To solve this problem, we need to calculate the shear modulus ($G$) of the material. The shear modulus is defined as the ratio of shear stress to shear strain. The formula for shear modulus is:

$G = \frac{\text{Shear Stress}}{\text{Shear Strain}}$

Given Data

• Shear force ($F$) = 2300 N
• Side length of the cube ($L$) = 0.50 m
• Shear angle ($\theta$) = 9.5°

Step 1: Calculate Shear Stress

Shear stress ($\tau$) is defined as force per unit area: $\tau = \frac{F}{A}$ Since the cube has a side length of 0.50 m, its cross-sectional area ($A$) is: $A = L \times L = 0.50 \, \text{m} \times 0.50 \, \text{m} = 0.25 \, \text{m}^2$ Thus, $\tau = \frac{2300 \, \text{N}}{0.25 \, \text{m}^2} = 9200 \, \text{N/m}^2$

Step 2: Calculate Shear Strain

Shear strain ($\gamma$) is the tangent of the deformation angle ($\theta$): $\gamma = \tan(\theta)$ Converting 9.5° to radians: $\theta = 9.5^\circ = \frac{9.5 \times \pi}{180} \approx 0.165 \, \text{radians}$ Now, $\gamma = \tan(0.165) \approx 0.166$

Step 3: Calculate the Shear Modulus

Now, we can substitute the values of shear stress and shear strain into the formula for $G$: $G = \frac{\tau}{\gamma} = \frac{9200 \, \text{N/m}^2}{0.166} \approx 55422.89 \, \text{N/m}^2$

Answer

The shear modulus of the material is approximately: $G \approx 55423 \, \text{N/m}^2$

Would you like a further breakdown of these steps, or do you have any questions?

Here are some related questions to expand your understanding:

1. What would happen to the shear modulus if the angle of deformation increased?
2. How is shear modulus different from Young's modulus?
3. How would the result change if the force applied was doubled?
4. Can the shear modulus be negative? Why or why not?
5. How does temperature affect the shear modulus of a material?

Tip: Remember that shear modulus applies to materials under parallel, opposing forces—often used to measure rigidity in materials.