Achild slides across a floor in a pair of rubber-soled shoes. the friction force acting on each foot is 20.0 n. the footprint area of each shoe sole is 14.0 cm2, and the thickness of each sole is 5.00 mm. find the horizontal distance by which the upper and lower surfaces of each sole are offset. the shear modulus of the rubber is 3.00 mn/m2.

The horizontal distance by which the upper and lower surfaces of each sole are offset is 2.38x10^-5 m

Explanation:

The shear modulus is the relationship between the stress and the shear strain. we can calculate it with the following equation:

G = (F/A)/(Δx/L), where F is the force, A is the area, Δx is the change in length and L is the length. Clearing Δx:

Δx = (F*L)/(A*G)

where F = 20N

L = 5 mm = 5x10^-3 m

A = 14 cm^2 = 0.0014 m^2

G = 3 MN*m^-2 = 3x10^6 N*m^-2

Replacing values:

Δx = (20*5x10^-3)/(0.0014*3x10^6) = 2.38x10^-5 m