In a classroom demonstration, a 790-N

physics professor lies on a “bed of nails.” The bed consists of a large number of evenly spaced, relatively sharp nails mounted in a board so that the points extend vertically outward from the board. While the professor is lying down, approximately 1950 nails make contact with his body.
a) What is the average force exerted by each nail on the professor's body?
b) If the area of contact at the head of each nail is 1.26 × 10^–6 m2, what is the average pressure at each contact?

To find the average force exerted by each nail on the professor's body, we can use the formula:

Average Force = Total Force / Number of Nails in Contact

In this case, the professor's weight provides the total force exerted on the nails. The weight is given as 790 N. Therefore, the total force is 790 N. The number of nails in contact is given as 1950.

a) Average Force = 790 N / 1950 = 0.405 N per nail

So, the average force exerted by each nail on the professor's body is approximately 0.405 Newtons.

To find the average pressure at each contact, we can use the formula:

Average Pressure = Total Force / Total Area

In this case, the total force exerted on the nails is still 790 N. The area of contact at the head of each nail is given as 1.26 × 10^–6 m².

b) Average Pressure = 790 N / (1.26 × 10^–6 m²)

To solve this equation, we need to convert the area to meters:

1.26 × 10^–6 m² = 1.26 × 10^–6 m²

Putting it all together:

Average Pressure = 790 N / (1.26 × 10^–6 m²) = 6.27 × 10^11 Pascal (Pa)

So, the average pressure at each contact is approximately 6.27 × 10^11 Pa.

(a) just divide 790N by 1950 nails to get the force in N/nail

(b) Now just change the number of nails to total area to get the pressure in N/m^2