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The cable supporting a \(2125-\mathrm{kg}\) elevator has a maximum strength of \(21,750 \mathrm{~N}\). What maximum upward acceleration can it give the elevator without breaking?

Short Answer

Expert verified
The maximum upward acceleration is approximately 0.431 m/s².

Step by step solution

01

Identify the Forces Acting on the Elevator

The elevator is acted upon by two forces: its weight (gravity) and the tension in the cable. The weight can be calculated as: \[ F_{ ext{gravity}} = m imes g \]where \( m = 2125 \, \text{kg} \) is the mass of the elevator, and \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.
02

Calculate the Gravitational Force

Using the equation from Step 1, calculate the gravitational force:\[ F_{ ext{gravity}} = 2125 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 20833.25 \, \text{N} \]
03

Calculate the Net Force Needed for Acceleration

Since the maximum tension that the cable can withstand is \( 21750 \, \text{N} \), the net force \( F_{ ext{net}} \) available for accelerating the elevator upward is:\[ F_{ ext{net}} = F_{ ext{tension}} - F_{ ext{gravity}} = 21750 \, \text{N} - 20833.25 \, \text{N} = 916.75 \, \text{N} \]
04

Calculate the Maximum Upward Acceleration

Using Newton's second law, \( F = m imes a \), solve for acceleration \( a \):\[ a = \frac{F_{ ext{net}}}{m} = \frac{916.75 \, \text{N}}{2125 \, \text{kg}} \approx 0.431 \, \text{m/s}^2 \]

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Key Concepts

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

Forces on an Elevator
When analyzing the forces acting on an elevator, it's essential to understand the two primary forces at play: the force due to gravity and the tension in the elevator cable. These forces work in opposite directions. Let's delve into what each force entails:

1. **Gravity** pulls the elevator downward. This force, often called the weight of the elevator, can be calculated using the formula: \[ F_{gravity} = m \times g \] where: - \( m \) is the mass of the elevator (in kilograms), and - \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). For our elevator, with a mass of \( 2125 \) kg, the gravitational force is found to be \( 20833.25 \) N.

2. **Tension** is the force exerted by the cable to hold or move the elevator. When the elevator moves upward, the cable tension needs to overcome the gravitational force and provide extra force for acceleration.

Understanding these forces helps in calculating how the elevator behaves, especially when determining safe operating conditions like maximum acceleration.
Tension in Cables
Cables are crucial in elevators, and understanding how much tension they can withstand is vital for safety. The tension is the force exerted by the cable on the elevator.

In this scenario, the tension needs to:
  • Neutralize the effect of gravity
  • Provide additional force for any upward acceleration

The maximum strength of the cable is given as \( 21750 \, \text{N} \). This means the cable can safely handle this amount of force without risking breakage. For the elevator to ascend safely:
  • The combined gravitational force \( (F_{gravity} = 20833.25 \, \text{N}) \)
  • Plus the needed force for acceleration (\( F_{net} \)) must not exceed the cable's maximum allowed tension of \( 21750 \, \text{N} \).

Therefore, the net force available to cause upward acceleration is \( 916.75 \, \text{N} \). This balance is crucial to ensure the elevator operates within the cable's limits.
Maximum Acceleration
The maximum acceleration an elevator can safely achieve while moving upwards depends on the net force available for acceleration. To determine this, we apply Newton's Second Law of Motion:

Newton's Second Law states that the force acting on an object equals its mass times its acceleration, or \( F = m \times a \). To solve for acceleration \( a \), we rearrange the equation:\[ a = \frac{F_{net}}{m} \] For our elevator with a mass of \( 2125 \) kg and a net force \( F_{net} \) of \( 916.75 \) N (after accounting for gravitational force), the maximum upward acceleration is:\[ a = \frac{916.75 \, \text{N}}{2125 \, \text{kg}} \approx 0.431 \, \text{m/s}^2 \]This means, with the given cable strength, the elevator can accelerate upwards at a maximum rate of approximately \( 0.431 \, \text{m/s}^2 \) without risking the mechanical integrity of the system. Keeping accelerations within safe limits ensures the elevator functions reliably and safely.

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Most popular questions from this chapter

A \(7650-\mathrm{kg}\) helicopter accelerates upward at 0.80 \(\mathrm{m} / \mathrm{s}^{2}\) while lifting a 1250 -kg frame at a construction site, Fig. \(59 .\) (a) What is the lift force exerted by the air on the helicopter rotors? (b) What is the tension in the cable (ignore its mass) that connects the frame to the helicopter? (c) What force does the cable exert on the helicopter?

At the instant a race began, a 65-kg sprinter exerted a force of \(720 \mathrm{~N}\) on the starting block at a \(22^{\circ}\) angle with respect to the ground. \((a)\) What was the horizontal acceleration of the sprinter? \((b)\) If the force was exerted for \(0.32 \mathrm{~s}\) with what speed did the sprinter leave the starting block?

A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be so that even low-powered cars can get up the hills without slowing down. A particular small car, with a mass of \(920 \mathrm{kg},\) can accelerate on a level road from rest to 21 \(\mathrm{m} / \mathrm{s}\) \((75 \mathrm{km} / \mathrm{h})\) in 12.5 \(\mathrm{s} .\) Using these data, calculate the maximum steepness of a hill.

(1I) A particular race car can cover a quarter-mile track \((402 \mathrm{m})\) in 6.40 \(\mathrm{s}\) starting from a standstill. Assuming the acceleration is constant, how many "g's" does the driver experience? If the combined mass of the driver and race car is \(535 \mathrm{kg},\) what horizontal force must the road exert on the tires?

What average force is needed to accelerate a 9.20-gram pellet from rest to \(125 \mathrm{~m} / \mathrm{s}\) over a distance of \(0.800 \mathrm{~m}\) along the barrel of a rifle?

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