Chapter 12: Problem 62
A mine elevator is supported by a single steel cable \(2.5 \mathrm{~cm}\) in diameter. The total mass of the elevator cage and occupants is \(670 \mathrm{~kg}\). By how much does the cable stretch when the elevator hangs by (a) \(12 \mathrm{~m}\) of cable and (b) \(362 \mathrm{~m}\) of cable? (Neglect the mass of the cable.)
Short Answer
Step by step solution
Identify Given Variables and Constants
Compute Cross-sectional Area of the Cable
Calculate the Force Exerted by the Elevator
Use Young's Modulus to Find Stretch for 12m Cable
Repeat Calculation for 362m Cable
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Elasticity
Stress-Strain Relationship
Material Properties
- Young's Modulus: Measures stiffness and dictates how much a material will deform under stress. For steel, this value is typically around \(2.10 \times 10^{11} \text{ N/m}^2\). Higher Young's Modulus values indicate stiffer materials that don't deform much with applied force.
- Density: Influences the weight of a material and the gravitational force it experiences. In the given exercise, the total mass of the elevator is considered while calculating stress.
- Tensile Strength: The maximum amount of tensile stress that a material can withstand before failure.