What is the compressive stress F/A on a woman's feet, in units of kilopascals(kPa) , if the woman's weight is spread out evenly over both soles of her shoes? Assume her mass is 59.0 kg and each shoe has an area of 11.7 cm2 in contact with the ground (as might be the case for high‑heeled shoes).

Since the woman's weight is spread out evenly over both soles of her shoes, the compressive stress on the woman's feet is 247.09kPa.

Given the data in the question;

Mass of woman; [tex]m = 59.0kg[/tex]

Area of each shoe; [tex]A_{e.s} = 11.7cm^2 = 1.17*10^{-3}m^2[/tex]

Since the woman's weight is spread out evenly over both soles of her shoes.

Force Applied on each shoe = [tex]\frac{1}{2}mg[/tex]

Where m is the mass of the g is acceleration due to gravity the woman's weight is spread out evenly over both soles of her shoes( [tex]9.8m/s^2[/tex] )

So,

Force Applied on each shoe = [tex]\frac{1}{2}*59.0kg\ *\ 9.8m/s^2[/tex]

Force Applied on each shoe = [tex]289.1 kg.m/s^2[/tex]

Now, compressive stress = Force / Area

We substitute in our values

Compressive stress = [tex]\frac{289.1kg.m/s^2}{1.17*10^{-3}m^2}[/tex]

Compressive stress = [tex]247094 N/m^2[/tex]

Compressive stress = [tex]247094 Pa[/tex]

Compressive stress = [tex]247094* \frac{1}{1000}kPa[/tex]

Compressive stress = [tex]247.09 kPa[/tex]

Therefore, since the woman's weight is spread out evenly over both soles of her shoes, the compressive stress on the woman's feet is 247.09kPa.

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