Solving Archimedes Problem: Rod in Pool

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SUMMARY

The problem involves a 6-meter rod with a specific gravity of 25/36, tied to a 5-meter rope in a 10-meter deep pool. The solution reveals that 1 meter of the rod remains above water when in equilibrium. Key equations include the buoyancy force equating to the tension in the rope plus the weight of the rod. Analyzing the rod as a free body and applying static equilibrium principles leads to the conclusion regarding the rod's position in the water.

PREREQUISITES
  • Understanding of buoyancy principles and Archimedes' principle
  • Knowledge of specific gravity and its implications on floating objects
  • Familiarity with static equilibrium and free body diagrams
  • Basic algebra for solving equations involving forces and densities
NEXT STEPS
  • Study the principles of buoyancy and Archimedes' principle in detail
  • Learn how to create and analyze free body diagrams for static systems
  • Explore the concept of specific gravity and its applications in fluid mechanics
  • Practice solving equilibrium problems involving multiple forces and tensions
USEFUL FOR

Students studying physics, particularly those focusing on fluid mechanics and statics, as well as educators seeking to enhance their understanding of buoyancy and equilibrium concepts.

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Homework Statement


A rod 6 meters in length has specific gravity 25/36. One end of the rod is tied to a 5 meter rope, which in turn is attached to the floor of a pool 10 meters deep. Find the length of the part of rod, which is out of water.

The answer is 1 meter.

Homework Equations


mg=V(imm)p(l)g=V(rod)p(rod)g

F(of buoyancy)=Tension + mg

The Attempt at a Solution


I don't really know what to do, but here's what I did anyway.

I tried using V(imm)*density of fluid=V(rod)*density of rod. But then I thought that there should be a tension component in the force equation as well. I further tried ignoring tension, and using the above mentioned equation to figure out how much of the rod would be outside. I got 2meters. I figured I could use trig to figure out the angle of inclination with the water level, but I couldn't really use it.
 
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erisedk said:

Homework Statement


A rod 6 meters in length has specific gravity 25/36. One end of the rod is tied to a 5 meter rope, which in turn is attached to the floor of a pool 10 meters deep. Find the length of the part of rod, which is out of water.

The answer is 1 meter.

Homework Equations


mg=V(imm)p(l)g=V(rod)p(rod)g

F(of buoyancy)=Tension + mg

The Attempt at a Solution


I don't really know what to do, but here's what I did anyway.

I tried using V(imm)*density of fluid=V(rod)*density of rod. But then I thought that there should be a tension component in the force equation as well. I further tried ignoring tension, and using the above mentioned equation to figure out how much of the rod would be outside. I got 2meters. I figured I could use trig to figure out the angle of inclination with the water level, but I couldn't really use it.

Hint: Analyze the rod as a free body. Since you know the density of the rod (or specific gravity, in this case), you should be able to determine if the rod floats. If the rod is capable of floating, you should be able to write equations of statics using the end tied to the rope as your reference. Knowing other facts, like the depth of the pool, you should be able to calculate how much of the rod sticks out of the pool when it is in equilibrium.
 

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