A child slides across a floor in a pair of rubber-soled...

A child slides across a floor in a pair of rubber-soled shoes. The frictional force acting on each foot is $20.0 \mathrm{~N}$. The footprint area of each shoe's sole is $14.0 \mathrm{~cm}^{2}$, and the thickness of each sole is $5.00 \mathrm{~mm}$. Find the horizontal distance by which the upper and lower surfaces of each sole are offset. The shear modulus of the rubber is $3.00 \times 10^{6} \mathrm{~N} / \mathrm{m}^{2}$

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Key Concepts

Frictional Force

The frictional force is the resisting force acting parallel to the contact surface between two bodies in relative motion. It quantifies how strongly the surfaces oppose sliding and directly influences the stress distribution over the contact area.

Contact Area

The contact area is the portion of a surface that is in actual direct contact with another surface. It is a crucial factor in determining the pressure or stress applied to a material, as the same force distributed over a smaller area results in higher stress.

Shear Stress

Shear stress is defined as the force per unit area acting parallel to the surface over which the force is distributed. It measures the intensity of the force that causes layers within a material to slide relative to each other.

Shear Strain

Shear strain describes the relative displacement or deformation of a material under shear stress, expressed as the ratio of the displacement between adjacent layers to the material’s thickness. It quantifies how much the material is deformed due to shearing forces.

Shear Modulus

The shear modulus is a material property that indicates its resistance to shear deformation. It is defined as the ratio of shear stress to shear strain within the elastic limit of the material, offering insight into the material's stiffness under shear forces.

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