A side wall of a wide open tank is provided with a narrowing tube (figure) through which water flows out. The cross-sectional area of the tube decreases from `S=3.0cm^2` to `s=1.0cm^2`. The water level in the tank is `h=4.6m` higher than that in the tube. Neglecting the viscosity of the water, find the horizontal component of the force tending to pull the tube out of the tank.

Explanation

Suppose the radius at A is R and it decreases uniformly to r at B where `S=piR^2` and `s=piR^2`. Assume also that the semi vectical angle at 0 is `alpha`. Then
`(R)/(L_2)=(r)/(L_1)=y/x`
So `y=r+(R-r)/(L_2-L_1)(x-L_1)`
where y is the radius at the point P distant x from the vertex O. Suppose the velocity with which the liquid flows out is `V` at A, v at B and u at P. Then by the equation of continuity
`piR^2V=pir^2v=piy^2u`
The velocity v of efflux is given by
`v=sqrt(2gh)`
and Bernoulli's theorem gives
`p_p+1/2rhou^2=p_0+1/2rhov^2`
where `p_p` is the pressure at P and `p_0` is the atmospheric pressure which is the pressure just outside of B. The force on the nozzle tending to pull it out is then
`F=int(p_(rho)-p_0)sintheta2piyds`
We have subtracted `p_0` which is the force due to atmospheric pressure the factor `sintheta` gives horizontal component of the force and `ds` is the length of the element of nozzle surface, `ds=dxsectheta` and
`tantheta=(R-r)/(L_2-L_1)`
Then `F=underset(L_1)overset(L_2)int 1/2(v^2-u^2)rho2piy(R-r)/(L_2-L_1)dx`
`=pirhounderset(r)overset(R)intv^2(1-(r^4)/(y^4))ydy`
`=pirhov^2 1/2(R^2-r^2+(r^4)/(R^2)-r^2)=rhogh((pi(R^2-r^2)^2)/(R^2))`
`=rhogh(S-s)^2//S=6*02N` on putting the values.
Note: If we try to calculate F from the momentum change of the liquid flowing out will be wrong even as regards the sign of the force.
There is of course the effect of pressure at S and s but quantitative derivation of F from Newton's law is difficult.