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

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SUMMARY

The discussion focuses on calculating the horizontal distance traveled by the sheared face of a rubber sole using the shear modulus of rubber, which is 3.0 x 106 Pa. The friction force is 25 N, the cross-sectional area is 20 cm2, and the height of the soles is 5.0 mm. The correct formula applied is F/A = S*(deltaX/n), leading to a calculated distance of 2.8 x 10-5 m or 0.028 mm. Participants emphasize the importance of unit conversion and understanding shear modulus in practical applications.

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  • Understanding of shear modulus in materials science
  • Basic knowledge of force and area calculations
  • Familiarity with unit conversions, particularly from cm2 to m2
  • Proficiency in applying the formula F/A = S*(deltaX/n)
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  • Research the implications of shear modulus on material performance
  • Learn about unit conversions in physics, especially for force and area
  • Explore practical applications of shear modulus in engineering
  • Study the effects of friction on motion in different materials
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Students and professionals in physics, materials science, and engineering who are interested in understanding the mechanics of rubber materials and their applications in real-world scenarios.

physics1234
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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.
 
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physics1234 said:
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).
 

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