# Question Video: Calculating the Dynamic Viscosity of a Fluid Physics

A thin plate of mass 2.5 g is pushed by a constant force 𝐹 = 0.50 mN,, moving at a constant speed over the surface of a viscous liquid that is 2.5 mN deep, as shown in the diagram. The speeds of the layers of the liquid between the top and bottom plates are shown in the diagram. The liquid in contact with the top and bottom plates moves at the same speed as the plates move. What is the dynamic viscosity of the liquid?

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### Video Transcript

A thin plate of mass 2.5 grams is pushed by a constant force 𝐹 equals 0.50 millinewtons, moving at a constant speed over the surface of a viscous liquid that is 2.5 millimeters deep, as shown in the diagram. The speeds of the layers of the liquid between the top and bottom plates are shown in the diagram. The liquid in contact with the top and bottom plates moves at the same speed as the plates move. What is the dynamic viscosity of the liquid?

Let’s start out by writing down the value of the force 𝐹 and then clearing space on screen to work. To find the dynamic viscosity, 𝜇, of this fluid, we’ll use the formula 𝜇 equals 𝐹 over 𝐴 times Δ𝑦 over Δ𝑣 sub 𝑥, where 𝐹 is the force applied on the top plate. 𝐴 is the area of that plate. Δ𝑦 is the height of each fluid layer. And Δ𝑣 sub 𝑥 is the change in speed between adjacent fluid layers.

We already know that the force on the top plate, 𝐹, equals 0.50 millinewtons. Let’s recall that the prefix milli- means 10 to the negative three. So we can write the force as 0.50 times 10 to the negative three newtons, which is equal to 5.0 times 10 to the negative four newtons. We’ve also been given the side lengths of the square top plate, so we can calculate its area by multiplying them together. Each side is 35 centimeters or 0.35 meters long, so the area 𝐴 equals 0.1225 meters squared.

Next, for Δ𝑦, we need to determine the height of each fluid layer. We were told that, in total, the fluid is 2.5 millimeters deep. From the diagram, we can count one, two, three, four, five different layers. So the height of each layer is given by 2.5 millimeters divided by five or 0.5 millimeters. Again recalling that milli- means 10 to the negative three, we have that Δ𝑦 equals 0.5 times 10 to the negative three or 5.0 times 10 to the negative four meters.

The last term we need in order to calculate the dynamic viscosity is Δ𝑣 sub 𝑥, the change in the speeds of any two adjacent layers. We can choose to calculate the change in speed between the second and third layers of fluid. So Δ𝑣 sub 𝑥 is given by 0.84 centimeters per second minus 0.60 centimeters per second, which equals 0.24 centimeters per second.

Before we move on though, let’s recall that centi- means 10 to the negative two. So, Δ𝑣 sub 𝑥 equals 0.24 times 10 to the negative two meters per second or 2.4 times 10 to the negative three meters per second.

Finally, we’re ready to substitute all these values into the formula to find 𝜇. Before we calculate though, it’s always a good idea to check out the units. Notice that this factor on the left has units of newtons per square meter. We can recall that this is equivalent to pascals, the SI-derived unit of pressure. So let’s make this substitution in the numerator. Next, in the factor on the right, we can see that the units of meters cancel from the numerator and denominator, leaving units of inverse seconds in the denominator, which is equivalent to just plain seconds in the numerator. Thus, the units associated with this entire expression are pascal seconds, which is a good sign, because those are the correct units for dynamic viscosity.

Now, plugging this into a calculator gives a result of 0.0008503 and so on pascal seconds. In scientific notation, that’s 8.503 times 10 to the negative four pascal seconds. And rounding this to one decimal place, we’ve reached our final answer. Thus, we’ve found that the fluid has a dynamic viscosity of 8.5 times 10 to the negative four pascal seconds.