A plate of metal 1.1mx1.1mx2mm is to be lifted up with a velocity of 0.1m/s through an infinitely extending gap 20mm wide containing an oil of sp. gr. 0.9 and viscosity 2.1NS/m2. Find the force required to lift the plate assuming the plate to remain midway in the gap. Assume the weight of the plate to be 30N.

Problem Statement

A metal plate with the following dimensions is being lifted through an oil-filled gap:

Plate dimensions: 1.1m × 1.1m × 2mm
Velocity of the plate: 0.1m/s
Specific gravity of oil: 0.9
Viscosity of oil: 2.1 Ns/m²
Gap width: 20mm (clearance on each side = 9mm = 0.009m)
Weight of plate: 30N

Determine the force required to lift the plate.

Solution

1. Calculate the Specific Weight of Oil (\(\gamma\))

The specific weight (\(\gamma\)) of oil is: \[ \gamma = \text{Specific Gravity} \times \gamma_{\text{water}} \] \[ \gamma = 0.9 \times 9810 \] \[ \gamma = 8829 \text{ N/m}^3 \]

2. Calculate the Upthrust on the Plate

The upthrust (buoyant force) is given by: \[ F_{\text{buoyant}} = \gamma \times \text{Volume of Plate} \] \[ = 8829 \times (1.1 \times 1.1 \times \frac{2}{1000}) \] \[ = 8829 \times 0.00242 \] \[ = 21.37 \text{ N} \]

3. Calculate the Viscous Force Acting on the Plate

The viscous force is given by: \[ F_{\text{viscous}} = 2 \times \mu \times \frac{du}{dy} \times A \] \[ = 2 \times 2.1 \times \frac{0.1}{0.009} \times (1.1 \times 1.1) \] \[ = 2 \times 2.1 \times 11.11 \times 1.21 \] \[ = 56.47 \text{ N} \]

4. Calculate the Total Force Required

\[ F_{\text{total}} = \text{Weight} – \text{Upthrust} + \text{Viscous Force} \] \[ = 30 – 21.37 + 56.47 \] \[ = 65.1 \text{ N} \]

Final Result:

Force required to lift the plate: 65.1N

Explanation

1. Components of Force:
The total force required to lift the plate consists of three main forces:

The weight of the plate (30N)
The upthrust (buoyant force) exerted by the oil, which reduces the effective weight of the plate (21.37N)
The viscous force resisting the motion of the plate, caused by the oil’s viscosity (56.47N)

2. Viscous Resistance:
The plate is moving through an oil-filled gap, and the oil’s viscosity resists this motion. This resistance is determined by the dynamic viscosity (\(\mu\)) of the oil and the velocity gradient (\(du/dy\)). Since the plate is centered in the gap, there are two contributions to the viscous force, one from each side of the plate.

3. Effect of Upthrust:
The oil provides a buoyant force due to the displacement of fluid, reducing the effective weight of the plate. This means that the force required to lift the plate is lower than if it were in air.

4. Importance of Understanding Forces in Fluids:
– This type of calculation is used in engineering applications where components move through viscous fluids, such as in lubrication systems and hydraulic machinery.
– Understanding viscous drag is essential in fluid dynamics for optimizing industrial processes and mechanical designs.

Physical Meaning

1. Engineering Applications:
– Used in lubrication analysis in mechanical systems.
– Helps in designing hydraulic machinery that involves movement through viscous fluids.

2. Industrial and Real-World Uses:
– Helps in understanding resistance in submerged structures.
– Applied in pipeline and filter design where fluid resistance is a key factor.