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 45 N and the dynamic coefficient of friction is 2.1. The pad is rectangular with the surface in contact with the wheel rim being 1.3 cm x 4.5 cm and the thickness of the pad between the rim and the holder being 1.3 cm.

The friction force drags the contact surface of the brake pad a distance of 1.4 mm from its natural, unstressed position in the direction of the wheel rotation and the normal force compresses the thickness of the pad by 1.1 mm.

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

p.s. I also have no idea what the friction coeffitient is for and wha is it 2.1?

could anyone help me with this question?

Explanation / Answer

well, Since I am also not clear on the question as you are, I can explain the way these question works. Hope it will help you.

First, Dymamic coefficient of friction can never be greater than 1. (If it is so, the friction generated will be greater than force applied, which is not possible, So considering it be 0.21 for time being)

Second, You need to calculate the young's modulus of the rubber.(Why..??, we'll see..)

Area = 1.3 cm * 4.5 cm = 5.85 cm2

Normal strain = 1.1mm / 1.3 cm = 0.0846 [Use proper units]

Shear strain = 1.4 mm / 4.5 cm = 0.03111

Youngs modulus = Normal stress/normal strain = (45/Area)/(0.0846)

Putting proper units and solving, Youngs modulus = 9*10^-3 N/m2

Shear stress = Shear force/Area

Since, Shear force = friction = coefficient*Normal force

Shear stress = Coefficient*Normal force / Area = Coefficient* Normal stress

Now, Normal stress = 45/Area = 7.7*10^-4 N/m2

So, Shear stress = 7.7*10^-4 * 0.21 = 1.62*10^-4 N/m2

Shear modulus = Shear Stress/Shear strain = 1.62*10^-4 / 0.03111 = 5.2*10^-3 N/m2

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