Question

A plate of area 2m^(2) is made to move horizontally with a speed of 2m//s by applying a horizontal tangential force over the free surface of a liquid. If the depth of the liquid is 1m and the liquid in contact with the bed is stationary. Coefficient of viscosity of liquid is 0.01 poise. Find the tangential force needed to move the plate.

Answer 1

Tangential force needed to move the plate = μA(Δv/Δx)

where μ = coefficient of viscosity = 0.01 P = 0.01 * 0.1 Pa. s = 0.001 Pa. s

          A = area = 2 $m^2$

          (Δv/Δx) = velocity gradient = $\frac{2-0}{1-0}$ = 2

Tangential force needed to move the plate = 0.001 * 2 * 2 = 0.004 N

Answer 2

Hence the tangential force needed to move the plate is |F| = 4 × 10^3 N

Explanation:

Velocity gradient=Δv / Δy = 2−0 / 1−0 = 2 m / s / m

From, Newton's law of viscous force.

|F| = ηA Δv / Δy

|F| =(0.01 × 10^−1) (2) (2)  

|F| = 4 × 10^3 N

So, to keep the plate moving,a force of 4 × 10^3 N must be applied.

Hence the tangential force needed to move the plate is |F| = 4 × 10^3 N