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A steel cable with cross-sectional area 3.00 \(\mathrm{cm}^{2}\) has an elastic limit of \(2.40 \times 10^{8}\) Pa. Find the maximum upward acceleration that can be given a \(1200-\mathrm{kg}\) elevator supported by the cable if the stress is not to exceed one-third of the elastic limit.

Short Answer

Expert verified
The maximum upward acceleration is approximately 10.19 m/s².

Step by step solution

01

Determine the Maximum Allowable Stress

The elastic limit of the steel cable is given as \(2.40 \times 10^{8} \ \text{Pa}\). The problem states that the stress should not exceed one-third of this limit. Therefore, the maximum allowable stress \( \sigma_{\text{max}} \) is \( \sigma_{\text{max}} = \frac{2.40 \times 10^{8}}{3} \ \text{Pa} = 0.80 \times 10^{8} \ \text{Pa} \).
02

Use the Formula for Stress

Stress \( \sigma \) is defined as the force applied per unit area, \( \sigma = \frac{F}{A} \), where \( F \) is the force and \( A \) is the cross-sectional area of the cable. The cross-sectional area given is \( 3.00 \ \text{cm}^2 = 3.00 \times 10^{-4} \ \text{m}^2 \). We set the maximum allowable stress equal to this formula: \( \frac{F}{A} = 0.80 \times 10^{8} \ \text{Pa} \).
03

Solve for the Maximum Force

Rearrange the stress formula to solve for force \( F \): \( F = \sigma_{\text{max}} \times A \). Substituting the values, \( F = 0.80 \times 10^{8} \ \text{Pa} \times 3.00 \times 10^{-4} \ \text{m}^2 = 2.40 \times 10^{4} \ \text{N} \). This is the maximum force that can be exerted without exceeding the allowable stress.
04

Account for Forces Acting on the Elevator

The force \( F \) must act against both the gravitational force on the elevator and provide the upward acceleration. The gravitational force is \( F_{\text{gravity}} = mg \), where \( m = 1200 \ \text{kg} \) and \( g = 9.81 \ \text{m/s}^2 \). Calculate \( F_{\text{gravity}} = 1200 \ \text{kg} \times 9.81 \ \text{m/s}^2 = 11772 \ \text{N} \).
05

Calculate the Net Force for Acceleration

The net force \( F_{\text{net}} \) used to accelerate the elevator is the remaining force after gravity: \( F_{\text{net}} = F - F_{\text{gravity}} = 2.40 \times 10^{4} \ \text{N} - 11772 \ \text{N} = 12228 \ \text{N} \).
06

Solve for Maximum Upward Acceleration

Use Newton's second law \( F_{\text{net}} = ma \) to solve for the maximum acceleration \( a \). Substitute \( F_{\text{net}} = 12228 \ \text{N} \) and \( m = 1200 \ \text{kg} \): \( 12228 \ \text{N} = 1200 \ \text{kg} \times a \). Solving for \( a \), we find \( a = \frac{12228}{1200} \ \text{m/s}^2 = 10.19 \ \text{m/s}^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elastic Limit
The elastic limit of a material is a crucial concept in understanding how materials respond to stress and strain. Simply put, the elastic limit is the maximum stress a material can withstand without undergoing permanent deformation. When a material is stressed beyond its elastic limit, it will not return to its original shape upon the removal of the stress. This is because the atoms within the material have been displaced to a point where they cannot return to their initial positions.

Understanding the elastic limit is essential for engineers and designers who must ensure that structures and materials can withstand the forces they encounter in everyday use. For example, in our problem, the elastic limit of a steel cable is given as \(2.40 \times 10^8\) Pa. However, to ensure safety and prevent the cable from exceeding this stress limit, the problem asks us to consider only one-third of this value. This conservative approach provides a buffer, ensuring that the material will remain within its elastic range under applied forces.
  • Elastic limit defines the threshold between elastic and plastic deformation.
  • Materials return to their original shape only when stressed within this limit.
  • It is vital for the safety and reliability of mechanical systems.
Stress Calculation
Stress calculation is fundamental in understanding how forces affect materials. Stress, denoted by \( \sigma \), is the force applied per unit area, and it is calculated by the formula \( \sigma = \frac{F}{A} \), where \( F \) is the force applied and \( A \) is the cross-sectional area over which the force is distributed.

In the case of the steel cable, the maximum allowable stress is limited to one-third of the elastic limit, thereby ensuring the cable's longevity and reliability. The cross-sectional area of the cable is converted into square meters from square centimeters for consistency in units, giving us \(3.00 \times 10^{-4}\) m². This conversion is crucial as stress is typically measured in Pascals (Pa), which are N/m².

Calculating stress ensures that materials are not overstressed, preventing structural failures. It allows engineers to design components that can withstand expected loads without exceeding safe stress levels.
  • Stress is calculated as force divided by area.
  • Units must be consistent for accurate calculations.
  • Stress calculation is essential for ensuring material performance and safety.
Newton's Second Law
Newton's Second Law is a cornerstone of classical mechanics, describing how forces affect the motion of objects. It states that the force acting on an object is equal to the mass of the object multiplied by its acceleration: \( F = ma \). This relation tells us that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

Applying this law, we can determine how quickly the elevator can accelerate upward without exceeding the specified stress limits of the cable. The total force exerted by the cable must account for both the force required to counteract gravity (weight of the elevator) and the force needed to accelerate the elevator. The equation is rearranged as \( a = \frac{F_{\text{net}}}{m} \), showing that the maximum acceleration is contingent on dividing the net force by the elevator's mass.

Understanding Newton's Second Law helps in designing systems with known parameters, allowing engineers to predict and control movements, ensuring safety and performance.
  • Force equals mass times acceleration: \( F = ma \).
  • Acceleration is influenced by net force and mass.
  • Essential for calculating movement and ensuring design constraints are met.

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