## Tuesday, November 26, 2013

### The Physics of Thanksgiving Footwear

Amidst the family, friends and turkey during the time of Thanksgiving, the act of shopping (due to our awesome lack of sales tax) and cold weather/snow activities (it has been pretty nippy over the past few days) are often on the minds of New Hampshirites at this time of year.

While shopping at TJMaxx with my friend, we encountered a pair of shoes that seemed physically questionable, so my friend tested them out. This caused me to think about the pressure that her foot was feeling in these shoes and other shoes that are worn during the holiday season.

If my petit friend weighs 100 lbs, then she has a mass of 100lbs/(1kg/2.21lbs)=45.5kg, and a weight of (45.5kg)(9.81 m/s2)=446N. Thus, she exerts a force downwards of 446N. Her foot is approximately 20cm long and 7cm wide.

According to an engineering blog from the University of South California, in a high heel, 90% of the shoe wearer’s weight is focused in the ball of the foot. Thus, there are (446N)(0.9)=401N of force on the ball of her foot. In her “Lady Gaga” shoes, extrapolated from measuring my own foot, the base of her shoe is 10cm long x 7cm wide, with a total area of (70cm)/(1m/100cm)2=0.007m2. Because pressure=force/area, and her weight is distributed between two feet, the ball of her foot feels
(401N)/(2x0.007m2)=3.2x104Pa of pressure.

If my friend were to go to Thanksgiving dinner in a more traditional heel, the base of the shoe actually covers less area than the “Lady Gaga” shoe that at first glance seems more physically questionable. The base of a more traditional heel covers a triangular area spanning from the ball of your foot to the tip of your toes. Estimating that the length of the ball of one’s foot to their toes is 6cm and the width of my friend’s foot is still 7cm, then: A=((1/2)x6cmx7cm)/(1m/100cm)2=0.0021m2. In a heel, 90% of her weight is felt on the ball of her foot, thus (401N)/(2x0.0021m2)=9.5x104Pa of pressure, almost 3x the amount of pressure felt by the ball of the foot in the previous shoe…ouch!

A more realistic footwear choice for the snowy weather we have been getting in New Hampshire, is the Bean Boot. In this boot, only the arch is not in contact with the ground and body weight is spread evenly from heel to ball of foot. Assuming that the heel is 5cmx7cm and the ball of the foot is supported by a rectangle of 11cmx7cm, the ball of her foot and heel feel, (446N)/(2x0.0112m2)=2.0x104Pa of pressure, almost 5X less pressure felt by just the ball of her foot in the traditional heel.

Because you wear a boot in snowshoes, your foot would feel the same amount of pressure as wearing boots alone; however, if my friend had to cross over a snowy lawn to get to Thanksgiving dinner, the snow would feel less pressure if she were wearing snowshoes. The smallest adult L.L. Bean snowshoes are 21”. Assuming they are 18cm wide, each one covers an area of (53.3cmx18cm)/(1m/100cm)2=0.096m2. The snowshoes have a mass of 4lbs each/(1kg/2.21lbs)=1.8kg, giving a weight of (1.8kg)(9.81 m/s2)=18N each and 36N total. Thus, the snow feels a pressure of
(446N+36N)/(2x0.096m2)=2.5x103Pa, a factor of 10 less than the pressure felt on my friend’s foot in all shoes.

However, New Hampshirites use a variety of methods to travel across snow. L.L. Bean cross-country skis are 166cm long and 7cm wide (the width of my friend’s foot). The snowshoes have a mass of 1lb each/(1kg/2.21lbs)=0.45kg, giving a weight of (0.45kg)(9.81 m/s2)=4.5N each and 9N total.  Thus, in if my friend wears skis, the snow feels a pressure of (446N+9N)/(2x0.12m2)=1.9x103Pa, even less than the snowshoes!

To conclude, traditional heels exert the most pressure on the smallest area of your foot and if you want to cross a snowy yard to reach your Thanksgiving dinner, traveling on a pair of cross-country skis is your best bet if you don’t want to sink into the snow and get your pants wet before a yummy dinner.