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An elevator cab that weighs \(27.8 \mathrm{kN}\) moves upward. What is the tension in the cable if the cab's speed is (a) increasing at a rate of \(1.22 \mathrm{~m} / \mathrm{s}^{2}\) and \((\mathrm{b})\) decreasing at a rate of \(1.22 \mathrm{~m} / \mathrm{s}^{2} ?\)

Short Answer

Expert verified
(a) 30947.40 N; (b) 24732.68 N.

Step by step solution

01

Understand the Problem

The exercise asks for the tension in the elevator cable while moving upward with two scenarios: an increase in speed (upward acceleration) and a decrease (deceleration). The elevator weight is given in kilonewtons (kN), which measures the gravitational force on it.
02

Identify Given Values

Given values include the weight of the elevator cab as 27.8 kN (which is equivalent to 27800 N) and the acceleration (increase and decrease) as 1.22 m/s².
03

Convert Elevator Weight to Mass

The elevator weight (force due to gravity) is 27800 N. Use the formula \( W = m \cdot g \) to find mass \( m \), where \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity). Calculate \( m = \frac{W}{g} = \frac{27800}{9.81} \approx 2832.82 \text{ kg} \).
04

Calculate Tension for Increasing Speed

When speed is increasing, the tension in the cable must support both the gravitational force and provide the upward acceleration. Use the formula \( T = m \cdot (g + a) \), where \( a = 1.22 \text{ m/s}^2 \). Substitute \( m = 2832.82 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( a = 1.22 \text{ m/s}^2 \) into the equation to find \( T = 2832.82 \cdot (9.81 + 1.22) \approx 30947.40 \text{ N} \).
05

Calculate Tension for Decreasing Speed

When the speed is decreasing, the tension needs to counteract the gravitational force and additionally work against the deceleration. Use the formula \( T = m \cdot (g - a) \). Substitute \( m = 2832.82 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( a = 1.22 \text{ m/s}^2 \) into the equation to find \( T = 2832.82 \cdot (9.81 - 1.22) \approx 24732.68 \text{ N} \).

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle in physics that explains how the motion of an object is affected by the forces acting upon it. In simple terms, it says that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The mathematical expression for this law is given by:
\[ F = m \cdot a \]
where \( F \) is the net force acting on the object, \( m \) is its mass, and \( a \) is its acceleration.

In the context of the elevator problem, we analyze the forces acting on the cab to determine the tension in the cable based on different scenarios of the cab's acceleration. By calculating the tension using the mass of the elevator and its acceleration, we can understand how the forces balance and vary when the elevator speeds up or slows down.
This basic principle of physics helps us determine not only the tension but also how much force must be exerted to achieve a certain acceleration.
Tension in Cables
Tension in a cable, especially in scenarios like an elevator, refers to the pulling force transmitted along the cable as it supports a load. The tension in a cable varies depending on the load it is supporting and how the load is moving.
  • When the elevator is accelerating upward, the tension in the cable needs to overcome both the gravitational force pulling downward and the additional upward acceleration.
  • Conversely, when the elevator is decelerating (or being slowed down), the cable tension has to counteract the gravitational pull while being reduced by the deceleration effect.
The formula for assessing tension, when the elevator accelerates, is:
\[ T = m \cdot (g + a) \]
And when it decelerates, it is:
\[ T = m \cdot (g - a) \]
where \( T \) is the tension, \( m \) is the mass, \( g \) is gravity, and \( a \) is the acceleration/deceleration. Understanding these aspects of tension are vital in ensuring efficient and safe elevator operation.
Gravitational Force
Gravitational force is the attractive force exerted by the Earth that pulls objects towards its center. For any object on Earth, this force acts downwards and can be calculated using the equation:
\[ W = m \cdot g \]
where \( W \) is the weight (or gravitational force), \( m \) is the mass of the object, and \( g \) stands for the acceleration due to gravity (approximately 9.81 m/s² on Earth).

In the elevator scenario, the gravitational force acts downward on the cab, giving us the initial weight of 27.8 kN. This force must be counteracted by the tension in the cable when the elevator moves. Knowing the gravitational force is crucial for calculating how much extra force is needed to move the elevator at a desired acceleration. It helps ensure the right amount of tension is applied to safely operate the elevator system.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects, focusing on parameters such as velocity, acceleration, and displacement, without regard to the forces that cause them.

In our elevator problem, we primarily consider acceleration. The elevator's speed changes at a rate of 1.22 m/s², either increasing or decreasing. Kinematics helps us describe these changes in motion, providing a clear framework to calculate movement and understand dynamics.
Using kinematic equations, we can relate different physical quantities and use known values (like acceleration) to deduce other aspects, like the necessary force adjustments or impacts on velocity over time.
By understanding kinematics, we gain insights into the details about the motion of the elevator, which also informs us about the tension needed in the cable to manage its movement accurately and safely.

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Most popular questions from this chapter

An interstellar ship has a mass of \(1.20 \times 10^{6} \mathrm{~kg}\) and is initially at rest relative to a star system. (a) What constant acceleration is needed to bring the ship up to a speed of \(0.10 c\) (where \(c\) is the speed of light, \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ) relative to the star system in \(3.0\) days? (b) What is that acceleration in \(g\) units? (c) What force is required for the acceleration? (d) If the engines are shut down when \(0.10 c\) is reached (the speed then remains constant), how long does the ship take (start to finish) to journey \(5.0\) light-months, the distance that light travels in \(5.0\) months?

The only two forces acting on a body have magnitudes of \(20 \mathrm{~N}\) and \(35 \mathrm{~N}\) and directions that differ by \(80^{\circ} .\) The resulting acceleration has a magnitude of \(20 \mathrm{~m} / \mathrm{s}^{2} .\) What is the mass of the body?

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A block of mass \(m_{1}=3.70\) \(\mathrm{kg}\) on a frictionless plane inclined at angle \(\theta=30.0^{\circ}\) is connected by \(\varepsilon\) cord over a massless, frictionless pulley to a second block of mass \(m_{2}=2,30 \mathrm{~kg}\) (Fig. \(5-52\) ). What are (a) the magnitude of the acceleration of each block, (b) the direction of the acceleration of the hanging block, and (c) the tension in the cord?

A \(2.00 \mathrm{~kg}\) object is subjected to three forces that give it an acceleration \(\vec{a}=-\left(8.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). If two of the three forces are \(\vec{F}_{1}=(30.0 \mathrm{~N}) \hat{\mathrm{i}}+(16.0 \mathrm{~N}) \hat{\mathrm{j}}\) and \(\vec{F}_{2}=-(12.0 \mathrm{~N}) \hat{\mathrm{i}}+(8.00 \mathrm{~N}) \hat{\mathrm{j}}\), find the third force.

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