Finding an increase in depth, pressure and thrust

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
4 replies · 2K views
Woolyabyss
Messages
142
Reaction score
1

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?
 

Discussion

Physics news on Phys.org
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.