- #1

ValeForce46

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- Homework Statement
- A container is made up of a vertical cylinder, diameter ##D=9.0 cm##, on which is placed an horizontal tube, diameter ##d=3.0 cm##, with axis at a distance ##l=5.0 cm## from the bottom of the cylinder. The other end of the horizontal tube is plugged and the container is filled with water until the height of ##h=50 cm## (see picture). Assuming that the surface is perfectly smooth (no friction), determine the value of the necessary force to keep the container still when the plug has been removed. [Advice: the lowering speed of the column of water should NOT be neglected.]

- Relevant Equations
- Bernoulli's Principle: ##p+ρgz+\frac{1}{2}ρv^2=constant##

I'll call ##v_2## the flow speed on the horizontal tube and ##v_1## the lowering speed of the column.

Even if the lowering speed isn't negligible, the problem says "...

__when__the plug has been removed", so can I consider the height unchanged?

In order to find the force, I need the pressure when the plug is off. Is it ##p'=p_{atm}+ρg(h-l)+\frac{1}{2}ρv_2^2## ?

First thing I know is ##v_1*A_1=v_2*A_2## (##A_1## and ##A_2## are the two section).

Second thing I know is ##p_{atm}+\frac{1}{2}ρv_2^2=p_{atm}+ρg(h-l)+\frac{1}{2}ρv_1^2## thanks to Bernoulli's principle.

So I can find ##v_2##.

The force should be

$$F=p'*A=[p_{atm}+ρg(h-l)+\frac{1}{2}ρv_2^2]*(\frac{D}{2})^2*π $$

Is something wrong?