A container with water, of total mass m , stands on a…

A container with water, of total mass $m$, stands on a horizontal surface and moves at constant velocity because water flows out through a small opening in its side. The water level is $h$ above the opening, and the opening has area $S$. What is the coefficient of friction between the container and the surface? Assume $h$ stays constant.

Explanation of Method

A container pushed by its own water jet.

Step 1: Outflow speed

\[ v = \sqrt{2gh} \ \text{(Torricelli)}, \]

\[ \dot m = \rho S v. \]

Step 2: Reaction force

\[ F = \dot m\,v = \rho S v^2 = 2\rho S g h. \]

Step 3: Constant velocity

\[ 2\rho S g h = \mu m g. \]

\[\mu = \frac{2\rho S h}{m}\]

$\rho S v^2 = 2\rho S g h$ $\mu m g$ $h$ $v = \sqrt{2gh}$

The jet leaving to the left pushes the container to the right with $\rho S v^2$. At constant velocity this reaction balances friction.

Result

\[\boxed{\,\mu = \frac{2\rho S h}{m}\,}\]

Remark

Check. The original result is correct. The reaction force equals the weight of a water column of height $2h$ over the area $S$: twice what the static pressure alone would suggest.