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Chapter 10: Problem 17

A marble column with a cross-sectional area of \(25 \mathrm{cm}^{2}\) supports a load of \(7.0 \times 10^{4} \mathrm{N} .\) The marble has a Young's modulus of \(6.0 \times 10^{10} \mathrm{Pa}\) and a compressive strength of $2.0 \times 10^{8} \mathrm{Pa} .$ (a) What is the stress in the column? (b) What is the strain in the column? (c) If the column is \(2.0 \mathrm{m}\) high, how much is its length changed by supporting the load? (d) What is the maximum weight the column can support?

Short Answer

Expert verified
Question: Calculate the stress, strain, length change, and maximum weight the marble column can support given the cross-sectional area of 25 cm², a load of 7.0 × 10^4 N, Young's modulus of 6.0 × 10^10 Pa, and compressive strength of 2.0 × 10^8 Pa. Answer: a) Stress in the column: σ = 2.8 × 10^7 Pa b) Strain in the column: ε ≈ 4.67 × 10^(-4) c) Length change in the column: ∆L ≈ 0.000933 m d) Maximum weight the column can support: F_max = 5.0 × 10^5 N

Step by step solution

01

Calculate the stress in the column

Using the stress formula (σ = F/A), we can determine the stress in the column: σ = (7.0 × 10^4 N) / (25 cm²) First, we need to convert the cross-sectional area into square meters: 25 cm² = 0.0025 m² Now, we can calculate the stress: σ = (7.0 × 10^4 N) / (0.0025 m²) = 2.8 × 10^7 Pa
02

Calculate the strain in the column

Using the strain formula (ε = σ/Y), we can determine the strain in the column: ε = (2.8 × 10^7 Pa) / (6.0 × 10^10 Pa) ≈ 4.67 × 10^(-4)
03

Calculate the length change in the column

Using the length change formula (∆L = ε × L), we can determine the length change in the column: ∆L = (4.67 × 10^(-4)) × (2.0 m) ≈ 9.33 × 10^(-4) m = 0.000933 m
04

Calculate the maximum weight the column can support

Using the maximum weight formula (F_max = σ_max × A), we can determine the maximum weight the column can support: F_max = (2.0 × 10^8 Pa) × (0.0025 m²) = 5.0 × 10^5 N So, the answers are: a) Stress in the column: σ = 2.8 × 10^7 Pa b) Strain in the column: ε ≈ 4.67 × 10^(-4) c) Length change in the column: ∆L ≈ 0.000933 m d) Maximum weight the column can support: F_max = 5.0 × 10^5 N

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