How Is Tension Calculated in a Submerged Object's Lifting Cable?

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SUMMARY

The tension in the lifting cable of a submerged object, such as a sealed safe with an average density of 5.3 grams/cc and a mass of 120 kg, is calculated using the principles of buoyancy and forces acting on the object. The buoyant force, which is equal to the weight of the displaced fluid, must be considered alongside the weight of the object. The correct formula involves calculating the pressure exerted by the fluid, which is derived from the density, gravity, and height of the object submerged. A free-body diagram is essential to visualize the forces acting on the safe.

PREREQUISITES
  • Understanding of buoyancy principles
  • Knowledge of free-body diagrams
  • Familiarity with basic physics equations related to force and pressure
  • Concept of density and its calculation
NEXT STEPS
  • Learn about Archimedes' principle and its applications in fluid mechanics
  • Study the calculation of buoyant force in various fluid scenarios
  • Explore the concept of net force and equilibrium in submerged objects
  • Investigate the relationship between density, volume, and mass in physical objects
USEFUL FOR

Students studying physics, particularly those focusing on fluid mechanics, engineers working with submerged structures, and anyone involved in maritime operations or crane mechanics.

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Homework Statement



A seagoing crane lifts a sealed safe from a sunken ship. The average density of the safe is 5.3 grams/cc and its mass is 120 kg. What is the tension in the lifting cable when the safe is still submerged?

Homework Equations



Pressure = density *gravity* height*
Buoyant force = mg

The Attempt at a Solution



I calculated 5300kg/m^2 * 9.81 m/s^2 * 120 kg

The above is what I got so far but I'm not sure I'm doing it right. Any help would be greatly appreciated.
 
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Draw a free-body diagram of the object. From the context of the problem, there must be no net force acting on the object. What forces act in which directions?

Note: the buoyant force is equal to the weight of the DISPLACED fluid, not necessarily equal to the weight of the submerged object.
 

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