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bsgdabest

A side wall of a wide open tank is provided with a narrowing tube through which water flows out, The cross sectional area of the tube decreases from S=3.0cm² to s=1.0cm²The water level in the tank is h=4.6m higher than that in the tube. Find the horizontal component of the force tending to pull the tube out of the tank. (neglect viscosity of water) ..this is from irodov..i am not able to solve it tried it many times..so please give the total solution without just giving the start..

anchitsaini

i think somebody else had also asked this question the other day
i had done this before(not on my own) but felt too tired to post the solution(it was a long one).

if nobody posts the soln i'll post the soln tonight

coolvignesh

hey anchit..
can u plz reply....
i'm not able to solve this on eout

anchitsaini

At A,B,C--

SV = S1u = sv (as volume flowing is same)
or

R²V =

y²u =

r²v ------------------1

where S,R,V are area of cross section, radius and velocity at point A
(As S=

R²)

similarly for B,C

and v² = 2gh

applying bernoullis theorem at B and C--

P_B +

u² / 2 =P_atmospheric +

v² / 2 ------------------2

horizontal component of force =

F=_{[ r ]}

^{[ R ]} (P_B - P_atmospheric )2

y dy

=_{[ r ]}

^{[ R ]} [

v² / 2 -

u² / 2] 2

y dy using 2

=

_{[ r ]}

^{[ R ]} (v² - u²)y dy

=

_{[ r ]}

^{[ R ]} v² ( 1 - r⁴/y⁴ )ydy using 1

=

v²_{[ r ]}

^{[ R ]} ( y - r⁴/y³ )dy

which on integration gives --

=

v²/2 * [y² + r⁴/y²] from r to R

=

v²/2 *

*(R² - r²)² / R² )

=

v²/2

²(R² - r²)² /

R²

=

v²/2 (S-s)² / S

=

gh(S-s)² / S as v² = 2gh

computer001

@ anchit
but how exactly does this force pull the tube outta the tank??

eistien

maybe the force with which the water rushes out is same as that the water exerts on the tube !!!

computer001

edited

coolvignesh

hey anchit...
thanx a lot yaar...
u're a genius!!!

coolvignesh

@ computer 001 and einstein
if you look at the tube... it's walls are oblique...
so, when water comes out, it exerts force on the walls normally..
that force is resolved to horizontal and vertical components...
the horizontal component is the force which pushes the tube out and vertical forces cancel out each other in this case..

computer001

edited:
my initial argument was wrong...
but @ coolvig:
no resolving of forces...
the total force itself will pull the pipe out...sorry for my initial mistake abt viscous force etc.

computer001

the part abt pulling the tube out..
it is very similar to force of water placed in a conical vessel...in tht case force due to water @ each pt has to be resolved so as to get the net force downwards..
in this case the force need not be resolved cuz the net force in this case itself is similar to the resolved downward force in case of the conical vessel mentioned above

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