Question

A child 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.

Answer 1

Modulus of rigidity $(\mu)$ $= \frac{Shear\>Stress(\tau)}{Shear\>Strain(\gamma)}$

$\Tau=\frac{Force}{Area}=\frac{20}{14×10^{-4}}=1.4286×10^4$

$\gamma=\frac{\Delta \>x}{l}$

Where $\Delta\>x=$ Sheer displacement

$l=$ length

$\mu$ is given a $3.00MN/m^2=3×10^6N/m^2$

From $\mu=\frac{\tau}{\gamma}$

$3×10^6=\frac{1.4286×10^3}{\gamma}$

$\implies \>\gamma=\frac{1.4286×10^4}{3×10^6}=4.760×10^{-3}$

$\therefore \>\frac{\Delta\>x}{l}=4.760×10^{-4}$

But $l=5×10^{-3}m$

$\Delta\>x=4.760×10^{-3}×5×10^{-3}$

$=2.38×10^{-5}m$