### Video Transcript

Water in a container is pushed
horizontally by a movable wall of the container, as shown in the diagram. The movable wall has sides of
length 𝐿 two equals 0.25 meters and 𝐿 three equals 0.75 meters. Within the water in the container
is a square metal sheet with sides of length 0.125 meters. The sheet is supported by a pole so
that all of its downward-facing surface is in contact with the liquid apart from a
small area that the pole fits through. The water pushes the sheet when the
container’s movable wall is moved. The force applied to the movable
wall is equal to the force applied to the water by the wall. What is the magnitude of the force
due to the motion of the movable wall that pushes upward on the downward-facing
surface of the metal sheet? Ignore the area of the hole where
the pole passes through the sheet.

Looking at our diagram, we see this
square horizontally oriented metal sheet submerged in water inside a container. One of the walls of the container,
this wall right here, is a movable wall, and we see that we’re exerting a force of
75 newtons on this wall. All of that force is transmitted
into the water and its contents. We’re told that as a result of this
force 𝐹 pushing on the movable wall, there is a force created that acts upward on
the downward-facing portion of our square metal sheet. We’ll call that force 𝐹 sub up,
and it’s that magnitude that we want to solve for.

Even though our problem is stated
in terms of forces, we can think of it at first in terms of pressure; that is, the
force spread out over this submerged portion of the movable wall creates a
pressure. That pressure is transmitted all
throughout the incompressible water in the container. One manifestation of that pressure
in the container is the generation of this force 𝐹 sub up. So let’s clear some space on screen
and start on our solution by solving for the pressure created at this movable
wall. Pascal’s principle tells us that
pressure is equal to force spread out over area. In our case, the pressure on this
wall is equal to 75 newtons divided by 𝐿 sub two times 𝐿 sub three. These are the dimensions of the
part of the movable wall that is submerged in water.

In our problem statement, we’re
told that 𝐿 two equals 0.25 meters and 𝐿 three is 0.75 meters. Substituting in these values, we
now have an equation we can solve for the pressure 𝑃 created by our force of 75
newtons. Before we do that though, let’s
realize that this pressure 𝑃 is perfectly transmitted all throughout the water in
the container. That means this pressure is exerted
on the underside of our square metal sheet and creates a force acting upward on
it. The key point here is that the
pressure created by the moving wall and the pressure exerted on the square metal
sheet are equal. Therefore, we can write this
equation that the pressure created by the moving wall is equal to the pressure on
the square metal sheet.

When it comes to the area of the
metal sheet, we know that this sheet is square. Its sides are the same length. Those side lengths are given to us
in the problem statement as 0.125 meters, so the area of the metal sheet is this
value squared. We can now rearrange this equation
to solve for 𝐹 sub up. Multiplying both sides by 0.125
meters quantity squared gives us this expression. When we go to calculate this force,
notice that the units of meters squared in the numerator and the units of meters
squared in the denominator will cancel. We’re left then just with units of
newtons in our final answer. This fraction equals exactly 6.25
newtons. This is the magnitude of the force
that pushes upward on the downward-facing surface of the metal sheet.

Let’s look now at part two of our
question.

What is the magnitude of the force
due to the motion of the movable wall that pushes downward on the upward-facing
surface of the metal sheet? Ignore the area of the hole where
the pole passes through the sheet.

Alright, so if this is a side-on
view of the square metal sheet — this is the sheet right here, and this is the pole
passing through it — then in part one of our question we calculated this force, we
call it 𝐹 sub up, acting upward on the downward-facing portion of the sheet. Now we want to do what’s in a sense
the opposite. We want to calculate the
downward-pointing force acting on the upward-facing surface of that metal sheet.

To figure this out, we actually
won’t need to do any calculations. That’s because, as we saw, the
pressure created by this movable wall on the fluid in the container spreads equally
throughout all that fluid. At any given point in the fluid,
say, this point right here, the pressure in all directions is the same. If the pressure were the same in
every direction, then there would be a net force on the fluid at that point, and it
would move.

Taking the same idea to our square
metal sheet, we can say that because this sheet is in equilibrium, because it’s not
in motion, we know that the upward-acting force acting on the downward-facing
portion of the sheet is the same as the downward-acting force acting on the
upward-facing portion. Those forces must be equal in
magnitude; otherwise, the sheet would start to accelerate. The magnitude of our
downward-acting force, then, is the same as the magnitude of our upward-acting
force, 6.25 newtons.