Irodov 1.327: Reaction from a Draining Slit – Amit & Deepali Sinha Foundation

Problem Irodov 1.327

A tall, wide, vertical cylindrical vessel of height $h = 75$ cm is filled with water while a narrow vertical slit in its side wall, running from the base up to a height $l = 50$ cm and of width $b = 1.0$ mm, is kept closed. Find the resultant force of reaction from the water jetting out through the whole slit at the instant it is opened.

The wide vertical vessel, filled to a height h, with the narrow slit of length l running up from the base in one side wall. — The wide vertical vessel, filled to a height $h$ , with the narrow slit of length $l$ running up from the base in one side wall.

Given

$h = 75$ cm
$l = 50$ cm
$b = 1.0$ mm

Find

Find

F

, the resultant reaction force immediately after the slit opens

Solution

Step 1Speed at a point along the slit

Measure $y$ upward from the base, with the slit occupying $0\le y\le l$ and the free surface at height $h$ . The vessel is wide, so at the instant the slit opens (before the level has time to fall) the depth of water above height $y$ is $h-y$ , and by Torricelli’s law

v(y) = \sqrt{2g(h-y)}. \tag{1}

Step 2Reaction from an element of the slit

The efflux across the slit: each arrow's length follows sqrt(2g(h-y)), fastest at the base, slowest at the slit's top. The elemental reactions, one per element dy at height y, are all parallel and add to a single force F pointing away from the slit. — The efflux across the slit: each arrow’s length follows $\sqrt{2g(h-y)}$ , fastest at the base, slowest at the slit’s top. The elemental reactions, one per element $dy$ at height $y$ , are all parallel and add to a single force $\vect F$ pointing away from the slit.

An element of height $dy$ has area $dA=b\,dy$ and carries momentum flux $\rho v(y)^2\,dA$ horizontally outward. By Newton’s third law the vessel feels a reaction of the same magnitude, directed horizontally back into the vessel. Every element’s reaction points the same way (perpendicular to the wall the slit is cut in), so the resultant is a simple sum; substituting (1) gives the elemental reaction as a function of $y$ alone:

\begin{align*}dF &= \rho v(y)^2\,b\,dy \\ &= \rho\bigl[2g(h-y)\bigr]b\,dy \\ &= 2\rho g b (h-y)\,dy.\end{align*}

Step 3Integrate over the slit

Summing every element’s reaction from the bottom of the slit ( $y=0$ ) to its top ( $y=l$ ) gives the total force.

\begin{align*}F &= \int_0^l 2\rho g b (h-y)\,dy \\ &= 2\rho g b\left[hy-\frac{y^2}{2}\right]_0^l \\ &= 2\rho g b\left(hl-\frac{l^2}{2}\right).\end{align*}

F = \rho g b l(2h-l)

(2)

Step 4Evaluate

With $\rho=1.00\times10^3$ kg/m $^3$ , $b=1.0\times10^{-3}$ m, $l=0.50$ m and $h=0.75$ m, so $2h-l=1.00$ m,

F = (1.00\times10^3)(9.8)(1.0\times10^{-3})(0.50)(1.00)\ \text{N}.

F = \rho gbl(2h-l) \approx 5\ \text{N}

(3)

Check

If the slit is narrow compared with its depth, $l\ll h$ , $F\approx 2\rho gbhl$ , matching a single narrow orifice of area $bl$ at depth $h$ carrying $\rho v^2 (bl) = \rho(2gh)(bl)$ : the full integral correctly reduces to the point-orifice result in that limit.

Remark

Writing $2h-l$ as $2(h-l/2)$ shows $F$ is exactly what a single compact orifice of area $bl$ would give sitting at the slit’s own mean depth $h-l/2$ . That works because $v(y)^2$ is linear in $y$ over the slit, so its average over the strip equals its value at the midpoint depth; a whole distributed slit collapses to one equivalent point orifice whenever the driving quantity varies linearly along it.