A conical cup of height b , semi-vertical angle α rests open end down on a flat surface as shown. The cup is filled to height H with liquid of density ρ and a small hole is punched at the apex of the cone. The upward lifting force on the cup is F = π ρ g [ b H 2 − n H 3 ] tan 2 α , then ' n ' is
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Can we do this in the following way: Force= Weight of water -(weight of water in the cylinder of radius (b-H)tan(alpha) and height H)
I did the question in the same way but the coefficient of H^3 was coming as 2/3 .
Can you please help??? @Aniket Sanghi
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Nope , as on the beside region also base exerts force which you have to take into account
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So,there is no way to avoid integration ?
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@A Former Brilliant Member – In my solution there is no integration!
Force = h d g . π r 2 − V d g
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@Aniket Sanghi – Can you please post your solution?....I am having difficulties understanding it
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@A Former Brilliant Member – I just now saw , the person wants us to neglect atmospheric pressure .
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@Aniket Sanghi – Thank you!!!!! And sorry for disturbing you so often.
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@A Former Brilliant Member – Never mind ! :) ! Always Welcome ! :)
@A Former Brilliant Member – I guess now it's clear?
@Aniket Sanghi Was there any need to neglect Atmospheric Pressure, it just cancels out.
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Firstly we have to neglect atmospheric pressure .
Then consider FBD of water .
Forces acting on it are :
net downward force by vertical walls.
Force of gravity
Pressure force exerted By base .
Now , force exerted by walls
F = PA - mg = ( H ρ g ) ( π b 2 t a n 2 α ) − 3 ( π ( b 3 − ( H − b ) 3 ) ( t a n 2 α ) ρ g )