Absolute pressure, buoyant force, tension problem

AI Thread Summary
The discussion revolves around calculating the absolute pressure, buoyant force, and tension in a cable for a submerged diving sphere. The absolute pressure at a depth of 0.8000 km was correctly calculated as approximately 8.145E+6 Pa. However, the initial calculation of the buoyant force was incorrect; the correct approach involves using the sphere's volume to determine the displaced water volume. The tension in the cable was calculated correctly, but participants emphasized the importance of clarity in unit representation. Overall, the calculations highlight the application of fundamental physics principles, including Archimedes' principle and geometric considerations for volume.
jenn91890
Messages
2
Reaction score
0

Homework Statement


A diving sphere has a mass of 150,000.0kg and an external diameter of 7.000 meters. The sphere is anchored with a cable on the bottom of an ocean, at a depth of .8000 km. The density of sea water is 1025.0kg/m3.
A) What is the absolute pressure at that depth in pascals?
B) What is the buoyant force felt by the sphere when it is completely submerged?
C) Calculate the tension of the cable.

Homework Equations


A) P = Ps+DgH
P=?
Ps=1.01E+5Pa
D=1.025E+3kg/m3
G=9.81m/s2
H=.8000km

B) P=F/A, A=pi R2
P=(from above)
BF=?
R=3.5m

C) WF=m*g
TF+WF=BF
m=150,000kg
g=9.81m/s2
TF=?
BF= (from above)

The Attempt at a Solution


A) P=(1.01E+5) + [(1.025E+3)*(9.81)*(800m)]
=8.145E+6Pa

B) 8.145E+6Pa = BF / (pi * 3.5)2
BF=3.135E+8N

C) WF=150000*9.81
WF=1471500
TF=313500000-1471500
TF=3.120E+8N

Is this correct? The numbers seem very high...
 
Last edited:
Physics news on Phys.org
Hello and welcome to PF!

Is the depth .8000 hm or .8000 km?

Your method for part (a) is correct. For part (b) you have not calculated the BF correctly. Consider Archimedes' principle.

Your approach to part (c) looks correct.
 
TSny said:
Hello and welcome to PF!

Is the depth .8000 hm or .8000 km?

Your method for part (a) is correct. For part (b) you have not calculated the BF correctly. Consider Archimedes' principle.

Your approach to part (c) looks correct.
km, I have corrected the typo.

for part (b) density=mass/displaced volume
so displaced volume = mass/density
so displaced volume = 150,000/1025
so displaced volume = 146.341
BF = density * displace volume * gravity
BF=1025 * 146.341 * 9.81
=1.471E+6
?
 
jenn91890 said:
so displaced volume = mass/density
so displaced volume = 150,000/1025
so displaced volume = 146.341

No, this isn't correct. 150,000 kg is the mass of the diving sphere, but 1025 kg/m3 is the density of sea water.

Try to think of another way to get the displaced volume. Hint: geometry.
 
Last edited:
jenn91890 said:
km, I have corrected the typo.

for part (b) density=mass/displaced volume
so displaced volume = mass/density
so displaced volume = 150,000/1025
so displaced volume = 146.341
BF = density * displace volume * gravity
BF=1025 * 146.341 * 9.81
=1.471E+6
?
Your calculations become a lot more intelligible if you state the units after each result. It's a habit you should cultivate now.
 
What is the volume of a sphere 7 m in diameter?

Chet
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top