A vertical gap 2.2 cm wide of infinite extent contains a fluid of viscosity 2.0 N s/m² and specific gravity 0.9. A metallic plate 1.2 m × 1.2 m × 0.2 cm is to be lifted up with a constant velocity of 0.15 m/s through the gap. If the plate is in the middle of the gap, find the force required. The weight of the plate is 40 N.

Force to Lift a Plate in Viscous Fluid

Problem Statement

Given Data

Total Gap Width: $2.2 \, \text{cm} = 0.022 \, \text{m}$
Fluid Viscosity, $\mu = 2.0 \, \text{N s/m}^2$
Fluid Specific Gravity, $S.G. = 0.9$
Plate Dimensions: $1.2 \, \text{m} \times 1.2 \, \text{m} \times 0.2 \, \text{cm}$
Plate Thickness: $0.2 \, \text{cm} = 0.002 \, \text{m}$
Plate Velocity, $u = 0.15 \, \text{m/s}$
Weight of Plate, $W_p = 40 \, \text{N}$

Solution

1. Determine the Gap on Each Side of the Plate

The plate is centered in the gap. The thickness of the fluid film, $dy$, on each side is:

$$ dy = \frac{\text{Total Gap Width} - \text{Plate Thickness}}{2} $$ $$ dy = \frac{0.022 \, \text{m} - 0.002 \, \text{m}}{2} = \frac{0.020 \, \text{m}}{2} $$ $$ dy = 0.01 \, \text{m} $$

2. Calculate the Total Shear Force

The shear force acts on both sides of the plate. First, we find the shear stress $\tau$ using Newton's law of viscosity, assuming a linear velocity profile.

$$ \tau = \mu \frac{du}{dy} $$ $$ \tau = 2.0 \, \frac{\text{N s}}{\text{m}^2} \times \frac{0.15 \, \text{m/s}}{0.01 \, \text{m}} = 30 \, \text{N/m}^2 $$

The surface area $A$ of one side of the plate is:

$$ A = 1.2 \, \text{m} \times 1.2 \, \text{m} = 1.44 \, \text{m}^2 $$

The shear force on one side, $F_{s1}$, is $\tau \times A$. Since the gap is the same on both sides, the total shear force $F_{s, \text{total}}$ is twice the force on one side.

$$ F_{s, \text{total}} = 2 \times (\tau \times A) $$ $$ F_{s, \text{total}} = 2 \times (30 \, \text{N/m}^2 \times 1.44 \, \text{m}^2) $$ $$ F_{s, \text{total}} = 2 \times 43.2 \, \text{N} = 86.4 \, \text{N} $$

3. Calculate the Buoyant Force (Upward Thrust)

The buoyant force is the weight of the fluid displaced by the plate. First, find the plate's volume $V_p$ and the fluid's weight density $\gamma_f$.

$$ V_p = 1.2 \, \text{m} \times 1.2 \, \text{m} \times 0.002 \, \text{m} = 0.00288 \, \text{m}^3 $$ $$ \gamma_f = S.G. \times \gamma_{\text{water}} = 0.9 \times 9810 \, \text{N/m}^3 = 8829 \, \text{N/m}^3 $$

The buoyant force $F_b$ is:

$$ F_b = \gamma_f \times V_p $$ $$ F_b = 8829 \, \text{N/m}^3 \times 0.00288 \, \text{m}^3 \approx 25.43 \, \text{N} $$

4. Determine the Total Force Required

The total force $F_{\text{req}}$ needed to lift the plate must overcome the downward plate weight $W_p$ and the downward viscous drag (total shear force $F_{s, \text{total}}$), while being assisted by the upward buoyant force $F_b$.

$$ F_{\text{req}} = F_{s, \text{total}} + W_p - F_b $$ $$ F_{\text{req}} = 86.4 \, \text{N} + 40 \, \text{N} - 25.43 \, \text{N} $$ $$ F_{\text{req}} = 100.97 \, \text{N} $$

Final Result:

The total force required to lift the plate is $ F_{\text{req}} = 100.97 \, \text{N} $.

Explanation of Forces

To lift the plate at a constant velocity, the upward applied force must perfectly balance all downward forces, while accounting for any upward-assisting forces. The forces involved are:

1. Viscous Drag (Shear Force):
As the plate moves, the fluid resists this motion, creating a shear stress on the plate's surfaces. This results in a downward drag force on both sides of the plate. Since the plate is centered, the drag force is equal on both sides.

2. Weight of the Plate:
The plate's own weight (given as 40 N) acts vertically downward due to gravity.

3. Buoyant Force:
According to Archimedes' principle, the fluid exerts an upward buoyant force on the plate equal to the weight of the fluid it displaces. This force assists in lifting the plate.

The final required force is therefore the sum of the two downward forces (drag and weight) minus the upward-assisting buoyant force.

Physical Meaning

This problem illustrates the interplay between three fundamental concepts in fluid mechanics:

1. Viscous Drag: The calculation demonstrates how a fluid's viscosity creates resistance to moving objects. This drag is critical in designing everything from lubricated bearings to aerodynamic vehicles.

2. Buoyancy: The upward thrust from the fluid reduces the net weight of the submerged object. This principle is essential for the flotation of ships and the operation of submarines.

3. Force Equilibrium: For the plate to move at a constant velocity (i.e., with zero acceleration), all forces acting on it must be in equilibrium. The calculation shows how to sum these vector forces to find the required external force to maintain this state of motion.