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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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Most popular questions from this chapter

Christy has a grandfather clock with a pendulum that is \(1.000 \mathrm{m}\) long. (a) If the pendulum is modeled as a simple pendulum, what would be the period? (b) Christy observes the actual period of the clock, and finds that it is \(1.00 \%\) faster than that for a simple pendulum that is \(1.000 \mathrm{m}\) long. If Christy models the pendulum as two objects, a \(1.000-\mathrm{m}\) uniform thin rod and a point mass located \(1.000 \mathrm{m}\) from the axis of rotation, what percentage of the total mass of the pendulum is in the uniform thin rod?
(a) What is the energy of a pendulum \((L=1.0 \mathrm{m}, m=\) $0.50 \mathrm{kg})\( oscillating with an amplitude of \)5.0 \mathrm{cm} ?$ (b) The pendulum's energy loss (due to damping) is replaced in a clock by allowing a \(2.0-\mathrm{kg}\) mass to drop \(1.0 \mathrm{m}\) in 1 week. What average percentage of the pendulum's energy is lost during one cycle?
In an aviation test lab, pilots are subjected to vertical oscillations on a shaking rig to see how well they can recognize objects in times of severe airplane vibration. The frequency can be varied from 0.02 to $40.0 \mathrm{Hz}\( and the amplitude can be set as high as \)2 \mathrm{m}$ for low frequencies. What are the maximum velocity and acceleration to which the pilot is subjected if the frequency is set at \(25.0 \mathrm{Hz}\) and the amplitude at \(1.00 \mathrm{mm} ?\)
Equipment to be used in airplanes or spacecraft is often subjected to a shake test to be sure it can withstand the vibrations that may be encountered during flight. A radio receiver of mass \(5.24 \mathrm{kg}\) is set on a platform that vibrates in SHM at \(120 \mathrm{Hz}\) and with a maximum acceleration of $98 \mathrm{m} / \mathrm{s}^{2}(=10 \mathrm{g}) .$ Find the radio's (a) maximum displacement, (b) maximum speed, and (c) the maximum net force exerted on it.
An anchor, made of cast iron of bulk modulus \(60.0 \times 10^{9} \mathrm{Pa}\) and of volume \(0.230 \mathrm{m}^{3},\) is lowered over the side of the ship to the bottom of the harbor where the pressure is greater than sea level pressure by \(1.75 \times 10^{6} \mathrm{Pa} .\) Find the change in the volume of the anchor.
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