Finding an increase in depth, pressure and thrust

AI Thread Summary
A cylindrical tank with a radius of 4 cm is partially filled with water, and a solid metal sphere of radius 3 cm is submerged, prompting calculations for depth, pressure, and thrust. The volume of water before the sphere is added is calculated as 16πh, while the volume after immersion includes the sphere's volume, leading to the equation 16π(x+h) = 16πh + 36π. The increase in depth, x, is determined to be 2.25 cm. Pressure is calculated using the formula for pressure in terms of depth, yielding 22.5g Pa, and the thrust on the base is found to be 360πg N. The discussion emphasizes the importance of including gravitational acceleration in calculations and suggests using variable notation for clarity.
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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