The plate in Fig below rests on top of the thin film of wate

The plate in Fig. below rests on top of the thin film of water, which is at a temperature of 25 C. When a small force Fis applied to the plate, the velocity profile across the thickness of the fluid can be described as u= (40y - 800) m/s, where y is in meters. Determine the shear stress acting on the fixed surface and on the bottom of the plate. The viscosity of water at 25 C is = 0.897 x 10 ^-3 N s/m² .

Solution

Shear stress is given by product of viscosity and velocity gradient per Newton\'s law of viscosity

in the given problem, u=40y-800

du/dy=d/dy(40y-800)=40

Shear stress = 40*0.897*10^-3=0.03588 Pa

Shear stress at fixed surface=shear stress at bottom of plate=0.03588 Pa