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Chapter 15: Problem 74

One vertical wall of an above-ground swimming pool is a regular trapezoid, with one base \(10 \mathrm{m}\) long on level ground and the other \(20 \mathrm{m}\) long at a height of \(3 \mathrm{m}\) above it. If the pool is filled to the top with water, what's the net fluid force on the wall? (Hint: Consider both the force exerted by the water on one side of the wall and the force exerted by the atmosphere on the other.)

Short Answer

Expert verified
The net fluid force on the wall is given by the difference of two forces: the total force exerted by the water on the wall and the net force due to atmospheric pressure. This is calculated through integration over the trapezoid's area for the water force and through subtraction of areas for the atmospheric force. The exact numerical value of \(F_{net}\) will depend on the specific value of \(\rho\), \(g\), and \(P_{atm}\) used.

Step by step solution

01

Understand the Pressure in Fluids

The pressure in fluids varies with depth due to the weight of the fluid above. At any point in a static fluid, the pressure is the same in all directions, and its value can be computed as \(P = \rho g h\), where \(P\) is pressure, \(\rho\) is fluid density, \(g\) is acceleration due to gravity, and \(h\) is the height above the point in question.
02

Calculate the Total Force on the Water-Side of the Wall

The total force on the water-side of the wall can be computed by integrating the pressure over the area of the wall. We break down the trapezoidal wall into infinitesimally small horizontal slabs of thickness \(\delta h\), running from height 0 (bottom) to 3m. The pressure at each such slab is \(P = \rho g h\) and the area is \(\delta A = (10 + (20-10)*(h/3)) \delta h\). So the force on each slab is \(dF = P \delta A = \rho g h (10 + (20-10)*(h/3)) \delta h\). Integrating from 0 to 3m, we get the total force \(F_{total\ water} = \int_0^3 \rho g h (10 + (20-10)*(h/3)) dh\).
03

Calculate the Total Force due to Atmospheric Pressure

The atmospheric pressure acts on both sides of the wall, but over different areas. On the water side, it acts over the whole trapezoidal area \(A_{water} = 1/2 * (10+20) * 3\). On the other side, it only acts over the top part of the wall of area \(A_{air} = 20*3\). Thus, the net force due to atmospheric pressure is \(F_{atm} = P_{atm} (A_{water} - A_{air})\), where \(P_{atm}\) is the atmospheric pressure.
04

Calculate Net Fluid Force

The net fluid force on the wall is the difference between the total water-side force and the atmospheric force, i.e., differentiating between the force exerted by the water on one side of the wall and the force exerted by the atmosphere on the other. So, \(F_{net} = F_{total\ water} - F_{atm}\). This gives the net fluid force on the wall of the pool.

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