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Water 80.0 m deep presses on a vertical dam wall 500 m wide, and the average pressure on that wall is 3.92 x 10^5 Pa. What force does the water exert on the dam?

A3.14 x 10^8 N
B1.57 x 10^10 N
C1.96 x 10^13 N
D4.90 x 10^7 N
Answer & Solution
Correct answer: B. 1.57 x 10^10 N
1. The force on the dam is the average pressure multiplied by the area of contact. 2. The wetted face of the dam is a rectangle 80.0 m tall by 500 m wide. 3. Area = (80.0 m)(500 m) = 4.00 x 10^4 m^2. 4. Force = (3.92 x 10^5 N/m^2)(4.00 x 10^4 m^2). 5. Multiplying gives 1.57 x 10^10 N. 6. The value 1.96 x 10^13 N is the weight of all the water held in the reservoir, and the force on the dam is only a tiny fraction of that. 7. The values 3.14 x 10^8 N and 4.90 x 10^7 N are too small by factors of about fifty and three hundred, which is what happens if the area is left in the wrong power of ten. _Source: OpenStax College Physics (CC BY 4.0), Ch 11 "Fluid Statics", section 11.4 Variation of Pressure with Depth in a Fluid_
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