What is the pressure at 80 m below the surface of the ocean.

AI Thread Summary
At a depth of 80 meters below the ocean surface, the pressure can be calculated using the formula p = P0 + ρGd, resulting in approximately 8.74 atm when using the correct seawater density. The initial calculation yielded 8.97 atm due to an incorrect assumption about the seawater density. For the tension in the rope, the buoyant force and the weight of the iron ball were calculated, leading to confusion about the correct values. The discussion highlighted the variability in seawater density and the importance of using the appropriate value for accurate results. Ultimately, clarification on density allowed the original poster to solve both parts of the problem correctly.
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Homework Statement



a) A sailor uses a rope to lower an iron ball of radius 30 cm a to a depth of 80 m below the surface of the ocean. What is the pressure (in atm) at that depth? b) Find the tension in the rope in the previous problem. The density of iron is 7.86 *103 Kg/m3.

Homework Equations



For part a) p=P0+ρGd

1 atm= 1.013*105 Pa

ρsea water= 1.03*103 kg/m3

For part b) Ʃy= F buoyant-mg-T=0

F buoyant=ρVg

ρ= m/V

The Attempt at a Solution



In part a) p=(1.013*105 Pa)+(1.03*103 kg/m3)(9.8 m/s2)(80m)= 9.08*105 Pa

9.08*105 Pa / 1.013*105 Pa = 8.97 atm

However, the answer is 8.74 atm what did I do wrong?? :confused::cry::frown:

In part b) I solved for T=F buoyant-mg

F buoyant= (7.86 *103 )(4π(0.3m)3/3)(9.8m/s2)= 8.70*103 N*m

For mg I used m=ρV = (7.86 *103 )(4π(0.3m)3/3)= 8.70*103 N*m

So T= 8.70*103 N*m - 8.70*103 N*m =0 ? The answer for part (b) is 7600 N :confused::cry::rolleyes: Help?
 
Last edited:
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You used 1030 kg/m3 for the density of the seawater. Could they have wanted you to use the density of water with no salt (1000 kg/m3)? The density of seawater varies quite a bit, depending on how much salt is in that part of the ocean. If they didn't specify the density of the water other than saying that it's "the ocean," I would guess they want you to use the standard density for water.
 
You were right! THANKS MAN! I could solve both of them! thanks again!
 

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