Minimum area of a shoe heel exerting pressure

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Homework Help Overview

The problem involves calculating the minimum area of a shoe heel that can exert pressure on a dance hall floor without causing damage. The context includes a specified pressure limit of 3.5 MPa and a lady's mass of 60 kg, leading to a discussion about the relationship between force, pressure, and area.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of area using the formula Area = Force/Pressure. There is uncertainty about the correctness of the calculations and the relationship between the calculated area and the answer provided in the book.

Discussion Status

Participants are actively engaging with the calculations and questioning the assumptions made in the problem. Some express confusion about the discrepancy between their results and the book's answer, while others suggest that there may be a misunderstanding regarding the area being calculated.

Contextual Notes

There is a mention of the book's answer being 8.4 x 10^-5 m², while one participant calculated 1.68 x 10^-4 m², leading to discussions about potential errors or assumptions in the problem setup.

Molly1235
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Homework Statement



The floor on a dance hall can take a pressure of 3.5 MPa before getting damaged. A lady of mass 60kg wears high heeled shoes. What is the smallest area of the sole of a heel she could wear such that it won't permanently damage the floor?

Homework Equations



Area = Force/Pressure

The Attempt at a Solution



3.5 MPa = 3500000 Pa

Force (weight) = 60 x 9.8 = 588N

But I don't know where to go from there...I tried just dividing Force by Pressure but got a different answer to that in the book. Thanks in advance to anyone that can help! :-)
 
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We can't tell what you did unless you show us your calculations.
 
... It will also help if you post the answer from the book.
 
haruspex said:
... It will also help if you post the answer from the book.

The book said 8.4 x 10^-5 m^2, but I did 588 (force) / 3500000 (pressure) = 1.68 x 10^-4.

I'm feeling there's a step I'm missing but I'm not sure...
 
Molly1235 said:
The book said 8.4 x 10^-5 m^2, but I did 588 (force) / 3500000 (pressure) = 1.68 x 10^-4.

I'm feeling there's a step I'm missing but I'm not sure...
You may have noticed that your answer is exactly twice that given in the book (as I suspected).
Why do you think that might be? Btw, I agree with your answer. Whoever wrote the book has outsmarted him/herself.
 
He outsmarted me at least, although that isn't too difficult in the general case. Damn! :biggrin:
 
haruspex said:
You may have noticed that your answer is exactly twice that given in the book (as I suspected).
Why do you think that might be? Btw, I agree with your answer. Whoever wrote the book has outsmarted him/herself.

Ah ok, so you don't think I've gone wrong? I was a little confused haha
 
Molly1235 said:
Ah ok, so you don't think I've gone wrong? I was a little confused haha
No, you haven't.
But:
Is that area the area under a SINGLE shoe?
 

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