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

[mailmas]

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?

 

[haruspex]

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?

 

[mailmas]

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?

 

[haruspex]

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?

 

[mailmas]

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?

 

[haruspex]

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.