Problem Statement
Two large plane surfaces are 2.4 cm apart. The space between the surfaces is filled with glycerine. What force is required to drag a very thin plate of surface area 0.5 square metre between the two large plane surfaces at a speed of 0.6 m/s, if:
(i) the thin plate is in the middle of the two plane surfaces, and
(ii) the thin plate is at a distance of 0.8 cm from one of the plane surfaces?
Take the dynamic viscosity of glycerine = \(8.10 \times 10^{-1} \, \text{N s/m}^2\).
Given Data
- Distance between two large surfaces, \(Y = 2.4 \, \text{cm} = 0.024 \, \text{m}\)
- Area of thin plate, \(A = 0.5 \, \text{m}^2\)
- Velocity of thin plate, \(u = 0.6 \, \text{m/s}\)
- Viscosity of glycerine, \( \mu = 8.10 \times 10^{-1} \, \text{N s/m}^2 = 0.81 \, \text{N s/m}^2\)
Solution
Case I: When the thin plate is in the middle of the two plane surfaces
In this case, the distance from the thin plate to each surface is equal.
1. Determine distances from surfaces
2. Calculate Shear Stress on the upper side (\(\tau_1\))
Using Newton's law of viscosity:
Here, \(du = u = 0.6 \, \text{m/s}\) (relative velocity between plate and stationary upper surface).
$$ \tau_1 = 0.81 \, \text{N s/m}^2 \times \left( \frac{0.6 \, \text{m/s}}{0.012 \, \text{m}} \right) $$ $$ \tau_1 = 0.81 \times 50 = 40.5 \, \text{N/m}^2 $$3. Calculate Shear Force on the upper side (\(F_1\))
4. Calculate Shear Stress on the lower side (\(\tau_2\))
Similarly, for the lower side:
5. Calculate Shear Force on the lower side (\(F_2\))
6. Calculate Total Force (F) for Case I
The total force required to drag the plate when it is in the middle is \(40.5 \, \text{N}\).
Case II: When the thin plate is at a distance of 0.8 cm from one of the plane surfaces
Let's assume the plate is 0.8 cm from the lower surface.
1. Determine distances from surfaces
Distance from lower surface, \(dy_2 = 0.8 \, \text{cm} = 0.008 \, \text{m}\)
Distance from upper surface, \(dy_1 = Y - dy_2 = 2.4 \, \text{cm} - 0.8 \, \text{cm} = 1.6 \, \text{cm} = 0.016 \, \text{m}\)
2. Calculate Shear Stress on the upper side (\(\tau_1\))
3. Calculate Shear Force on the upper side (\(F_1\))
4. Calculate Shear Stress on the lower side (\(\tau_2\))
5. Calculate Shear Force on the lower side (\(F_2\))
6. Calculate Total Force (F) for Case II
The total force required to drag the plate when it is 0.8 cm from one surface is approximately \(45.56 \, \text{N}\).
Explanation
1. Principle of Shear Stress:
The problem involves calculating the shear force exerted by a viscous fluid on a moving plate. This force arises from the fluid's resistance to shear deformation, governed by Newton's law of viscosity, \(\tau = \mu \frac{du}{dy}\).
2. Velocity Gradient:
The velocity gradient (\(\frac{du}{dy}\)) is crucial. It represents how rapidly the fluid's velocity changes across the gap. Since the plate is thin and moving at a constant speed, and the large surfaces are stationary, the velocity profile in each gap is assumed to be linear.
3. Force Calculation:
The total force required to drag the plate is the sum of the shear forces acting on its upper and lower surfaces. Each shear force is calculated by multiplying the shear stress by the plate's surface area.
4. Effect of Plate Position:
The position of the plate significantly affects the required force. When the plate is closer to one surface (Case II), the velocity gradient in that smaller gap becomes steeper, leading to a higher shear stress and thus a larger shear force on that side. This imbalance results in a greater total force compared to when the plate is centered (Case I).
Physical Meaning
1. Viscous Drag:
The force calculated represents the viscous drag experienced by the plate as it moves through the glycerine. This drag is a direct consequence of the fluid's viscosity, which resists the relative motion between fluid layers.
2. Importance of Gap Size:
The results highlight the inverse relationship between shear stress and the gap thickness. A smaller gap leads to a larger velocity gradient, resulting in a higher shear stress and consequently a greater force required to maintain the plate's speed. This principle is fundamental in lubrication and fluid damping systems.
3. Energy Dissipation:
The work done to drag the plate against this viscous force is converted into heat, representing energy dissipation within the fluid due to internal friction. This is a common phenomenon in fluid mechanics, particularly in situations involving shear flow.
4. Design Implications:
Understanding how plate position affects drag force is vital in engineering design, such as in the design of bearings, hydraulic actuators, and other machinery where components move relative to each other through a viscous fluid. Minimizing drag can improve efficiency and reduce wear.