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Chapter 9: Problem 3

What is the average pressure on the soles of the feet of a standing 90.0 -kg person due to the contact force with the floor? Each foot has a surface area of \(0.020 \mathrm{m}^{2}\).

Short Answer

Expert verified
Answer: The average pressure exerted on the soles of the feet of the standing person is 22.1 kPa.

Step by step solution

01

Calculate the weight of the person

To find the weight of the person, we can use the formula \(F = mg\). Given the mass \(m = 90.0\,\mathrm{kg}\) and the acceleration due to gravity \(g = 9.81\,\mathrm{m/s^2}\), the weight of the person is: \(F = 90.0\,\mathrm{kg} \times 9.81\,\mathrm{m/s^2} = 882.9\,\mathrm{N}\).
02

Calculate the total surface area of both feet

Now, let's calculate the total surface area of both feet. Each foot has an area of \(0.020\,\mathrm{m}^{2}\). Since there are two feet, the total area is: \(A_\text{total} = 2 \times 0.020\,\mathrm{m}^{2} = 0.040\,\mathrm{m}^{2}\).
03

Calculate the average pressure

With the weight of the person (which is the force exerted on the floor) and the total surface area of both feet, we can now calculate the average pressure exerted on the floor using the formula \(P = \frac{F}{A}\). Substituting the weight and total area, we get: \(P = \frac{882.9\,\mathrm{N}}{0.040\,\mathrm{m}^{2}} = 22072.5\,\mathrm{Pa} = 22.1\,\mathrm{kPa}\) (rounded to one decimal place). The average pressure exerted on the soles of the feet of the standing person is \(22.1\,\mathrm{kPa}\).

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