Force of water against a dam gate

[brotherbobby]

Homework Statement

The upper edge of a gate in a dam runs along the water surface. The gate is 2.00 m high and 4.00 m wide and is hinged along a hor izontal line through its center as shown in the diagram below. Calculate the torque about the hinge arising from the force due to the water.

Relevant Equations

1. Pressure due to a liquid at a depth $d$ : $P=\rho gd$
2. Force as a result of pressure $P$ : $F=PA$, where $A$ is the area of the surface. Here the force $F$ is constant at all points on the surface.
3. Torque arising due to force about an origin : $\tau = Fx$ where $x$ is the distance of the point where the force acts from the origin.

Question : I start by putting the image of the problem from the book. The water surface is given to run along the (top) edge of the bridge.

Attempt : The crux of this problem lies in the fact that the pressure of water against the gate will vary along its depth $h$ as $\rho gh$. This makes the use of calculus inevitable. Before I proceed further, I'd like to draw an image of how the gate looks when viewed from one side and outside the dam, as shown below. Note that the bridge is hinged along the line OP. I suppose I have to calculate the torques on each half separately.

(a) The top half : Let's imagine a thin layer of water of thickness $dx$ acting at a distance $x$ from the hinge (shown in light blue). The pressure on the gate due to this layer $P = \rho g\left(\frac{h}{2}-x\right)$, noticing that the layer is at a depth of $h/2 - x$ from the top. The (infinitesimal) force on this layer : $dF = \rho g\left(\frac{h}{2}-x\right)bdx$, since the area of this layer is $bdx$. The torque due to this layer about the "midrib" OP is $d\tau=\rho g\left(\frac{h}{2}-x\right)bxdx$, as the layer is distant $x$ from the midrib. Hence the total torque due to water in the upper half above the gate : $$\tau_{\text{up}}= \rho g\int_0^{h/2} \left(\frac{h}{2}-x\right)bxdx=\dfrac{\rho gbh^3}{48}\; \text{(agrees with book answer, hence details suppressed)}$$
(b) Bottom half : (Now here I run into trouble). Proceeding as (a) above, we imagine a thin layer of water of thickness $dx$ acting at a distance $x$ from the hinge (shown in light blue). The pressure on the gate due to this layer $P = \rho g\left(\frac{h}{2}+x\right)$, noticing that the layer is at a depth of $h/2 + x$ from the top. The (infinitesimal) force on this layer : $dF = \rho g\left(\frac{h}{2}+x\right)bdx$, since the area of this layer is $bdx$. The torque due to this layer about the "midrib" OP is $d\tau=\rho g\left(\frac{h}{2}+x\right)bxdx$, as the layer is distant $x$ from the midrib. Hence the total torque due to water in the upper half above the gate : $$\tau_{\text{down}}= \rho g\int_{h/2}^{h} \left(\frac{h}{2}+x\right)bxdx=\rho gb\left[\frac{hx^2}{4}+\frac{x^3}{3}\right]_{h/2}^{h}= \rho gb\left[\frac{h^3}{4}+\frac{h^3}{3}-\frac{h^3}{16}-\frac{h^3}{24}\right] = \frac{23}{48} \rho gbh^3$$ We can put values later for calculation, but the second answer for $\tau_{\text{down}}$ does not match with that of the book.

The issue : The book does the second integral also from 0 to $h/2$ (like the first one for $\tau_{\text{up}}$) rather than from $h/2\; \text{to}\; h$ , as I have done. It ends up with the result $\frac{5}{48}\rho gbh^3$. My reasoning is that the gate height (or depth) be taken as a whole, even when considering each torque separately.

Request : A hint or a suggestion as to integral limits would be helpful.

 

[SilverSoldier]

Well, I believe you must reconsider the range of values through which the variable $x$ you chose (which is the distance that the thin layer of water is below the level of $OP$, and not from the surface of the water itself), actually varies through, in describing the entire column of water.

 

[brotherbobby]

Well, I believe you must reconsider the range of values through which the variable $x$ you chose (which is the distance that the thin layer of water is below the level of $OP$, and not from the surface of the water itself), actually varies through, in describing the entire column of water.

Yes an elementary error, as I now realize. Since the variable $x$ is taken from the line OP, which should the values of $x$ be chosen from some other line? Thanks.

 

[haruspex]

Yes an elementary error, as I now realize. Since the variable $x$ is taken from the line OP, which should the values of $x$ be chosen from some other line? Thanks.

There's no need to break it into two integrals, and it does not matter where you take the zero height to be as long as you are consistent.
I couldn't understand your question above.

 

[rude man]

I'd haver said you do need two integrals: one for the net torque above the hinge and another below.

 

[haruspex]

I'd haver said you do need two integrals: one for the net torque above the hinge and another below.

There's no discontinuity in the derivative. The change of sign of the torque arm should take care of things.

 

[SilverSoldier]

which should the values of x be chosen from some other line?

Had $x$ been chosen to be the distance to a "thin line of water" as measured from a different fixed line, such as maybe the line $AB$, then the limits of integration should vary through the distance measured to the uppermost "thin line of water" from $AB$, and the distance measured to the lowermost "thin line of water", also from $AB$, which in this case means from $\dfrac{h}{2}$ to $h$.

 

[jbriggs444]

Had $x$ been chosen to be the distance to a "thin line of water" as measured from a different fixed line, such as maybe the line $AB$, then the limits of integration should vary through the distance measured to the uppermost "thin line of water" from $AB$, and the distance measured to the lowermost "thin line of water", also from $AB$, which in this case means from $\dfrac{h}{2}$ to $h$.

So if we integrate force times moment arm from x=0 to x=h/2 we have the torque on the top part.
And if we integrate force times moment arm from x=h/2 to x=h we have the torque on the bottom part.

But the quantity being integrated (force times moment arm) is the same for both integrals. So the sum of the two is the same thing as the integral from x=0 to x=h. Which is one way of seeing the point being made by @haruspex

 

[brotherbobby]

So if we integrate force times moment arm from x=0 to x=h/2 we have the torque on the top part.
And if we integrate force times moment arm from x=h/2 to x=h we have the torque on the bottom part.

But the quantity being integrated (force times moment arm) is the same for both integrals. So the sum of the two is the same thing as the integral from x=0 to x=h. Which is one way of seeing the point being made by @haruspex

Yes, the best thing for me (the OP of this thread) to do is to carry out the integral once along the height (or depth) of the gate to calculate the torque acting on it.

 

[rude man]

. The change of sign of the torque arm should take care of things.

AKA two separate integrals.
BTW what derivatives?
(PM me your solution if you want.)

 

[rude man]

Yes, the best thing for me (the OP of this thread) to do is to carry out the integral once along the height (or depth) of the gate to calculate the torque acting on it.

And your answer is ... ?

 

[haruspex]

AKA two separate integrals.

No, one integral. If we take the water as being on the right, force as positive left and torque as positive anticlockwise then at any given vertical displacement y from the axis the torque is like Fy.dy. This is true where y is negative (producing a clockwise torque) and where y is positive. (Note that dy is positive throughout.) The integral across the full range produces the correct total.

You would need two integrals if the integrand were F|y|, which would have a discontinuous derivative. I'm not saying continuity of the derivative is a foolproof test, but it is a good guide.

 

[rude man]

No, one integral. If we take the water as being on the right, force as positive left and torque as positive anticlockwise then at any given vertical displacement y from the axis the torque is like Fy.dy. This is true where y is negative (producing a clockwise torque) and where y is positive. (Note that dy is positive throughout.) The integral across the full range produces the correct total.

You would need two integrals if the integrand were F|y|, which would have a discontinuous derivative. I'm not saying continuity of the derivative is a foolproof test, but it is a good guide.

I think you'e settimg x=0 at the half-way mark. Then I agree one integral automatically separates the two torques to get the net torque.

But I think it's less intuitive than to think of two torques pushing against each other. So I set x=0 at the water line and computed the two torques separately.
I also can't get excited about the derivative being discontinuous; it does not affect the computation.
The key thing is where x=0 is set and I agree one integral does the job nicely and elegantly from a mathematical viewpoint if it's set half-way into the water.

 

[haruspex]

The key thing is where x=0 is set

No, as I mentioned in post #4, it does not matter where you take the zero height. You just have to be consistent.

 

[rude man]

If you set x=0 at the waterline then all torques are of the same polarity.
$p = \rho ~g~x$
Then
$dF = p~ L~ dx$
$d\tau = x dF$
and $\tau$ is of the same sign thruout 0 < x < h.
Something has to change and I claim it's effectively making x=0 at the half-way depth.

 

[haruspex]

If you set x=0 at the waterline then all torques are of the same polarity.
$p = \rho ~g~x$
Then
$dF = p~ L~ dx$
$d\tau = x dF$
and $\tau$ is of the same sign thruout 0 < x < h.
Something has to change and I claim it's effectively making x=0 at the half-way depth.

But if you set x=0 there then x is not the torque arm. The torque arm will be x-h/2, or somesuch.
As I wrote, you have to be consistent.

 

[rude man]

But if you set x=0 there then x is not the torque arm. The torque arm will be x-h/2, or somesuch.
As I wrote, you have to be consistent.

I claim taht no matter what manipulations you engage in it has to effectively be making x=0 at the half-way depth.

 

[haruspex]

I claim taht no matter what manipulations you engage in it has to effectively be making x=0 at the half-way depth.

Of course, but that does not mean you have to measure x from the axis. You can set the zero point where you like provided you write the correct expression for the torque arm.
It was somewhat garbled, so I may have misinterpreted it, but I took that to be the question being asked in post #3.

 

[rude man]

Of course, but that does not mean you have to measure x from the axis. You can set the zero point where you like provided you write the correct expression for the torque arm.
It was somewhat garbled, so I may have misinterpreted it, but I took that to be the question being asked in post #3.

OK. I didn't pay a lot of attention to most of the posts. Should have.

 

[Delta2]

No, as I mentioned in post #4, it does not matter where you take the zero height. You just have to be consistent.

The OP text says that the wooden door is hinged along the line OP so doesn't this force us to take x=0 at the line OP?
We would have to take into account the torque from the reaction force at the hinge otherwise.

 

[haruspex]

The OP text says that the wooden door is hinged along the line OP so doesn't this force us to take x=0 at the line OP?
We would have to take into account the torque from the reaction force at the hinge otherwise.

Where you choose to measure x from and what line you take as the axis for moments are, in principle, separate choices.

 

[Delta2]

Where you choose to measure x from and what line you take as the axis for moments are, in principle, separate choices.

Ehm, I am not sure I understand this, can you give an example where these two choices are different, and how the calculations can be correct ?

EDIT: Depends what you mean by "axis for moments". I took it that you mean the line such that all forces with point of application on this axis, they have moment zero. Hence it would be x=0 for the arm of all these forces.

 

[haruspex]

Ehm, I am not sure I understand this, can you give an example where these two choices are different, and how the calculations can be correct ?

EDIT: Depends what you mean by "axis for moments". I took it that you mean the line such that all forces with point of application on this axis, they have moment zero. Hence it would be x=0 for the arm of all these forces.

You are probably looking for something more Earth-shattering than I intend.

If we take x as the height of an element of depth dx above axis OP then the force on the element is $\rho gb(h/2-x)dx$. The moment arm about OP is $x$, so the torque is $\rho gb(h/2-x)x.dx$. The integration range is $(-h/2,h/2)$.

If we take x as the depth of the element below AB then the force on the element is $-\rho gb.dx$. The moment arm about OP is $x-h/2$, so the torque is $-\rho gbx(x-h/2)dx$. The integration range is $(0,h)$.

In each case we get $-h^3/12$.

Yet again, we could take moments about AB, but as you wrote, we would need to introduce the force from the axle and write a linear force equation to find its value.

 

[Delta2]

Yes ok I see now, I confused the arm of the force with the coordinate of the point of application of the force, these two usually are the same though, cause that's usually a convenient choice for the axis of moments to be the axis x=0.

 

[brotherbobby]

There's no need to break it into two integrals, and it does not matter where you take the zero height to be as long as you are consistent.
I couldn't understand your question above.

I am the creator of this thread (OP) and am sorry for coming in late. The question in my post #1 above has already been answered in the response post#2 below. However, a different matter came up which has been the cause of some thinking. In order to clear the waters, I'd like to review the problem matter to begin with, if you don't mind. Thanks to @haruspex, @Delta2 and @rude man for your comments which gave me cause to think that this problem (finding the torque) can be done via a $\text{single integral}$ about a reference line than $\text{two separate ones}$.

Problem statement (review) : The upper edge of a gate in a dam runs along the water surface. The gate is 2.00 m high and 4.00 m wide and is hinged along a horizontal line through its center as shown in the diagram. Calculate the torque about the hinge arising from the force due to the water.

Solution (review) : I took the cross section of the gate imagining the water pushing it from the back. I considered two thin layers of water above and below the hinge separately, calculating the torque they would apply on the wooden gate about the hinge. I integrated this infinitesimal torque from the hinge upwards and downwards to either end of the gate to calculate the net torque on the gate. The total answer comes out to be equal to $\boxed{\tau = \dfrac{\rho gbh^3}{12}}$. The answer matches with that in the book and the case could have been closed.

The issue (review) : A suggestion was made by user @haruspex (post #4) that there was no need to take two separate integrals about the hinge of the gate. A single integral from a reference line along the gate, say the top edge and running along the height of the bridge would do just as well. The strategy baffled me because the line $x=0$, the "midrib" of the gate, is given to be the hinge. Wouldn't the terms of the integral from the top change as to whether one was above or below the hinge? I have decided to carry the integral out below to put these doubts to rest.

Attempt : I take the reference line to be the top edge of the gate, $\textbf{AB}$. Let's consider a (first) thin layer of water (shown in blue - 1) at a distance $x$ below the line of reference. The (gauge) pressure of water on the gate at that point is $P(x) = \rho gx$. The (small) force of water due to this pressure $dF = \rho gx b dx$, where the (small) area of this layer of water $dA = bdx$. Hence the torque that this layer of water exerts on the gate about the hinge across the middle of the gate is $d\tau = \rho bgx (h/2-x) dx$, we the distance of this thin layer of water from the hinge is $h/2 - x$. We note that this torque comes out of the page, often denoted by the sign $\bullet$. $\\[5pt]$
What about the second layer of water, distant at some other $x$ from the line of reference? We find that for this layer, we have $x>h/2$, hence the expression for torque would read $d\tau = \rho bgx (x-h/2) dx$. Moreover, though the torque would also be directed out of the gate, its amount would be higher than that for the layer of water above the hinge. The net effect would be a torque that would rotate the part of the gate below the hinge "out" and that above the hinge "in". The value of this torque : $$d\tau =\rho bg\left[\int_0^{h/2} x (h/2-x) dx-\int_{h/2}^{h}x(x-h/2)dx\right] = \rho bg\left[ \left|hx^2/4- x^3/3\right|_{0}^{h/2} - \left|x^3/3-hx^2/4\right|_{h/2}^{h} \right]$$ $$\Rightarrow \tau = \rho bg\left[ \cancel{h^3/16}-\bcancel{h^3/24}-\left( h^3/3-h^3/4-\bcancel{h^3/24}+\cancel{h^3/16} \right) \right] = \boxed{\frac{-\rho gbh^3}{12}}$$, which is the same as the answer found earlier, except for the minus sign.

Outstanding : It must be noted that the issue raised by @haruspex remains unresolved. Even though only one reference line was used ($\textbf{AB}$), two different integrals still had to be made use of. More, the integrals had to be "tailored" to suit the value of the torque about the midrib of the bridge. The method here, though done in one step, wasn't fundamentally different to the one in my original post #1, but I could be mistaken.

All comments and responses welcome. Thank you for your time.

 

[haruspex]

The value of this torque : $$d\tau =\rho bg\left[\int_0^{h/2} x (h/2-x) dx-\int_{h/2}^{h}x(x-h/2)dx\right]$$
Even though only one reference line was used ($\textbf{AB}$), two different integrals still had to be made use of.

Did you try doing it as a single integral?
$$\tau =\rho bg\int_0^{h} x (h/2-x) dx$$
All the way down the gate, the depth from the surface is x and the height of the element above the hinge is (h/2-x).

 

[brotherbobby]

Did you try doing it as a single integral?
$$\tau =\rho bg\int_0^{h} x (h/2-x) dx$$
All the way down the gate, the depth from the surface is x and the height of the element above the hinge is (h/2-x).

Yes I checked - I am not getting the correct answer. Let me write the details below.

Attempt : If the torque about the hinge were this integral $$d\tau = \rho bg\int_0^h x(h/2-x) dx = \rho bg \left[x^2h/8 - x^3/3 \right]_0^h = \rho bgh^3 \left[ 1/8 - 1/3 \right] = \boxed{-5\rho gbh^3/24},$$ which is not the correct answer.

Please let me know where am going wrong. Thank you for your time.

 

[haruspex]

Yes I checked - I am not getting the correct answer. Let me write the details below.

Attempt : If the torque about the hinge were this integral $$d\tau = \rho bg\int_0^h x(h/2-x) dx = \rho bg \left[x^2h/8 - x^3/3 \right]_0^h = \rho bgh^3 \left[ 1/8 - 1/3 \right] = \boxed{-5\rho gbh^3/24},$$ which is not the correct answer.

Please let me know where am going wrong. Thank you for your time.

How do you get that divisor of 8?

 

[brotherbobby]

How do you get that divisor of 8?

Thank you. Sorry about that. Let me do it correctly again. $$d\tau = \rho bg\int_0^h x(h/2-x) dx = \rho bg \left[x^2h/4 - x^3/3 \right]_0^h = \rho bgh^3 \left[ 1/4 - 1/3 \right] = \boxed{-\rho gbh^3/12},$$
which is the correct answer.

Thank you, but I must confess, I am still confused. The location of the thin layer of water above the hinge is $h/2-x$ as measured from the hinge and that below the hinge is $x-h/2$, also measured from the hinge. Surely they are opposite in sign. How does the integral work its way around it?

 

[haruspex]

Thank you, but I must confess, I am still confused. The location of the thin layer of water above the hinge is $h/2-x$ as measured from the hinge and that below the hinge is $x-h/2$, also measured from the hinge. Surely they are opposite in sign. How does the integral work its way around it?

You need to use the scalar displacement, a signed quantity, not the distance.
E.g. think in terms of the height of the element above the hinge. For an element above the hinge that's positive; for an element below the hinge it is negative and leads to a torque of the opposite sign.
If you look at your two integrals in post #25 you should see how to combine them into the single integral. The two integrands are identical and the ranges abut.