Let’s do one more example. We have a diver who descends too quickly to allow his ears to equilibrate.
Let’s suppose that we know the area of an eardrum is 100 square millimeters. How deep could this diver
go before the water contacting his eardrum exceeds 3 newtons of force pushing his eardrum inwards?
We could also say if the internal sinus pressure remained at 1 atmosphere because of the speed
of his descent. In other words, the sinuses that we have right now are at an equilibrated pressure.
So, if you went down too quickly, that sinus pressure might not have time to change.
So, give this problem a shot and see what you come up with. Now, we’re going to try it here.
If you did this problem, what you should see is that we have air from the sinuses inside the eardrum
while water is pushing on the eardrum from the outside. It’s going to be at some pressure.
So, what we know is that the pressure will be equal to the force per unit area where this is the maximum
allowable force that’s given to us in this problem divided by the area. In this case, it’s the area
of the eardrum. We also know that the pressure is equal to ρgh from the water.
We can solve this for h since we’re trying to find the depth that this person could go to.
So, h will be equal to the maximum force divided by the area of the ear times the density of the water
times the gravitational acceleration. So, let’s solve this and see what we get. We have 3 newtons,
it was what we were asked for, divided by the area of the ear. So, this is something that’s tricky.
We need the cross sectional area of the ear but we can’t use this 100 square millimeters
because what we need to do is convert these units to be in units of meter squared.
So, doing this very quickly, we have 1 meter in 1000 millimeters. Again, we have to square this
whole quantity to make sure that our units of millimeters will properly cancel. So doing this,
we have 100 in the numerator. The square millimeters units will cancel. We have three zeroes here
being squared so we have six zeroes in the denominator. So, this is over 10 to the 6th.
In other words, this is 1 times 10 to the minus 4 and now, we have it in square meters.
So, this unit analysis thing becomes very important especially when we’re going into the fluids chapter.
So, plugging in this for the area, we now have 10 to the minus four, 1 times 10 to the minus four,
same thing. The density of water which again is 1000 kilograms per cubic meter.
Just writing those units there because we want to make sure we’re using the right density for water.
Then we have g and we can approximate this as 10. Again, for efficiency's sake in finding
the right answer quickly. Then we have simply three divided by many factors of zero.
We have 3 times 10 or let’s just do the whole thing. If 3, 10 to the minus 4 is competing with
these four zeroes, in other words, we have 10 to the minus 4 times 10 to the 4 from 1000 times 10.
So, these will go away to be just 1. Our answer, since we used SI units everywhere
and we’re careful to do so will just be 3 meters, so a little more than 10 feet.
So, what we see in this problem is that if you dive very quickly before your sinuses have any chance
to equilibrate, you will experience a few newtons of force acting on your eardrums
before anything has a chance to equilibrate already at 3 meters down, around 10 feet.
So, this is an example of how we can think about gauge pressure and pressures competing
and pressures per unit area. We’ve talked about Archimedes’ law and a few laws of basic hydrostatics,
as well as Pascal’s principle which gave us the basics for hydraulics, as well as a way to understand
that the pressure throughout an entire fluid is always going to be the same.
Next time, we’re going to get into some ideal hydrodynamics and then some more
applied hydrodynamics and some more complicated and more realistic physical scenarios.
Thanks for watching.