Question

(a) A man weighing 714 N lies on a bed of nails. Find the average pressure exerted by a nail on his skin if there are 1060 nails and the flattened point on each nail has a radius of 0.500 mm. (b) Calculate the pressure exerted on the skin by an acupuncture needle of radius 0.100 mm applied with a force of 4.40 N.

Answer 1

Solution) (a) W = 714 N

Here Weight (W) = force (F)

N = 1060 nails

r = 0.500 mm = 0.500×10^(-3) m

Average Pressure , P = ?

P = F/(NA)

A = (pi)(r^2)

A = (pi)(0.5×10^(-3))^2 = 0.785×10^(-6) m^2

P = F/(NA)

P = (714)/(1060×0.785×10^(-6))

P = 858069.94 N/m^2

P = 8.58×10^(5) N/m^2

(b) r = 0.1 mm = 0.1×10^(-3) m

A = (pi)(r^2)

A = (pi)(0.1×10^(-3))^2

A = 3.14×10^(-8) m^2

F = 4.40 N

Pressure , P = ?

P = F/A

P = (4.40)/(3.14×10^(-8))

P = 1.40×10^(8) N/m^2