y=1/9 x^2 - 2
And the intersections with the surface are at y=0:
0=1/9 x^2 - 2
x^2 = 18
x = +/- 3√2
F = Pressure x Area
But Pressure is also a function of depth y (or d, or h, whatever):
Pressure = w * g * y ("w" = rho, usually)
So the incremental force at depth y, due to an infinitesimal strip dy deep and x-(-x)=2x wide, is:
dF = dP * dA
= (w * g * y) * (2x dy)
It's easier to parameterize for y than x.
So if we know y, x is given by:
y = 1/9 x^2 - 2
x = 3 √(y+2)
dF = (w * g * y) * (6 √(y+2) dy)
dF = (6wg) y sqrt(y+2) dy
Then integrating between y=-2 and y=0 :
F = (6wg) ∫y √(y+2) dy between y=-2 and 0
Make a substitution z = y+2, thus dz = dy
F = (6wg) ∫((z-2) √z dz) between z=0 and +2
= (6wg) ∫ z^(3/2) -2z^(1/2) dz)
= (6wg) [-2/5 z^(5/2) + 2(2/3) z^(3/2)]
= (wg) [-12/5 z^(5/2) + 8z^(3/2)]
= (wg) [-12/5 2^(5/2) + 8*2^(3/2)]
= (wg2√2) [8 - 12/5 * 2]
= (wg2√2) [8 - 4.8]
= (wg) 6.4√2
sahsjing, you should say your approach is the same as mine, because I posted the approach first :)
- sahsjingLv 71 decade ago
My approach might contain some errors. The thing is you didn't post the picture. I assume the gate has parabola shape, opens up and most part of the gate is above the water surface .Is this right? But I don't understand why " there are points (-6,4) and (6,4) on the parabolic gate"? These two points should be above the water surface, aren't they?
If you can't post the picture, would you just post the answer? I am 100% sure I can figure it out.
If p = wh, then
F = ∫PdA
= ∫w(2-y) 2xdy [y: 0...2], (dA = 2xdy)
= ∫w(2-y) 6√y dy [y: 0...2], (x = 3√y)
= 9.0501 w
The density is a function of height. Therefore, your idea doesn't work. By the way, what do you think the parabola opens up or down?
Your approach is the same as mine. You missed a sign only. Check it. You should get the same answer as I showed here.
I am very glad that finally you got the same answer as I had showed here. :)
By the way, you really don't need to modify the equation, which makes the integral more complicated.
- PretzelsLv 51 decade ago
All you need to do is integrate over the pressure of the water over the submerged area of the gate. I will do everything algebraically and put numbers in last. Things are much clearer that way and, if I misunderstood some point, it is easy for you to correct 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 the water is k. k is not specified but for dimensions in feet it is 62.4 lb/ft^3 for meters it is 1000 kg/m^3
The water pressure is proportional to the density so it is a straight line fnction with a value of b*k at y = 0. It increases with increasing depth (decreasing y) at a rate of k. These two point give the slope and intercept of the pressure function. The pressure as a function of y is:
p = b*k - k*y = k(b - y)
The force on a differential area is the product of the pressure and the area:
dF = p dA = k(b - y) dy dx
All that is left is to set up the limits and integrate. y ranges from 0 to b and x ranges the width of the gate, from -sqrt(y/a) to +sqrt(y/a). Since the x limits are a function of y, you will integrate on x first. the first integration yields: k(b - y) x dy
Applying the limits gives: 2 k (b - y) sqrt(y/a) dy
Integrating this on y (and simplifying) gives: 4ky(b/3 - y/5) sqrt(y/a)
Applying the limits of 0-b:
F = 4kb(b/3 - b/5)sqrt(b/a) = 8kb^2 sqrt(b/a)/15
Putting in the numbers: F = 8 k 2^2 sqrt(18)/15 = 9.051 k
The final result depends on your selection of units to choose k.
- 1 decade ago
Wouldn't you just find the area of the gate and multiply that by the density?
The area is int (4 - x^2/12) dx, from -6 to 6.