Finally, now that we’ve gone over Archimedes’ principle, let’s touch on Pascal’s law for the remainder
of this hydrostatics discussion. Pascal’s law basically says that if you have some pressure in a fluid,
that pressure has got to be the same throughout the entire fluid. So, if I have a pressure in a system
like I’ve shown here, we might have one small piston, one small area and one very large area.
I can ask about the pressure at each location. The pressure by Pascal’s law is the same throughout
the entire liquid. So, the pressure at point 1 and the pressure at point 2 are the same.
What this means is the force per unit area, which is the definition of pressure is the same
at both of these locations. Using this simple principle, we can come up with a very great machine
because we can have equal forces per unit area in both sides. In other words, the force per unit area
in this small area where I’m pushing my hand will be equal to the force per unit area
of the large right hand side. Since the force per unit area is the same in both cases but the areas
are drastically different, where one area is very small and the other one is very big, that means
my force can be very, very small. I can get a great resultant force lifting up a very heavy object
in the place that has a much greater area. This is the basis of hydraulics. We use the difference in force
between two locations to gain a mechanical advantage by making one area greater than another
such that equal distribution of pressure will come to our advantage in applying a great force
to the greater area since the force per unit area is the same and we have a much greater area
on the right around which to distribute this force. We can solve for specifically force, F2,
just by rearranging Pascal’s principle here, Pascal’s law. By multiplying both sides, we see that we have
the force that’s lifting this big object like a house will be equal to whatever force I apply
times the ratio of the areas. So, if I make area 2 much, much greater than area 1, I can apply
a great force lifting this house. Let’s apply Pascal’s law that we just went through to a simple example
in which a boy puts his finger in a dam to try to hold back a huge amount of water.
Suppose the water is 20 meters high. We know the cross sectional area of the end of his finger
is half a square centimeter. We can ask how much force is on his finger from all this water.
Try this problem on your own again using Pascal’s principle, that the pressure throughout the liquid
is the same and we can find the hydrostatic pressure from the water. If you did this problem,
it hopefully looks something like this. We have a very high dam. So, you have a lot of water in here
with a height of 20 meters. At 20 meters, this is where we’re applying a force where we have a small hole
whose area is a half a square centimeter. The question is how much force, how much force
would be applied at this location. Well, we know that the pressure is something that we can find
down at the bottom of this. So, the pressure from the water is equal to ρgh. This is again
just the pressure from the water, the gauge pressure. This is something we can find because we know h
and we know the density of water. So, we have 1000. Again, this is in units of kilograms per cubic meter
times gravity which we’ll approximate as 10, so we can solve this problem very quickly
and then the height of 20 meters. Seeing all the factors of zero here, you can see that this is 2 times 10
to the 5th from 1, 2, 3, 4, 5 zeroes. This is a pressure so it will be in pascals. We could also write this
and it might often be written as 2 times 10 to the 2nd kilopascals or 200 kilopascals.
So, we know the pressure. We found the pressure. The question is if we know the pressure,
can we find the force? Well, this isn’t too bad because we know the definition of pressure
is the force per unit area. So, to find the force, we just need to multiply the pressure by whatever area
is relevant for this problem. We know that. We know that the area will be given by 0.5 centimeters squared.
So, let’s do this very quickly. We see that the pressure is 2 times 10 to the 5th pascals.
Now, we have to do something tricky which is multiply by an area but this area is not in meters.
So, we can’t put it directly with our units of meters. So, let’s change those units really quickly.
We have 1/2 or 0.5 centimeters squared. Let’s change those units by saying that there are 100 centimeters
in one meter and remembering that since this is squared, we need to square this quantity.
So now, we have the 0.5, let’s be consistent here and just call this 0.5, divided by four zeroes
because we have two zeroes here and then we square the quantity. So, this is times 10 to the minus 4.
This is now in squared meters. So, this is on a side to make sure that we get the area and the right units.
So, we’re ready to put that in here. So, we now have 0.5 times 10 to the minus 4. This is meter squared.
Pascals times meter squared will be units of force because we know that the force per unit area
is equal to pressure. 2 times 0.5 is just half of 2 which is 1. 10 to the five times 10 to the minus 4
leaves us with one factor of 10. Again, we have this now in terms of newtons, in terms of force.
So, this is the force on the boy’s finger, 10 newtons which is hard to tell whether that’s a lot or not
but it’s actually the weight of less than 1 kilogram since we know that the force, the gravitational force
of an object is mass times gravity. If gravity is approximately 10, this is the weight of again
less than 1 kilogram. So, it’s not terribly heavy but it might be slightly uncomfortable.