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A wooden pole 25 mm X 25 mm X 4 m attached to a steel weight at the bottom floats in sea water so that 0.6 m length of pole is exposed. Calculate the steel weight attached. Sp. Gravity of sea water, wood and steel may be taken as 1.025, 0.60 and 7.85 respectively.?
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A wooden pole 25 mm X 25 mm X 4 m attached to a steel weight at the bo...
Calculation of Steel Weight Attached to a Wooden Pole in Sea Water
Given:
- Dimensions of wooden pole: 25 mm X 25 mm X 4 m
- Length of pole exposed: 0.6 m
- Specific Gravity of sea water: 1.025
- Specific Gravity of wood: 0.60
- Specific Gravity of steel: 7.85
To find:
- Steel weight attached to the wooden pole
Approach:
- Determine the volume of the wooden pole submerged in water
- Calculate the weight of the wooden pole submerged in water
- Calculate the buoyant force acting on the wooden pole
- Determine the volume of the steel weight required to counteract the buoyant force
- Calculate the weight of the steel weight required
Calculation:
1. Volume of wooden pole submerged in water:
- Cross-sectional area of wooden pole = 25 mm X 25 mm = 625 mm² = 0.000625 m²
- Volume of wooden pole submerged in water = 0.6 m X 0.000625 m² = 0.000375 m³
2. Weight of wooden pole submerged in water:
- Density of wood = Specific Gravity of wood X Density of water
- Density of water = 1000 kg/m³ (at 4°C)
- Density of wood = 0.60 X 1000 = 600 kg/m³
- Weight of wooden pole submerged in water = Density of wood X Volume of wooden pole submerged in water X Acceleration due to gravity
- Weight of wooden pole submerged in water = 600 kg/m³ X 0.000375 m³ X 9.81 m/s² = 2.21 N
3. Buoyant force acting on the wooden pole:
- Buoyant force = Weight of water displaced
- Volume of water displaced = Volume of wooden pole submerged in water
- Density of sea water = Specific Gravity of sea water X Density of water
- Density of sea water = 1.025 X 1000 = 1025 kg/m³
- Weight of water displaced = Density of sea water X Volume of wooden pole submerged in water X Acceleration due to gravity
- Weight of water displaced = 1025 kg/m³ X 0.000375 m³ X 9.81 m/s² = 3.78 N
- Buoyant force acting on the wooden pole = Weight of water displaced = 3.78 N
4. Volume of steel weight required to counteract the buoyant force:
- Density of steel = Specific Gravity of steel X Density of water
- Density of steel = 7.85 X 1000 = 7850 kg/m³
- Volume of steel weight required = Buoyant force / Density of steel / Acceleration due to gravity
- Volume of steel weight required = 3.78 N / 7850 kg/m³ / 9.81 m/s² = 0.000000048 m³
5. Weight of steel weight required:
- Weight of steel weight required = Density of steel X Volume of steel weight required X Acceleration due to gravity
- Weight of steel weight required = 7850 kg/m³ X 0.000000048 m³ X 9.81 m/s² = 0.000036 N
Therefore, the steel weight attached to the wooden
Given:
- Dimensions of wooden pole: 25 mm X 25 mm X 4 m
- Length of pole exposed: 0.6 m
- Specific Gravity of sea water: 1.025
- Specific Gravity of wood: 0.60
- Specific Gravity of steel: 7.85
To find:
- Steel weight attached to the wooden pole
Approach:
- Determine the volume of the wooden pole submerged in water
- Calculate the weight of the wooden pole submerged in water
- Calculate the buoyant force acting on the wooden pole
- Determine the volume of the steel weight required to counteract the buoyant force
- Calculate the weight of the steel weight required
Calculation:
1. Volume of wooden pole submerged in water:
- Cross-sectional area of wooden pole = 25 mm X 25 mm = 625 mm² = 0.000625 m²
- Volume of wooden pole submerged in water = 0.6 m X 0.000625 m² = 0.000375 m³
2. Weight of wooden pole submerged in water:
- Density of wood = Specific Gravity of wood X Density of water
- Density of water = 1000 kg/m³ (at 4°C)
- Density of wood = 0.60 X 1000 = 600 kg/m³
- Weight of wooden pole submerged in water = Density of wood X Volume of wooden pole submerged in water X Acceleration due to gravity
- Weight of wooden pole submerged in water = 600 kg/m³ X 0.000375 m³ X 9.81 m/s² = 2.21 N
3. Buoyant force acting on the wooden pole:
- Buoyant force = Weight of water displaced
- Volume of water displaced = Volume of wooden pole submerged in water
- Density of sea water = Specific Gravity of sea water X Density of water
- Density of sea water = 1.025 X 1000 = 1025 kg/m³
- Weight of water displaced = Density of sea water X Volume of wooden pole submerged in water X Acceleration due to gravity
- Weight of water displaced = 1025 kg/m³ X 0.000375 m³ X 9.81 m/s² = 3.78 N
- Buoyant force acting on the wooden pole = Weight of water displaced = 3.78 N
4. Volume of steel weight required to counteract the buoyant force:
- Density of steel = Specific Gravity of steel X Density of water
- Density of steel = 7.85 X 1000 = 7850 kg/m³
- Volume of steel weight required = Buoyant force / Density of steel / Acceleration due to gravity
- Volume of steel weight required = 3.78 N / 7850 kg/m³ / 9.81 m/s² = 0.000000048 m³
5. Weight of steel weight required:
- Weight of steel weight required = Density of steel X Volume of steel weight required X Acceleration due to gravity
- Weight of steel weight required = 7850 kg/m³ X 0.000000048 m³ X 9.81 m/s² = 0.000036 N
Therefore, the steel weight attached to the wooden
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A wooden pole 25 mm X 25 mm X 4 m attached to a steel weight at the bottom floats in sea water so that 0.6 m length of pole is exposed. Calculate the steel weight attached. Sp. Gravity of sea water, wood and steel may be taken as 1.025, 0.60 and 7.85 respectively.?
Question Description
A wooden pole 25 mm X 25 mm X 4 m attached to a steel weight at the bottom floats in sea water so that 0.6 m length of pole is exposed. Calculate the steel weight attached. Sp. Gravity of sea water, wood and steel may be taken as 1.025, 0.60 and 7.85 respectively.? 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 wooden pole 25 mm X 25 mm X 4 m attached to a steel weight at the bottom floats in sea water so that 0.6 m length of pole is exposed. Calculate the steel weight attached. Sp. Gravity of sea water, wood and steel may be taken as 1.025, 0.60 and 7.85 respectively.? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wooden pole 25 mm X 25 mm X 4 m attached to a steel weight at the bottom floats in sea water so that 0.6 m length of pole is exposed. Calculate the steel weight attached. Sp. Gravity of sea water, wood and steel may be taken as 1.025, 0.60 and 7.85 respectively.?.
A wooden pole 25 mm X 25 mm X 4 m attached to a steel weight at the bottom floats in sea water so that 0.6 m length of pole is exposed. Calculate the steel weight attached. Sp. Gravity of sea water, wood and steel may be taken as 1.025, 0.60 and 7.85 respectively.? 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 wooden pole 25 mm X 25 mm X 4 m attached to a steel weight at the bottom floats in sea water so that 0.6 m length of pole is exposed. Calculate the steel weight attached. Sp. Gravity of sea water, wood and steel may be taken as 1.025, 0.60 and 7.85 respectively.? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A wooden pole 25 mm X 25 mm X 4 m attached to a steel weight at the bottom floats in sea water so that 0.6 m length of pole is exposed. Calculate the steel weight attached. Sp. Gravity of sea water, wood and steel may be taken as 1.025, 0.60 and 7.85 respectively.?.
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A wooden pole 25 mm X 25 mm X 4 m attached to a steel weight at the bottom floats in sea water so that 0.6 m length of pole is exposed. Calculate the steel weight attached. Sp. Gravity of sea water, wood and steel may be taken as 1.025, 0.60 and 7.85 respectively.?, a detailed solution for A wooden pole 25 mm X 25 mm X 4 m attached to a steel weight at the bottom floats in sea water so that 0.6 m length of pole is exposed. Calculate the steel weight attached. Sp. Gravity of sea water, wood and steel may be taken as 1.025, 0.60 and 7.85 respectively.? has been provided alongside types of A wooden pole 25 mm X 25 mm X 4 m attached to a steel weight at the bottom floats in sea water so that 0.6 m length of pole is exposed. Calculate the steel weight attached. Sp. Gravity of sea water, wood and steel may be taken as 1.025, 0.60 and 7.85 respectively.? theory, EduRev gives you an
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