Math Problem Statement

A 20 meter tall flagpole and is equal to a strong cylinder with a 4 cm diameter. The wind bends the pole with a horizontal force of 900 N at the top of the pole. The pole flexes and has a displacement of 0.500mm. What is the shear strain, the shear stress, and the shear modulus?

Solution

To solve this problem, we need to calculate the shear strain, shear stress, and shear modulus based on the information given.

1. Shear Stress

Shear stress (τ\tau) is defined as the force applied per unit area. The formula for shear stress is:

τ=FA\tau = \frac{F}{A}

Where:

  • FF is the applied force (900 N).
  • AA is the cross-sectional area of the pole (which is circular).

The cross-sectional area of a cylinder is given by the formula:

A=π(d2)2A = \pi \left(\frac{d}{2}\right)^2

Where dd is the diameter of the cylinder (4 cm = 0.04 m).

Thus, the cross-sectional area is:

A=π(0.042)2=π×(0.02)21.2566×103m2A = \pi \left(\frac{0.04}{2}\right)^2 = \pi \times (0.02)^2 \approx 1.2566 \times 10^{-3} \, \text{m}^2

Now, calculate the shear stress:

τ=9001.2566×103717,700Pa(orN/m2)\tau = \frac{900}{1.2566 \times 10^{-3}} \approx 717,700 \, \text{Pa} \, (\text{or} \, \text{N/m}^2)

2. Shear Strain

Shear strain (γ\gamma) is defined as the displacement of the material divided by the original length. The formula for shear strain is:

γ=ΔxL\gamma = \frac{\Delta x}{L}

Where:

  • Δx\Delta x is the displacement (0.500 mm = 0.0005 m).
  • LL is the original length of the flagpole (20 m).

Thus, the shear strain is:

γ=0.000520=2.5×105\gamma = \frac{0.0005}{20} = 2.5 \times 10^{-5}

3. Shear Modulus

The shear modulus GG (also known as the modulus of rigidity) relates shear stress to shear strain. The formula for shear modulus is:

G=τγG = \frac{\tau}{\gamma}

Substitute the values of shear stress and shear strain into this formula:

G=717,7002.5×10528.71×109Pa=28.71GPaG = \frac{717,700}{2.5 \times 10^{-5}} \approx 28.71 \times 10^9 \, \text{Pa} = 28.71 \, \text{GPa}

Final Results:

  • Shear Stress (τ\tau): 717,700 Pa or 717.7 kPa
  • Shear Strain (γ\gamma): 2.5×1052.5 \times 10^{-5}
  • Shear Modulus (G): 28.71 GPa

Would you like more details or have any further questions about this problem?

Here are some related questions you may find interesting:

  1. How does shear modulus relate to material deformation?
  2. What would happen to the shear strain if the pole was twice as long?
  3. How is shear stress different from normal stress?
  4. How does the material type affect the shear modulus?
  5. What is the relationship between shear modulus and Young's modulus?

Tip: When calculating stress, always ensure you are using consistent units (e.g., meters for length and Newtons for force) to avoid errors in your calculations.

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Math Problem Analysis

Mathematical Concepts

Shear Stress
Shear Strain
Shear Modulus
Cylindrical Geometry

Formulas

Shear Stress (τ) = F / A
Shear Strain (γ) = Δx / L
Shear Modulus (G) = τ / γ

Theorems

Shear Stress and Strain Relationship
Modulus of Rigidity

Suitable Grade Level

Grades 11-12