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I have a circular swimming pool with a diameter of $4.5\;\text{m}$. The water height usually is about 1m. Somewhere in this pool is a single floodlight, circular aswell, diameter $25\;\text{cm}$.

I would like to calculate how much pressure this floodlight is exposed to (the pool is above ground).

My first approach was to measure how deep the floodlight is: bottom of it is $70\;\text{cm}$ beneath the surface.

-------    pool top
       |
-------|   water surface
       |
       |
       <   floodlight upper end
       <
       <   floodlight lower end
       |   
_______|   pool bottom

The distance pool bottom to floodlight lower end is $30\;\text{cm}$, floodlight lower endto water surface $70\;\text{cm}$. Diameter, as mentioned, $25\;\text{cm}$.

The water below the floodlight should be safe to ignore, so I assumed the pool is $70\;\text{cm}$ high and the floodlight is located right at the bottom. A pool with $4.5\;\text{m}$ diameter and $70\;\text{cm}$ high water contains about 11.1 tons of water ($11.13\;\text{m}^3$).

With a base area of $15.9\;\text{m}^2$ and a lateral surface of $9.9\;\text{m}^2$ this makes $25.8\;\text{m}^2$ of surface that the water is in contact with and that it can put pressure on (of course more pressure at the bottom than the top).

If I assume that the pressure per $\;\text{cm}^2$ is equal across the whole surface, this would equal to $\frac{11100\;\text{kg}}{258000\;\text{cm}^2} * 491\;\text{cm}^2 = 21,12\;\text{kg}$ mass on the whole speaker - but it's obvious that I can't calculate it like that.

Given that the floodlight is located at the bottom of my fictional $70\;\text{cm}$ high swimming pool, I could just calculate the amount of mass per $\;\text{cm}^2$ floor $(\frac{11100\;\text{kg}}{159000\;\text{cm}^2} = 0.07\;\frac{\text{kg}}{\;\text{cm}^2})$ and could generalize that "as the floodlight is close to the floor, mass per $\;\text{cm}^2$ would be equal enough in order to just calculate $0.07\frac{\;\text{kg}}{\;\text{cm}^2} * 491\;\text{cm}^2 = 34.3\;\text{kg}$".

Disclaimer: I know that pressure is measured in $\frac{\;\text{N}}{\;\text{m}^2}$, but as we're talking about stationary water on earth, the difference is just taking the gravity into account or not - if I can calculate $\;\text{kg}$ or $\;\text{N}$, I can calculate the other one as well.

So the question is: Can I generalize like above or am I on the wrong track? If so, how can I calculate the pressure on the floodlight?

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closed as off-topic by Kyle Oman, ACuriousMind, Kyle Kanos, yuggib, Ryan Unger Jun 26 '15 at 16:11

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  • $\begingroup$ possible duplicate of Water pressure on slanted wall $\endgroup$ – Kyle Oman Jun 24 '15 at 17:15
  • $\begingroup$ Rule of thumb: roughly an additional $10$kPa (about $0.1$ atmospheres) per metre of water depth, based on $g$ times the density of water. Don't forget the air above the swimming pool (about $100$kPa or $1$ atmosphere). The issue is going to be making the floodlight waterproof more than pressure at $45-70$cm. $\endgroup$ – Henry Jun 24 '15 at 17:17
  • $\begingroup$ Yeah I stumbled upon that rule, but does it also apply for a closed environment like a swimming pool with walls? I always figured thats more applicable for open environments like oceans etc (without walls). Following only that rule of thumb it would be completely irrelevant how big the pool is - that somehow feels wrong to me. $\endgroup$ – liz Jun 24 '15 at 17:25
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You should ignore the sides of the pool when calculating the pressure. This is because weight of the water must be fully balanced by the upwards force exerted by the pool's bottom. The isotropy of the water then ensures that an exactly equal amount of force is exerted regardless of the direction. (Newton's third law is satisfied because the force on one wall is exactly cancelled by the opposite wall.)

The upshot of this is that you can safely use the highschool formula, $p=\rho g \,\Delta h$, regardless of which way your surface is facing. Furthermore, as Mike mentions, you can safely take the pressure at the centre of the spotlight, and symmetry and linearity will do the rest of the work for you.

This means that the pressure at $\Delta h=(70-12.5)\,\mathrm{cm}=57.5\,\mathrm cm$ is $p=5.6\,\mathrm{kPa}$ above atmospheric pressure, for a total force of $276\,\mathrm{N}=28\,\mathrm{kg}·g$.

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Just assume the pressure at the center applies over the whole floodlight. It's actually less at the top and more at the bottom, but since the pressure is linear in depth, and the floodlight is vertically symmetric, the pressure differences between the top and bottom should exactly cancel.

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You can take the integral of the pressure, and add up all the pressures on this disc, to find the total force. However, the integral really does work out to be very, very close to the assumption of uniform pressure for water at over one diameter deep.

The integral works out to:

$$\rho g\pi r^2(h + \frac{r^2}{4h})$$

The last term is pretty small when r << 4h

So, you are on the right track with your method of force measurement. Don't worry about the variation of pressure with height - but realize that the force will try to rotate the lamp - so you can't think symmetry will cancel everything out.

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What i think is that since the floodlight is 25cm in diameter a distance of 12.25 Could also be subtracted to get a better approximaion since 25+30=55 is almost half of a meter and can make a difference.……………………………………………………… .You are going right though.

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    $\begingroup$ I also said the same thing $\endgroup$ – Sikander Jun 24 '15 at 17:54

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