Problem on the required force to keep the gate of a dam at equilibrium

[Kaushik]

Homework Statement

A rectangular gate, $3m * 1m$ , stands vertical with the water on one side of it, hinged at the middle. What is the force F required to be applied at the bottom to keep the gate in equilibrium?

Relevant Equations

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$\int \!dτ = \int \!P.dA.x$
Here, I am taking torque of the part above the axis$(τ_1)$ ,which is clockwise, and the torque of the part below the axis $(τ_2)$, which is anticlockwise, separately.

Now, let me consider a thin strip $dx$ at a height $x$ from the axis.
$τ_1 = \int_0^{\frac{1}{2}} \!ρg(x-\frac{1}{2})dA.3x$ $\tag{1}$
Considering, a thin strip $dx$ at a depth $x$ from the axis.
$τ_2 = \int_0^{-\frac{1}{2}} \!ρg(x+\frac{1}{2})dA.3x$ $\tag{2}$
Net torque due to water is given by,
$τ_{water} = τ_1 - τ_2$ (considering the anticlockwise torque to be negative direction)
Then,
$τ_{water} = \frac{F}{2}$
But this does not yield the answer. It seems like taking $0$ to $-\frac{1}{2}$ as the limit is the reason why I am getting an incorrect answer. But why is it so? What is wrong when I take $-\frac{1}{2}$? I took $-$ve as the points below the axis are negative while the points above the axis are $+$ve, considering the axis as the origin?

 

[BvU]

thin strip $dx$ at a height $x$ from the axis

So what is the $-{1\over 2}$ doing there ?
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[Kaushik]

So what is the $-{1\over 2}$ doing there ?
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So taking the axis as the origin, the part below it is negative while the part above it is $+$ve. That was my intention. Hence, I took it as $-$ve.

When I asked my friend about it, he asked me to take the magnitude. But I don't get the reason. Why shouldn't we take $-\frac{1}{2}$?

 

[BvU]

If you use -1/2 your integrand is negative and the $\tau_1$ comes out negative. Would that mean it is pointing anticlockwise ?

 

[BvU]

Note that what you want to use is a difference in height to calculate the pressure. Pressure better be positive.

 

[Kaushik]

If you use -1/2 your integrand is negative and the $\tau_1$ comes out negative. Would that mean it is pointing anticlockwise ?

Using -1/2, I am getting +ve τ. According to my convention +ve is for clockwise. But τ has to be anticlockwise for the lower part. Is this what you are saying?

 

[Kaushik]

Note that what you want to use is a difference in height to calculate the pressure. Pressure better be positive.

So taking -1/2 has limit does not fit for pressure.
For e.g., for -1/2, when in pressure expression, then the value is $ρg(0)$ which is not the case. It should be $ρg(2*0.5)$. So I should consider my limit as 1/2 and then later add a negative sign to it to show that it is anticlockwise. Is it?

 

[BvU]

Make a sketch (side view) and try to catch the whole thing in one integral from -1/2 to +1/2 . It's easy !

 

[Kaushik]

Make a sketch (side view) and try to catch the whole thing in one integral from -1/2 to +1/2 . It's easy !

Let me give it a try!

 

[Kaushik]

I got $ρg/4$ using the limit as -1/2 to 1/2.

 

[BvU]

You get the picture

Nitpicking:

• the 4 in $\rho g/4$ has a dimension
• with that dimension, $\rho g/4$ is a torque
• the exercise asks for a force

but that's the easy part

 

[Kaushik]

You get the picture

Nitpicking:

• the 4 in $\rho g/4$ has a dimension
• with that dimension, $\rho g/4$ is a torque
• the exercise asks for a force

but that's the easy part

• Dimension of 4: $[M^4L^0T^0]$
• $\rho g/4 = F/2 \implies F = 5 * 10^3 N$

 

[haruspex]

• Dimension of 4: $[M^4L^0T^0]$

No. How do you get that?

 

[Kaushik]

No. How do you get that?

Oops! Is it $[M^0L^{-4}T^0]$?

 

[haruspex]

Oops! Is it $[M^0L^{-4}T^0]$?

Yes.

 

[BvU]

• Dimension of 4: $[M^4L^0T^0]$ [corrected to $L^{-4}$]
• $\rho g/4 = F/2 \implies F = 5 * 10^3 N$

same comment: the ${1\over 2}$ has a dimension too

(urgent) advice: work with symbols, substitute numbers+dimensions at the very last moment and do the math (numerical with the numbers, algebraic with the dimensions)

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[Kaushik]

same comment: the ${1\over 2}$ has a dimension too

(urgent) advice: work with symbols, substitute numbers+dimensions at the very last moment and do the math (numerical with the numbers, algebraic with the dimensions)

## ##

Dimension of $\frac{1}{2}$ : $[M^0L^1T^0]$

Why is it helpful to first solve it in general and then substitute?

 

[BvU]

• Allows you to check dimensions
• Less prone to errors, easier to fix if an error is discovered
• Often you can strike variables common in numerator and denominator (if not immediately, then sometimes in a subsequent part of a multi-part exercise)
• easier to calculate
• Physics teachers tend to be lenient with numerical calculation errors if the physics formula is OK
• gives a feeling of dependencies (cf $1/r^3$ with $1/8$)
• and so on

 

[haruspex]

• Allows you to check dimensions
• Less prone to errors, easier to fix if an error is discovered
• Often you can strike variables common in numerator and denominator (if not immediately, then sometimes in a subsequent part of a multi-part exercise)
• easier to calculate
• Physics teachers tend to be lenient with numerical calculation errors if the physics formula is OK
• gives a feeling of dependencies (cf $1/r^3$ with $1/8$)
• and so on

I would add:
Makes it easier for others to follow your process
Makes it easier for others to check your algebra without constantly having to check the arithmetic too.
Allows sanity checks, like plugging in extreme or special values where the correct result would be obvious.
If you plug in numbers early you may find later that you should have kept more digits.
If plugging in numbers only at the end you will often be able to do it all in one sequence in your calculator. Internally that keeps many more digits, so you get a much more accurate answer.

And wrt checking dimensions, that is not just for physics questions where the variables naturally have dimension. I have often used it in purely algebraic contexts, where it may be possible to assign notional dimensions to the variables in the initial equations in a consistent way and check it is all still consistent at the end. If it is not, you can "binary chop" to track down the error: check it half way through the working; if ok there check 3/4 of the way through...