Shear modulus of rubber

In summary, a child slides across a floor in a pair of rubber-soled shoes with a friction force of 25 N. The cross-sectional area of each foot is 20 cm2 and the height of the soles is 5.0 mm. Using the equation F/A=S*(deltaX/n) with S=3.0 x 10^6 Pa, the horizontal distance traveled by the sheared face of the sole is calculated to be 2.8*10^-5m or .028 mm.
  • #1
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.
 
Physics news on Phys.org
  • #2
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).
 
  • #3


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.
 

1. What is the shear modulus of rubber?

The shear modulus of rubber is a measure of the material's resistance to deformation under shear stress. It is also known as the modulus of rigidity, and is represented by the letter G in equations. The value of shear modulus for rubber can vary depending on factors such as temperature, strain rate, and the type of rubber.

2. How is the shear modulus of rubber measured?

The shear modulus of rubber is typically measured using a rheometer, which applies shear stress to a rubber sample and measures the resulting strain. The shear modulus can also be calculated by dividing the shear stress by the shear strain. The units for shear modulus are typically in pascals (Pa) or megapascals (MPa).

3. What factors can affect the shear modulus of rubber?

The shear modulus of rubber can be affected by various factors such as temperature, strain rate, and the type of rubber. As temperature increases, the shear modulus of rubber tends to decrease. Higher strain rates can also result in a decrease in shear modulus. The type of rubber used can also greatly affect the shear modulus, as different rubber materials have varying properties and structures.

4. How does the shear modulus of rubber compare to other materials?

The shear modulus of rubber is generally lower than that of other materials such as metals or plastics. This is due to the rubber's unique properties of high elasticity and ability to deform under stress. However, the shear modulus of rubber can still vary greatly depending on the type and composition of the rubber material.

5. What are the practical applications of understanding the shear modulus of rubber?

Understanding the shear modulus of rubber is important in various industries such as automotive, aerospace, and construction. It allows engineers to design and select the appropriate rubber materials for specific applications based on their desired level of stiffness and deformation under stress. It can also help in predicting the performance and durability of rubber components in different environments.

Back
Top