A dam holds back water that is 80.0 m deep at the wall, so the average depth of the water in contact with the wall is 40.0 m. Taking the density of water as 1.00 x 10^3 kg/m^3 and g as 9.80 m/s^2, what is the average pressure on the dam due to the water?

1. Pressure due to the weight of a fluid is pressure = h rho g. 2. Pressure rises linearly with depth, so the average pressure over the wall is the pressure at the average depth. 3. The average depth is stated as 40.0 m, which is half of the 80.0 m depth at the wall. 4. Substituting: pressure = (40.0 m)(1.00 x 10^3 kg/m^3)(9.80 m/s^2). 5. This gives 3.92 x 10^5 N/m^2, which is 392 kPa. 6. The value 784 kPa is the trap for using the full 80.0 m depth instead of the average depth. 7. The value 196 kPa halves the average depth a second time, and 3.92 kPa is the same figure with two powers of ten lost. _Source: OpenStax College Physics (CC BY 4.0), Ch 11 "Fluid Statics", section 11.4 Variation of Pressure with Depth in a Fluid_