Irodov 1.326: Recoil from Two Opposite Jets – Amit & Deepali Sinha Foundation

Problem Irodov 1.326

A wide vertical vessel filled with water has two identical small holes, one in each of two opposite walls, at heights differing by $\Delta h = 51$ cm, each hole of cross-sectional area $S = 0.50$ cm $^2$ . Both holes drain water simultaneously. Find the resultant force of reaction exerted on the vessel by the two escaping jets.

Two holes of equal area S, one in each of two opposite walls of the filled vessel, each a gap in its own wall, their heights differing by Δ h. — Two holes of equal area $S$ , one in each of two opposite walls of the filled vessel, each a gap in its own wall, their heights differing by $\Delta h$ .

Given

$S = 0.50$ cm $^2$
$\Delta h = 51$ cm

Find

Find

F

, the resultant reaction force from the two jets

Solution

Step 1Efflux speed from each hole

Let the free surface sit at height $H$ above the base, and let the upper hole (labeled $1$ ) and lower hole (labeled $2$ ) sit at heights $z_1$ and $z_2$ , with $z_1-z_2=\Delta h$ . The vessel is wide, so each speed follows Torricelli’s law from the depth of water above that hole,

v_1 = \sqrt{2g(H-z_1)}, \qquad v_2 = \sqrt{2g(H-z_2)}. \tag{1}

Step 2The speed-squared difference reduces to

\Delta h

alone

The two depths of water are related directly to the height difference between the holes,

(H-z_2)-(H-z_1) = z_1-z_2 = \Delta h.

Squaring each speed in (1) and subtracting carries this same difference straight through, so the speed difference below depends on $\Delta h$ alone:

\begin{align*}v_2^2-v_1^2 &= 2g(H-z_2)-2g(H-z_1) \\ &= 2g\bigl[(H-z_2)-(H-z_1)\bigr]\end{align*}

v_2^2-v_1^2 = 2g\Delta h

(2)

Step 3Reaction force from each jet, and their resultant

The same vessel with the two jets drawn: the lower hole 2 carries the faster jet v_2, the arrows drawn in that proportion, and the resultant reaction F pushes the vessel toward the upper hole's side. — The same vessel with the two jets drawn: the lower hole $2$ carries the faster jet $v_2$ , the arrows drawn in that proportion, and the resultant reaction $\vect F$ pushes the vessel toward the upper hole’s side.

Each jet carries momentum flux $\rho S v_i^2$ away from the vessel, so by Newton’s third law the vessel feels a reaction of that same magnitude directed opposite its own jet, i.e. horizontally back into the vessel. The two holes are on opposite walls, so the two reactions are collinear: the weaker jet ( $1$ , upper) pushes the vessel toward the lower hole’s side, while the stronger jet ( $2$ , lower) pushes it back toward the upper hole’s side, harder. The net force is therefore the difference of the two momentum fluxes,

F = \rho S v_2^2 - \rho S v_1^2 = \rho S(v_2^2-v_1^2), \tag{3}

directed toward the side of the upper (weaker) hole.

Step 4Substitute the speed difference

Substituting (2) into (3),

F = \rho S \cdot 2g\Delta h.

F = 2\rho g S\Delta h

(4)

Step 5Evaluate

With $\rho=1.00\times10^3$ kg/m $^3$ , $S=0.50\times10^{-4}$ m $^2$ and $\Delta h=0.51$ m,

F = 2(1.00\times10^3)(9.8)(0.50\times10^{-4})(0.51)\ \text{N}.

F = 2\rho gS\Delta h \approx 0.50\ \text{N}

(5)

Check

If $\Delta h\to0$ the two jets carry equal speed and the two reactions cancel exactly, $F\to0$ , as they must for two identical holes at the same depth.

Remark

$F$ depends only on the hole separation $\Delta h$ , not on $H$ or the holes’ absolute depths, so the whole problem could equally well be posed with the vessel only half full, or brimming: sliding both holes up or down together by the same amount never changes the answer, since neither hole’s own depth below the free surface ever enters the formula.