Calculate the average pressure exerted on the palm of a shot-putter's hand by the shot if the area of contact is 50.0 cm250.0 \textrm{ cm}^2 and he exerts a force of 800 N on it. Express the pressure in  N/m2\textrm{ N/m}^2 and compare it with the 1.00×106 Pa1.00 \times 10^6 \textrm{ Pa} pressures sometimes encountered in the skeletal system.
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Final Answer

1.60×105 N/m21.60 \times 10^5 \textrm{ N/m}^2

This pressure is 16.0% of that sometimes found in the skeletal system.

Solution video

OpenStax College Physics for AP® Courses, Chapter 11, Problem 21 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. We're going to find the pressure exerted on a shot-putter’s hand while they're throwing the shot put and so we need to go force divided by area to find that pressure. The force we're told is 800 newtons and the area is 50 centimeters squared which we have to convert into meters squared by multiplying by one meter for every 100 centimeters. We multiply by that twice because it's centimeters squared. We get 0.005 meters squared. So the force divided by area is 800 newtons divided by 0.005 meters squared which is 1.60 times ten to the five newtons per meter squared. We could have written Pascals there instead of newtons per meter squared if we wanted. Now comparing this to pressures often found in the skeleton, we take that 1.6 times ten to the five and divide it by one times ten to the six newtons per square meter and that gives us 0.16. So that means this pressure the shot-putter is experiencing in the palm of her hand is 16 percent of what is sometimes found in the skeletal system.