# 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 ?

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• smci
Lv 7

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)

So

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

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sahsjing, you should say your approach is the same as mine, because I posted the approach first :)

math55,

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 = â«PdA

= â«w(2-y) 2xdy [y: 0...2], (dA = 2xdy)

= â«w(2-y) 6ây dy [y: 0...2], (x = 3ây)

= 9.0501 w

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bictor717,

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?

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smci,

Your approach is the same as mine. You missed a sign only. Check it. You should get the same answer as I showed here.

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smci,

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.

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.