Question

A child slides across a floor in a pair of rubber-soled shoes. The frictional force acting on each foot is 20.0 N. The footprint area of each shoe sole is 11.0 cm2, and the thickness of each sole is 5.10 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 ? 106 N/m2.

Answer 1

We presume that the frictional force opposing the movement of forward force are in balance and the the child is not in acceleration. Therefore the forward force F is equal to the frictional force in magnitude.

The shearing modulus G is given bY

G = [F/A]/[x/ l], where F is the shearing force forward = 20 N in the case, A is the area are of foot = 11 cm^2= 11/10^-4 sq meter. and x is the horizontal liar shear. l is the thickness of the sole = 5.10mm = 0.0051m. given G = 3.0*10^6 Pa.

x = F*l/(AG) = 20*0.0051/(11*10^(-4)*3*10^6) = 3.09090*10^(-5) meter.

= 0.00310 mm Approximately.