A 75-Kg athlete does a single-hand handstand. If the area of the hand in contact with the floor is 125 cm2, what pressure is exerted on the floor?

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- Jul 25th 2006, 04:30 PM #1

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- Jul 25th 2006, 07:36 PM #2

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Originally Posted by**Celia**

then the gravitational force on him is $\displaystyle 75.g \mbox{ Newtons} \approx 750 \mbox{ N}$. The area

is $\displaystyle 125 \mbox{ cm^2} \approx 0.0125 \mbox{ m^2}$.

Therefore the pressure on his hand is $\displaystyle 750/0.0125 \mbox{ N/m^2} = 60000 \mbox{ Pascal}$.

RonL

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