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 23 N and the dynamic coefficient of friction is 2.9. The pad is rectangular with the surface in contact with the wheel rim being 1.4 cm x 3.9 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

friction force = coefficient of frcition * normal force = 2.9 * 23 = 66.7 N

so.. shear force = 66.7 N

shear stress = force / area = 66.7 * 10^4 / ( 1.4 * 3.9 ) = 122161.17216 pa

shear strain = tan inverse [ 1.9 / ( 14 - 1.3 )] = 0.148504888 radians ..

so.. shear modulus = shear stress / shear strain = 122161.17216 / 0.148504888 = 822607.0792 Pa

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