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A uniform \(25.0 \mathrm{~kg}\) chain \(2.00 \mathrm{~m}\) long supports a \(50.0 \mathrm{~kg}\) chandelier in a large public building. Find the tension in (a) the bottom link of the chain, (b) the top link of the chain, and (c) the middle link of the chain.

Short Answer

Expert verified
The tension is 490 N at the bottom, 735 N at the top, and 612.5 N in the middle.

Step by step solution

01

Understanding Forces on the Chain

Start by understanding that the chain has its own weight and is also supporting the weight of the chandelier. The weight of the chandelier is given by its mass times the gravitational acceleration: \( W_c = m_c \cdot g = 50.0 \, \mathrm{kg} \cdot 9.8 \, \mathrm{m/s^2} \). The weight of the chain, similarly, is \( W_h = m_h \cdot g = 25.0 \, \mathrm{kg} \cdot 9.8 \, \mathrm{m/s^2} \).
02

Calculating Tension at the Bottom Link

At the bottom link of the chain, the tension is only due to the weight of the chandelier. Therefore, the tension in the bottom link \( T_{bottom} \) is equal to \( W_c = 50.0 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} = 490 \, \mathrm{N} \).
03

Calculating Tension at the Top Link

The top link supports the entire weight of both the chandelier and the chain. Thus, the tension at the top link \( T_{top} \) is \( W_c + W_h = 490 \, \mathrm{N} + 245 \, \mathrm{N} = 735 \, \mathrm{N} \).
04

Calculating Tension at the Middle Link

The middle link of the chain supports the weight of the chandelier and half of the chain. The weight of half of the chain is \( 0.5 \times W_h = 122.5 \, \mathrm{N} \). Hence, the tension \( T_{middle} \) is given by \( W_c + 0.5 \times W_h = 490 \, \mathrm{N} + 122.5 \, \mathrm{N} = 612.5 \, \mathrm{N} \).

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Key Concepts

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

Understanding Uniform Chain
A uniform chain has consistent mass distribution along its length. This means that the mass per unit length remains constant throughout its entire length. For example, if you have a chain that is 2 meters long and weighs 25 kg, then every meter of the chain weighs the same.

When dealing with problems involving uniform chains, it's crucial to recognize its uniformity, as it simplifies calculations. In this case, we know that for a chain weighing 25 kg and 2 meters long, the weight per meter is 12.5 kg/m, which will be helpful when calculating individual sections of the chain. This allows you to compute the weight for different sections of the chain easily, such as in problems that ask about tensions at different points along the chain.
Steps for Weight Calculation
Weight calculation is foundational in physics, especially in problems involving tension and forces. In our scenario, each mass involves its weight, which is the force exerted by gravity on it.

Here's how you calculate weight:
  • Identify the mass: Knowing the mass of the object is the first step.
  • Find gravitational acceleration: On Earth, this is typically taken as 9.8 m/s².
  • Calculate the weight: Weight is simply the product of the mass and the acceleration due to gravity, given by the formula: \( W = m \cdot g \).
In our example, the chandelier weighs 490 N, calculated from its mass (50 kg) and the gravitational pull (9.8 m/s²). Similarly, the chain's total weight is 245 N.
Basics of Gravitational Force
Gravitational force is a key player in understanding why objects have weight. It's the attractive force exerted by the Earth on any mass. Every object with mass experiences this force pulling it towards the center of the Earth.

In physics problems, gravitational force often appears as weight (the force due to gravity on an object). It's calculated using the formula \( F_g = m \cdot g \), where \( m \) is mass and \( g \) is the acceleration due to gravity (9.8 m/s² on Earth).
  • Gravitational force is always directed towards the Earth's center.
  • It acts the same on all masses, regardless of their position relative to each other in a uniform gravitational field.
This fundamental force is what underpins the calculation of tensions in a suspended uniform chain where each link supports a portion of the overall weight.
Link Tension Calculation
Tension in a chain refers to the force transmitted through the links when it is subjected to forces such as the weight of a suspended object. When finding tension in chains or ropes, we consider each link or section individually, since the tension varies along the length.

To calculate tension in a chain like our example:
  • At the bottom link: The tension equals the weight of the chandelier because that's all the force the bottom link needs to support.
  • At the top link: The tension here equals the combined weight of the chandelier and the entire chain. This is because it supports everything below it.
  • At the middle link: You calculate for the chandelier's weight plus half of the chain's weight, as it's the midpoint.
Understanding this process helps make complex physics problems more manageable by breaking them down into simpler concepts, like calculating the forces acting on a uniform chain at different points.

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Most popular questions from this chapter

Two dogs pull horizontally on ropes attached to a post; the angle between the ropes is \(60.0^{\circ} .\) If \(\operatorname{dog} A\) exerts a force of \(270 \mathrm{~N}\) and \(\operatorname{dog} B\) exerts a force of 300 \(\mathrm{N},\) find the magnitude of the resultant force and the angle it makes with \(\operatorname{dog} A\) 's rope.

Compared with the force her neck exerts on her head during the landing, the force her head exerts on her neck is A. the same. B. greater. C. smaller. D. greater during the first half of the landing and smaller during the second half of the landing.

(a) What is the mass of a book that weighs \(3.20 \mathrm{~N}\) in the laboratory? (b) In the same lab, what is the weight of a dog whose mass is \(14.0 \mathrm{~kg} ?\)

Forces on a dancer's body. Dancers experience large forces associated with the jumps they make. For example, when a dancer lands after a vertical jump, the force exerted on the head by the neck must exceed the head's weight by enough to cause the head to slow down and come to rest. The head is about \(9.4 \%\) of a typical person's mass. Video analysis of a \(65 \mathrm{~kg}\) dancer landing after a vertical jump shows that her head slows down from \(4.0 \mathrm{~m} / \mathrm{s}\) to rest in a time of \(0.20 \mathrm{~s}\) What is the magnitude of the average force that her neck exerts on her head during the landing? A. \(0 \mathrm{~N}\) B. \(60 \mathrm{~N}\) C. \(120 \mathrm{~N}\) D. \(180 \mathrm{~N}\)

A person throws a 2.5 lb stone into the air with an initial upward speed of \(15 \mathrm{ft} / \mathrm{s}\). Make a free-body diagram for this stone (a) after it is free of the person's hand and is traveling upward, (b) at its highest point, (c) when it is traveling downward, and (d) while it is being thrown upward but is still in contact with the person's hand.

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