It is hard to imagine that there can be enough air between a book and a table so that there is a net upward (buoyant) force on the book despite the large downward force on the top of the book. About how many air molecules are there between a textbook and a table, if there is an average distance of about 0.006 mm between the uneven surfaces of the book and table?
Additionally, assume the book's area (width x height) to be 138 in^2 (converted suitably to SI units).
Given from previous problems, density of air is 1.3 kg per m^3
Fbuoy = pVg
where Fbuoy is the buoyant force
p is the density
V is the volume
and g is gravity
The Attempt at a Solution
Converted 138 in^2 to appropriate SI units. Came out to 0.08903208 m^2.
V = (.006*10^-3 m) * (0.08903208 m^2) = 5.3419248*10-7 m^3
Fbuoy = (1.3 kg/m^3)*(5.3419248*10-7 m^3)*(9.8 m/s^2) = 6.805612195*10-6 N
I'm not sure if I did that right and if I did, I don't have an equation to convert N to number of air molecules.
I'm pretty sure I need Avogadro's number but I don't see how it ties in to converting.