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

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AI Thread Summary
The discussion revolves around calculating the tension in a tethered bomb submerged 4 meters below the ocean surface. The participant initially miscalculates the force by considering the weight of a water column instead of focusing on the buoyant force. They are guided to apply Archimedes' principle to find the buoyant force acting on the bomb. The correct approach involves using the buoyant force formula and setting up an equation for net force to solve for tension. Ultimately, the tension can be determined by balancing the buoyant force against the weight of the bomb.
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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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