A fire hydrant delivers water of density $\rho$ at a volume rate$L$. The water travels vertically upwards through the hydrant and then does $90{^\circ}$ turn to emerge horizontally at speed$v$. The pipe and nozzle have uniform cross-section throughout. The force exerted by water on the corner of the hydrant is [image:10_image32.emf]

Question

Dr. Abubacker · Accepted Answer

(c) In time $\mathrm{\Delta}t$, momentum of water entering the hydrant
${\overrightarrow{\mathbf{p}}}_{1} = (\rho\ L\mathrm{\Delta}t)v\widehat{\mathbf{j}}$
Momentum of water while leaving the hydrant in time $\mathrm{\Delta}t$ is ${\overrightarrow{\mathbf{p}}}_{2} = (\rho\ L\mathrm{\Delta}t)v( - \widehat{\mathbf{i}}\mathbf{)}$
Change in momentum in time $\mathrm{\Delta}t$ is
$\mathrm{\Delta}\ \overrightarrow{\mathbf{p}} = {\overrightarrow{\mathbf{p}}}_{2} - {\overrightarrow{\mathbf{p}}}_{1} = \rho\ L\mathrm{\Delta}tv\ ( - \widehat{\mathbf{i}} - \widehat{\mathbf{j}})$
$\left| \mathrm{\Delta}\overrightarrow{\mathbf{p}} \right| = \rho\ L\ \mathrm{\Delta}tv\ \sqrt{( - 1)^{2} + ( - 1)^{2}}$
$= \sqrt{2}\ \rho\ L\mathrm{\Delta}tv$
Force exerted by water, $F = \frac{|\mathrm{\Delta}\ \overrightarrow{\mathbf{p}}|}{\mathrm{\Delta}t} = \sqrt{2}\ \rho Lv$

Explanation

Reviewed by Dr. Abubacker

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