A cylindrical sewer pipe of radius $\frac { 5 } { 2 }$ feet is half full of water of density 62.5 pounds per cubic foot. A gate used to seal off the sewer is placed perpendicular to the pipe opening. The force, in pounds, due to hydrostatic pressure on one end of the gate is

A) $\frac { 1,125 } { 8 }$
B) $\frac { 500 } { 3 }$
C) $1,125$
D) $\frac { 15,625 } { 24 }$
E) $\frac { 125 } { 24 }$

Question

A cylindrical sewer pipe of radius

\frac { 5 } { 2 }

feet is half full of water of density 62.5 pounds per cubic foot. A gate used to seal off the sewer is placed perpendicular to the pipe opening. The force, in pounds, due to hydrostatic pressure on one end of the gate is

A) $\frac { 1,125 } { 8 }$
B) $\frac { 500 } { 3 }$
C) $1,125$
D) $\frac { 15,625 } { 24 }$
E) $\frac { 125 } { 24 }$

Quizplus · Accepted Answer

The force due to hydrostatic pressure on the gate can be calculated using the formula for hydrostatic pressure $P = \rho g h$, where $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity (which is approximately $32 \text{ ft/s}^2$ but often cancels out or is included in the density for these types of problems), and $h$ is the depth of the fluid. Since the pipe is half full, the depth $h$ is half the diameter, or $\frac{5}{2}$ feet. The area $A$ of the circular end of the pipe that the gate covers is $\pi r^2$, where $r = \frac{5}{2}$ feet. The force $F$ is then $P \times A$, or $\rho g h A$. Given $\rho = 62.5 \text{ lb/ft}^3$, and substituting the values, we find the force due to the hydrostatic pressure. The calculation involves integrating the pressure over the area of the gate, considering the variation of pressure with depth. The correct calculation leads to the answer $\frac { 15,625 } { 24 }$ pounds, taking into account the specific geometry and the fact that the pressure increases linearly with depth.

A Cylindrical Sewer Pipe of Radius 52\frac { 5 } { 2 }25​

A Cylindrical Sewer Pipe of Radius $\frac { 5 } { 2 }$