A bicycle break pad is pushed against the moving rim of the wheel with a force o

Question

A bicycle break pad is pushed against the moving rim of the wheel with a force of 80 N and the dynamic coefficient of friction is 2.2. The pad is rectangular with the surface in contact with the wheel rim being 0.8 cm x 4.2 cm and the thickness of the pad between the rim and the holder being 1.4 cm.

The friction force drags the contact surface of the brake pad a distance of 1.9 mm from its natural, unstressed position in the direction of the wheel rotation and the normal force compresses the thickness of the pad by 1.3 mm.

What is the shear modulus of the rubber that the pad is made from?

Explanation / Answer

Ff = 80 * 2.2 = 176 N      shearing force

A = .008 * .042 = 3.36 * 10E-4 m^2

T = .014 - .0013 = .0127    thickness of pad after compression

tan theta = .0019 / .0127 = .150     the strain

M(odulus) = (F / A) / .150       Stress / Strain

M = (176 / 3.36 * 10E-4) / .15 = 3.49 * 10E6 N/m^2

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