### Video Transcript

A thin plate of area 0.25 square meters is pushed horizontally by a constant force of 150 micronewtons across the surface of water that has a dynamic viscosity of 8.9 times 10 to the negative four pascal seconds. What is the rate of change of the speed of the water beneath the plate with the vertical distance from it?

Let’s say that this sketch shows our thin plate being pushed horizontally while on top of a series of layers of water. Due to this force, we’ll call it 𝐹, of 150 micronewtons, the plate will move to the right, as we’ve drawn it, with some speed. And this will cause the layers of water beneath the plate to move to the right as well. The farther away a given layer of water is from the plate vertically, we’ll call this distance Δ𝑦, the greater the difference between the speed of that layer of water and the speed of the thin plate, we’ll call this difference Δ𝑣, will be.

Our question asks us to solve for the rate of change of the speed of the water beneath the plate with the vertical distance from the plate. We can symbolize that rate as Δ𝑣 divided by Δ𝑦. This rate of change, Δ𝑣 divided by Δ𝑦, actually shows up in an equation we can recall. This equation says that the shear stress applied to some material, represented by the letter 𝜏, is equal to that material’s dynamic viscosity, sometimes just called viscosity for short represented by 𝜇, multiplied by the rate of change of speed of this material over the distance away from where the shear stress is applied.

A shear stress 𝜏 is a force that’s spread out over an area. Importantly, this is not the same as a pressure, which would be a force acting perpendicularly to some area, but rather it’s a force that acts across the surface of the given area. In our case, we have a shear stress acting on the water because our force acts on a thin plate which moves across the surface of the water.

Combining these equations that both involve shear stress, we can write that the shear stress, a shearing force spread out over an area, is equal to the viscosity of the material involved multiplied by Δ𝑣 divided by Δ𝑦. As we’ve seen, in our case, Δ𝑣 divided by Δ𝑦 is the rate that we want to solve for. Since that’s so, let’s divide both sides of our equation by the dynamic viscosity. That value then cancels on the right. Switching the left and right sides of our equation, we now have that Δ𝑣 divided by Δ𝑦 equals 𝐹 over 𝜇 times 𝐴.

The force 𝐹 is 150 micronewtons. The water’s dynamic viscosity is given as 8.9 times 10 to the negative four pascal seconds. And the area of our thin plate is 0.25 square meters. Before we calculate this fraction, we’ll want to convert the units of our numerator from micronewtons into newtons. We can recall that one micronewton equals 10 to the negative six or one one millionth of a newton. And therefore 150 micronewtons equals 150 times 10 to the negative six newtons.

Before we calculate this fraction, let’s notice that the units in the numerator are now newtons, while those in the denominator are pascal seconds meter squared. If we recall that a pascal is defined as a newton per meter squared, then we see that we could rewrite newtons per meter squared as simply pascals so that the units of pascals will cancel out in numerator and denominator. Therefore, the only units left in this expression when we calculate it will be inverse seconds.

These might seem like strange units, but let’s consider that we’re calculating a change in speed, which could have units, say, of meters per second, divided by a change in position, which could, say, have distance units of meters. If we multiplied numerator and denominator of these units by one over meters, then we would find that the unit of meters cancels out entirely, and we’re left just with units of one over seconds or inverse seconds. So then with our unit of inverse seconds on the right-hand side, it looks like we’re on the right track.

To two decimal places, this fraction is equal to 0.67 inverse seconds. This is the rate of change of the speed of the water beneath the plate with the vertical distance from the plate.