Math Problem Statement

A marble column with a cross-sectional area of 27.2 cm2 supports a load of 7.0 × 104 N. The marble has a Young’s modulus of 60 GPa and a compressive strength of 200 MPa.

If the column is 2.00 m high, how much is its length changed by supporting the load?

m

Solution

To determine how much the length of the marble column changes under the given load, we’ll use the formula for deformation under compression:

ΔL=FLAE\Delta L = \frac{F \cdot L}{A \cdot E}

where:

  • FF is the applied force (load),
  • LL is the original length of the column,
  • AA is the cross-sectional area,
  • EE is the Young’s modulus of the material.

Let's go through each step:

1. Convert Units

We have:

  • A=27.2cm2=27.2×104m2=0.00272m2A = 27.2 \, \text{cm}^2 = 27.2 \times 10^{-4} \, \text{m}^2 = 0.00272 \, \text{m}^2,
  • E=60GPa=60×109PaE = 60 \, \text{GPa} = 60 \times 10^9 \, \text{Pa},
  • F=7.0×104NF = 7.0 \times 10^4 \, \text{N},
  • L=2.00mL = 2.00 \, \text{m}.

2. Calculate the Deformation

Using the formula:

ΔL=(7.0×104)×2.000.00272×60×109\Delta L = \frac{(7.0 \times 10^4) \times 2.00}{0.00272 \times 60 \times 10^9}

Step-by-Step Calculation

  1. Compute AEA \cdot E: 0.00272×60×109=1.632×108N/m20.00272 \times 60 \times 10^9 = 1.632 \times 10^8 \, \text{N/m}^2

  2. Compute FLF \cdot L: (7.0×104)×2.00=1.4×105Nm(7.0 \times 10^4) \times 2.00 = 1.4 \times 10^5 \, \text{N} \cdot \text{m}

  3. Calculate ΔL\Delta L: ΔL=1.4×1051.632×1088.58×104m=0.858mm\Delta L = \frac{1.4 \times 10^5}{1.632 \times 10^8} \approx 8.58 \times 10^{-4} \, \text{m} = 0.858 \, \text{mm}

Answer

The length of the column changes by approximately 0.858 mm under the load.

Would you like more details on the calculation process?

Related Questions

  1. How would the deformation change if the cross-sectional area were doubled?
  2. What would happen to the deformation if the Young’s modulus of marble was 120 GPa?
  3. How does Young's modulus relate to the material’s compressive strength?
  4. What is the significance of the compressive strength in this context?
  5. What would be the deformation if the column height was halved?

Tip:

For materials under stress, remember that Young’s modulus quantifies stiffness—higher values mean less deformation under a given load.

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Math Problem Analysis

Mathematical Concepts

Physics
Material Deformation
Stress and Strain
Elastic Modulus

Formulas

ΔL = (F × L) / (A × E)

Theorems

Hooke's Law

Suitable Grade Level

Grades 11-12