A negligibly thin horizontal plate PQ has a length 3 m and width 1 m. It is being pulled along its length at a speed of 1 m/s in between two static parallel plates as shown in the figure. The gap of 6 cm between the plates is filled with a Newtonian fluid of dynamic viscosity \( \mu = 0.2 \, {N-s/m}^2 \). The thin plate is located at 3 cm from the top surface. The velocity distribution between the thin plate and the static plates is linear.
The steady force required to pull the plate is __________ N (answer in integer).
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In Couette flow problems, the velocity gradient is often assumed to be linear, making it easier to calculate the shear stress and the force required to move the plate.
Step 1: Identify the given parameters. The problem provides the following information: Length of the plate \( L = 3 \) m Width of the plate \( W = 1 \) m Speed of the moving plate \( v = 1 \) m/s Distance from the top static plate to the moving plate \( h_1 = 3 \) cm \( = 0.03 \) m Distance from the moving plate to the bottom static plate \( h_2 = 6 \) cm \( - 3 \) cm \( = 3 \) cm \( = 0.03 \) m Dynamic viscosity of the fluid \( \mu = 0.2 \, {N-s/m}^2 \) Step 2: Calculate the viscous force on the top surface of the moving plate. The velocity gradient between the top static plate (velocity 0) and the moving plate (velocity 1 m/s) is \( \frac{1 - 0}{0.03} = \frac{100}{3} \, {s}^{-1} \). The shear stress on the top surface is \( \tau_1 = 0.2 \times \frac{100}{3} = \frac{20}{3} \, {N/m}^2 \). The area of the top surface is \( A_1 = 3 \times 1 = 3 \, {m}^2 \). The viscous force on the top surface is \( F_1 = \tau_1 A_1 = \frac{20}{3} \times 3 = 20 \, {N} \). Step 3: Calculate the viscous force on the bottom surface of the moving plate. The velocity gradient between the moving plate (velocity 1 m/s) and the bottom static plate (velocity 0) is \( \frac{1 - 0}{0.03} = \frac{100}{3} \, {s}^{-1} \). The shear stress on the bottom surface is \( \tau_2 = 0.2 \times \frac{100}{3} = \frac{20}{3} \, {N/m}^2 \). The area of the bottom surface is \( A_2 = 3 \times 1 = 3 \, {m}^2 \). The viscous force on the bottom surface is \( F_2 = \tau_2 A_2 = \frac{20}{3} \times 3 = 20 \, {N} \). Step 4: Calculate the total steady force required. The total force required is the sum of the viscous forces: \( F_{{total}} = F_1 + F_2 = 20 \, {N} + 20 \, {N} = 40 \, {N} \).
