How Does Shear Modulus Affect the Distance Traveled by a Rubber Sole?

[physics1234]

A child slides across a floor in a pair of rubber-soled shoes. The friction force acting on each foot is 25 N, the cross-sectional area of each foot is 20 cm2, and the height of the soles is 5.0 mm. Find the horizontal distance traveled by the sheared face of the sole. The shear modulus of the rubber is 3.0 x 10^6 Pa.

I used F/A=S*(deltaX/n) plugging in 20/14=3x10^6(deltaX/.5) but didn't get the right answer.

 

[PhanthomJay]

A child slides across a floor in a pair of rubber-soled shoes. The friction force acting on each foot is 25 N, the cross-sectional area of each foot is 20 cm2, and the height of the soles is 5.0 mm. Find the horizontal distance traveled by the sheared face of the sole. The shear modulus of the rubber is 3.0 x 10^6 Pa.

I used F/A=S*(deltaX/n) plugging in 20/14=3x10^6(deltaX/.5) but didn't get the right answer.

I don't undersrtand where you are getting your numbers. F is 25N; A is 20cm^2; G is 3*10^6 N/m^2, and L is 5 mm.

Convert the length units to meters.

25/0.0020 = 3(10^6)x/0.005
x = 2.8*10^-5m
x = .028 mm
Hmm, seems small, better check (I don't have a good feel for pascals, having been born and bred in the USA).

 

[kyle8921]

I would like to clarify that the equation used, F/A = S*(deltaX/n), is correct. However, it appears that there may have been a mistake made while plugging in the values. The correct calculation should be:

25 N / (20 cm^2) = (3.0 x 10^6 Pa) * (deltaX / 0.5 cm)

Solving for deltaX, we get:

deltaX = (25 N * 0.5 cm) / (20 cm^2 * 3.0 x 10^6 Pa)

= 6.25 x 10^-8 m

Therefore, the horizontal distance traveled by the sheared face of the sole is 6.25 x 10^-8 meters or 0.0625 micrometers. It is always important to double check calculations when working with scientific equations to ensure accuracy.