What is the tension in the tether of a bomb tethered to the ocean floor?

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Homework Help Overview

The problem involves a bomb tethered to the ocean floor, specifically focusing on calculating the tension in the tether. The scenario includes a 300 kg sphere submerged 4.00 m below the surface, with considerations for buoyancy and the forces acting on the object.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of pressure equations and buoyancy principles. There is an exploration of the relationship between the forces acting on the submerged object, including the weight force and buoyant force. Questions arise regarding the relevance of initial calculations and the correct application of Archimedes' principle.

Discussion Status

The discussion is active, with participants questioning the initial approach and clarifying concepts related to buoyancy and tension. Some guidance has been provided regarding the correct application of Archimedes' principle and the formulation of the net force equation.

Contextual Notes

Participants are navigating the complexities of buoyancy and tension in a submerged context, with some confusion about the initial calculations and their relevance to the problem at hand.

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Homework Statement


A bomb is tethered to a cable on the ocean floor. Imagine one is a 300 kg sphere with a radius of 0.600 m. If it is 4.00 m below the surface and the tether is low mass enough to be negligible, what is the tension in the tether?

Homework Equations


Pbot = Ptop + density*g*h
P = F/A
Acircle = (pi)(r)^2

The Attempt at a Solution



Pbot = 101300 + (1030)(9.81)(4)
Pbot = 141717.2 Pa
F/A = 141717.2 Pa
F = 141717.2(pi)(.6)^2
= 160197 N
Is this the force I am looking for?
 
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mailmas said:
Is this the force I am looking for?
No. That would be the weight of matter in a column radius 0.6m extending from 4m below the surface of the water, all the way to the top of the atmosphere!

You have heard of Archimedes, right?
 
haruspex said:
No. That would be the weight of matter in a column radius 0.6m extending from 4m below the surface of the water, all the way to the top of the atmosphere!

You have heard of Archimedes, right?
Yeah but honestly don't really understand it. So the force I found is the force which acts upwards to an object that is being submerged? Then I should look at the weight force and the difference between those forces must be the force due to tension?
 
mailmas said:
the force I found is the force which acts upwards to an object that is being submerged?
No, forget the force you found. It is not relevant to the question.
Can you state Archimedes' principle?

Edit: just noticed you correctly applied it to find the buoyant force in another thread.
So what is the buoyant force here?
 
haruspex said:
No, forget the force you found. It is not relevant to the question.
Can you state Archimedes' principle?

Edit: just noticed you correctly applied it to find the buoyant force in another thread.
So what is the buoyant force here?
FBuoyancy = density*volume*g
(1030)(4/3)(pi)(.6)^3(9.81)

Then Fnety = 0 = FBuoyancy - Tension - mg
Could I solve for tension this way?
 
mailmas said:
FBuoyancy = density*volume*g
(1030)(4/3)(pi)(.6)^3(9.81)

Then Fnety = 0 = FBuoyancy - Tension - mg
Could I solve for tension this way?
Yes.
 
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