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A 20 mm wide gap between two vertical plane surfaces is filled with an oil of specific gravity 0.85 and dynamic viscosity 2.5 N.s/m². A metal plate 1.25 m x 1.25 m x 2 mm thick and weighing 30 N is placed midway in the gap. Determine the force required to lift the plate with a constant velocity of 0.18 m/s.?
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A 20 mm wide gap between two vertical plane surfaces is filled with an...
Problem Statement: Determine the force required to lift a metal plate with a constant velocity of 0.18 m/s from a 20 mm wide gap filled with oil of specific gravity 0.85 and dynamic viscosity 2.5 N.s/m². The metal plate has dimensions of 1.25 m x 1.25 m x 2 mm and weighs 30 N.

Solution:

Step 1: Calculation of the buoyancy force:
Buoyancy force is the force exerted by the fluid on the object immersed in it. It is given by:
Buoyancy force = Weight of the displaced fluid
Weight of the displaced fluid = Volume of the displaced fluid x Density of the fluid x Acceleration due to gravity
Volume of the displaced fluid = Area of the plate x Distance moved by the plate = 1.25 * 1.25 * 0.002 m³
Density of the fluid = Specific gravity x Density of water = 0.85 * 1000 kg/m³
Acceleration due to gravity = 9.81 m/s²
Buoyancy force = 1.25 * 1.25 * 0.002 * 0.85 * 1000 * 9.81 = 204.96 N

Step 2: Calculation of the viscous drag force:
Viscous drag force is the force exerted by the fluid on the object due to its viscosity. It is given by:
Viscous drag force = Dynamic viscosity x Velocity x Area of the plate / Distance between the plates
Viscosity of the fluid = 2.5 N.s/m²
Velocity of the plate = 0.18 m/s
Area of the plate = 1.25 * 1.25 = 1.5625 m²
Distance between the plates = 20 / 1000 = 0.02 m
Viscous drag force = 2.5 * 0.18 * 1.5625 / 0.02 = 70.3125 N

Step 3: Calculation of the force required to lift the plate:
Force required to lift the plate = Weight of the plate + Buoyancy force + Viscous drag force
Weight of the plate = 30 N
Force required to lift the plate = 30 + 204.96 + 70.3125 = 305.2725 N

Therefore, the force required to lift the metal plate with a constant velocity of 0.18 m/s is 305.2725 N.
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A 20 mm wide gap between two vertical plane surfaces is filled with an oil of specific gravity 0.85 and dynamic viscosity 2.5 N.s/m². A metal plate 1.25 m x 1.25 m x 2 mm thick and weighing 30 N is placed midway in the gap. Determine the force required to lift the plate with a constant velocity of 0.18 m/s.?
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A 20 mm wide gap between two vertical plane surfaces is filled with an oil of specific gravity 0.85 and dynamic viscosity 2.5 N.s/m². A metal plate 1.25 m x 1.25 m x 2 mm thick and weighing 30 N is placed midway in the gap. Determine the force required to lift the plate with a constant velocity of 0.18 m/s.? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A 20 mm wide gap between two vertical plane surfaces is filled with an oil of specific gravity 0.85 and dynamic viscosity 2.5 N.s/m². A metal plate 1.25 m x 1.25 m x 2 mm thick and weighing 30 N is placed midway in the gap. Determine the force required to lift the plate with a constant velocity of 0.18 m/s.? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 20 mm wide gap between two vertical plane surfaces is filled with an oil of specific gravity 0.85 and dynamic viscosity 2.5 N.s/m². A metal plate 1.25 m x 1.25 m x 2 mm thick and weighing 30 N is placed midway in the gap. Determine the force required to lift the plate with a constant velocity of 0.18 m/s.?.
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