Question

A side wall of a wide open tank is provided with a narrowing tube (Fig. 1.88) through which water flows out. The cross-sectional area of the tube decreases from S = 3.0 cm² to s = 1.0 cm². The water level in the tank is h = 4.6 m 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.

Answer 1

Suppose the radius at A is R and it decreases uniformaly to r at B where S = πR² and s = πr² . Assume also that the semi-vertical angle at 0 is α. Then

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

where P_p is the pressure at P and P_o is the atmospheric pressure which is the pressure just outside of B. The force on the nozzle tending to pull it out is then

We have subtracted P_o which is the force due to atmosphenic pressure the factor sinθ gives horizontal component of the force and ds is the length of the element of nozzle surface, ds = dx sec θ and 

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.