For simplicity each filament is assumed to be a long rectang

For simplicity, each filament is assumed to be a long rectangular plate 1 mu m long and 10 nm wide. The gap between the filaments is 10 nm. Further assume that the no-slip condition applies, (requires Monday\'s lecture). If the viscosity of cytoplasm is equal to that of water (0.0011 Pa s) and the filaments are sliding past each other at 10 mu m/s, what is the shear force on either filament? If a sarcomere consists of 1000 thin filaments in parallel, all moving identically, what is the total shear force on all of the thin filaments? (that is 500 thin filaments from one z-disk and 500 thin filaments for the other z-disk). What is the shear force (at equilibrium) exerted on all of the thick filaments between these two z-disks?

Solution

a)

Shear stress = u*du/dy

= 0.0011 * 10*10^-6 / 10*10^-9

= 1.1 N/m^2

Area = 10 nm * 1 um

= 10*10^-9 * 1*10^-6

= 1*10^-14 m^2

Shear force = Shear stress*area

= 1.1*1*10^-14

= 1.1*10^-14 N

b)

Total shear force on 1000 filaments = 1000* 1.1*10^-14 = 1.1*10^-11 N

c)

Shear force on thick filaments = 1.1*10^-11 N