Consider a large square plate with sides length l = 1 meter moving on a thin film of oil of thickness h, l≫ℎ ( h=0.0001meters). if the plate is moved with velocity u = 0.1 m/s, and the viscosity of the oil is μ= 0.968/kgm. s(i. e., sae 30), and you ignore any edge effects. what is the viscous shear stress τyx and drag force (d= τyx. area )?

The viscous shear stress is 968 Pa (or can be written as 968 ) and the drag force is 968 N.

Explanation:

Viscous shear stress

For a parallel flow of a Newtonian fluid, the shear stress is proportional to the gradient of the velocity,

Considering a large square plate moving on a thin film of thickness h, the velocity profile is

Thus its derivative will be just

So replacing on the viscous shear stress formula we get

We can then replace the given information

Evaluating we get

The viscous shear stress is 968 Pascals or .

Drag force

Using the given equation for the drag force we have

And since we have a large square plate of sides L we can write the area as

So the drag force is

Replacing values

We get

The drag force is 968 N.