Area of contact between shoes?

So analysing this question I can conclude that the answer is equal to 34.5, I am a physics professor at the university of Cambridge so trust me

http://www.ocr.org.uk/Images/59850-question-paper-unit-g481-mechanics.pdf

Question 1cii pls

Hello.

There is no absolute answer here. The question is looking at your estimating ability based on your knowledge of an average everyday 'guesstimate' using your knowledge of measurements and also knowing how to calculate pressure.

For the physics part, we know that pressure = force per unit area and is either stated in N/m², Nm^-2 or Pascals.

F = ma where a is the acceleration due to gravity (10m/s² rounded up) and you estimated this in the previous part b)I of this question.

The surface area in question must also be estimated. It's no coincidence that in the old Imperial measurements 1 foot length was a standardised concept of an average mans foot length! In SI unit terms this is around 30 - 35cm with shoes looking at the person on the picture. Call it 32cm.

Again, the width of his foot (with shoes) is probably 10 to 14cm. Call it 12cm.

His shoes look flat (no heels) so we can estimate his shoe contact area with the ground as 12cm x 32cm = 384cm².

However, his shoes are not rectangular so we can round down somewhat and he has two shoes, so total surface area in contact with the ground is:

2 x 384 = 768cm² (round down to account for rounded toes and heel) which you can round to say 650cm²

Finally we need to convert to m² and plug that into the pressure expression above.

Surface area = 650/10,000 = 0.065m² (1m² = 100cm x 100cm)

which makes the pressure:

Pressure = force/area = 750N/0.065 = 11500 Pascals or 11500N/m² (I used 75kg for his mass).

This is my estimate. The mark scheme will have a wide range and I would think anywhere between say 10000N/m² and 14000N/m² would be OK. The key here is how you went about getting the estimate and also the formula you used so showing your full working is vital.