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 63 N and the dynamic coefficient of friction is 1.5. The pad is rectangular with the surface in contact with the wheel rim being 1 cm x 3.1 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 1.1 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 mm.

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

Explanation / Answer

we know,
shear modulus = shear stress/shear starin

= (F/A)/(delta_x/L)

= (63/(0.01*0.031))/(0.0011/0.014)

= 2.59*10^6 N/m^2 or Pa

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