Two large parallel plates are separated by a small gap of distance 2 mm. See figure below. A fluid with dynamic viscosity µ = 10-3 Pa.s fills the gap between the plates. The surface area of each side of the plates is 2 m2. The top plate has a velocity 2.0 m/s in the
x-direction, while the bottom plate has velocity 0.7 m/s in the x-direction. A negative velocity signifies a velocity in the negative x-direction.
Determine the magnitude of the shear force (Fτ) on each plate in Newtons.
To determine the magnitude of the shear force on each plate, we can use the equation for shear force:
Fτ = μ * A * (du/dy)
Where:
- Fτ is the shear force
- μ is the dynamic viscosity of the fluid
- A is the surface area of each plate
- du/dy is the velocity gradient in the direction perpendicular to the plates
In this case, the velocity gradient can be calculated as (u2 - u1) / (y2 - y1), where:
- u2 is the velocity of the bottom plate (0.7 m/s)
- u1 is the velocity of the top plate (2.0 m/s)
- y2 is the distance from the bottom plate to the top plate (2 mm or 0.002 m)
- y1 is the distance from the top plate to the bottom plate (0)
Substituting the values into the formula, we get:
Fτ = (10^-3 Pa.s) * (2 m^2) * ((0.7 m/s - 2.0 m/s) / (0.002 m))
Simplifying the equation, we have:
Fτ = -0.65 N
Therefore, the magnitude of the shear force on each plate is 0.65 N.