Finding an increase in depth, pressure and thrust

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Homework Help Overview

The problem involves a cylindrical tank partially filled with water and the immersion of a solid metal sphere. Participants are exploring the effects of the sphere's addition on the depth of water, pressure at the base, and thrust on the base, within the context of fluid mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the volume of water before and after the sphere is added, questioning the calculations related to the volume and the resulting depth increase. There are attempts to derive the new depth and pressure based on the volume changes.

Discussion Status

Some participants have provided calculations and suggested methods for determining the increase in depth and pressure. There is an ongoing examination of the correctness of the volume equations and the conversion of units. Multiple interpretations of the problem setup are being explored, with no explicit consensus reached.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can use or the methods they can apply. There is also a mention of using variables for the tank and sphere radii to facilitate unit checking.

Woolyabyss
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Homework Statement


A cylindrical tank of radius 4cm is partly filled with water.A solid metal sphere of radius 3cm is lowered into the water by means of a thin wire until it is totally immersed. Find the increase in
(i) the depth
(ii) the pressure at a point on the base
(iii) the thrust on the base(in terms of π )


Homework Equations


Pressure = (depth)(g)(density)

The Attempt at a Solution


(i)
h = depth of water before sphere added

Volume of water before sphere added = 16(π )(h)

volume after sphere added = 16(π)(h) + (4/3)(π)(64)

pressure = (1)gh (since the density of water is 1/cm^3)

I'm not sure what to do next I'd presume I have to get the depth(h) somehow?
 
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Woolyabyss said:
1. volume after sphere added = 16(π)(h) + (4/3)(π)(64)



Check that second term on the right ...
 
Oh sorry V = 16πh + (4/3)π(27) = 16πh + 36π

I'm still not sure how to get the height
 
rude man said:
Check that second term on the right ...

Would this be correct?
(i)
V = volume after sphere is added

V = 16πh +36π

but V also = 16π(x+h) ... where x is the additional depth the water has risen)

16π(x+h) = 16πh + 36π

16πx + 16πh = 16πh +36π ... 16πh cancels

16πx = 36π

x = 36/16 =2.25 cm

(ii) convert to kg/m^3 and meters so our answer will be in pascals

1000g(0.0225+h) - 1000gh = 1000g(0.0225) +1000gh - 1000gh = 22.5g Pa

(iii) F = 22.5g(16π) = 360πg N

(these are the answers my book gives)
 
Last edited:
Woolyabyss said:
Would this be correct?
(i)
V = volume after sphere is added

V = 16πh +36π

but V also = 16π(x+h) ... where x is the additional depth the water has risen)

16π(x+h) = 16πh + 36π

16πx + 16πh = 16πh +36π ... 16πh cancels

16πx = 36π

x = 36/16 =2.25 cm

(ii) convert to kg/m^3 and meters so our answer will be in pascals

1000g(0.0225+h) - 1000gh = 1000g(0.0225) +1000gh - 1000gh = 22.5g Pa

(iii) F = 22.5g(16π) = 360πg N

(these are the answers my book gives)

OK but you should include g = 9.8 in your answer.

I also advise using Rt and Rs for radius of the tank and sphere instead of the number, until the very end. That way you can check units term-by-term as you go along.
 

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