Estimate the pressure exerted on a floor by a) one pointed heel of area = 0.45 cm ^2, and b) one wide heel of area 16 cm ^2, Fig. 10–48. The person wearing the shoes has a mass of 56 kg.

one pointed heel of $area = 0.45 \textrm{ cm}^2$ , and
one wide heel of area $16 \textrm{ cm}^2$ , Fig. 10–48. The person wearing the shoes has a mass of 56 kg.

Figure 10-48.

$6.1 \times 10^6 \textrm{ Pa}$
$1.7 \times 10^5 \textrm{ Pa}$

Giancoli 7th Edition, Chapter 10, Problem 9 solution video poster

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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Pressure is force divided by area and so the force exerted upwards on one of the heels is gonna be half the person's weight so one-half mg divided by the area of the heel. So pressure is one-half times 56 kilograms times 9.8 newtons per kilogram divided by 0.45 centimeter squared— area of the pointy heel— times by 1 meter for every 100 centimeters squared so that the centimeters squared cancels leaving us with meters squared and we end up with 6.1 times 10 to the 6 pascals. And then when you have 16 centimeters squared on the bottom, you end up with 1.7 times 10 to the 5 pascals so it decreases the pressure by a whole order of magnitude: 10 to the 6 versus 10 to the 5.

COMMENTS

By klgully on Wed, 6/17/2015 - 2:25 AM

for part a: How is it 6.1 x10^6 every time I do the calculation I get 6.1 x10^4. Can you offer me some explanation/help understanding please?

By Mr. Dychko on Wed, 6/17/2015 - 2:29 AM

Hi klgully,

Are you remembering to square the $\dfrac{1}{100}$ factor in the denominator? That would put your answer off by a factor of 100.

Cheers,
Mr. Dychko

By klgully on Wed, 6/17/2015 - 2:46 AM

Thank you! Just had this explained to me...love this site and thanks for your helpful replies

By Mr. Dychko on Wed, 6/17/2015 - 2:49 AM

You're welcome, and thanks for the very nice feedback!

By jr59j on Fri, 12/1/2017 - 7:55 PM

Why is it 1/2 of m*g

By Mr. Dychko on Sun, 12/3/2017 - 11:52 PM

Hello, thanks for the question. The lady has two heels, so the weight supported by one heel is half the lady's weight, which works out to $\dfrac{1}{2}mg$ .

Cheers,
Mr. Dychko

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