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A 75.0 -kg wrecking ball hangs from a uniform heavy-duty chain having a mass of 26.0 \(\mathrm{kg}\) . (a) Find the maximum and minimum tension in the chain. \((b)\) What is the tension at a point three-fourths of the way up from the bottom of the chain?

Short Answer

Expert verified
(a) Maximum Tension: 989.8 N; Minimum Tension: 735 N. (b) Tension at three-fourths up: 798.7 N.

Step by step solution

01

Identify the Forces

Identify the forces acting on the system. The chain has a weight due to gravity and the wrecking ball also has a weight. The tension in the chain will vary along its length from the maximum at the top to the minimum at the bottom.
02

Calculate Weight of Wrecking Ball

The weight of the wrecking ball is given by the formula \( W = mg \), where \( m = 75.0 \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Thus, \( W = 75.0 \times 9.8 = 735 \text{ N} \).
03

Calculate Weight of the Chain

The total weight of the chain can also be found by \( W_{chain} = mg \), where \( m = 26.0 \text{ kg} \). Therefore, \( W_{chain} = 26.0 \times 9.8 = 254.8 \text{ N} \).
04

Determine Maximum Tension

The maximum tension occurs at the top of the chain and must support the weight of both the chain and the wrecking ball. Therefore, \( T_{max} = W + W_{chain} = 735 + 254.8 = 989.8 \text{ N} \).
05

Determine Minimum Tension

The minimum tension occurs at the very bottom of the chain, right above the wrecking ball. At this point, the tension equals the weight of the wrecking ball alone: \( T_{min} = 735 \text{ N} \).
06

Find Tension Three-Fourths Up the Chain

First, calculate the weight of the chain below the three-fourths point. This segment of the chain has a mass of \( \frac{1}{4} \times 26.0 = 6.5 \text{ kg} \). The weight of this segment is \( 6.5 \times 9.8 = 63.7 \text{ N} \). Therefore, the tension at this point is the weight of the wrecking ball plus the weight of the chain below it: \( T = 735 + 63.7 = 798.7 \text{ N} \).

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Key Concepts

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

Physics Problems
Physics problems often involve analyzing forces and understanding how different physical quantities interact. In this problem, we consider a wrecking ball hanging from a chain. We first need to identify all the forces involved, predominantly the force due to gravity on the wrecking ball and chain. Each exercise in physics requires breaking down these components:
  • The mass and weight of the wrecking ball, given its mass.
  • The mass and weight of the chain itself.
  • The varying tension along the chain, from its maximum to minimum values.
Physics problems can appear complex at first, but by systematically analyzing the forces and using known physical laws, they become manageable. In this case, understanding how gravity acts on different objects within a system helps determine the tension at various points in the chain. Remember, the key to mastering physics problems is practice and breaking down the problem into smaller parts.
Mechanics
Mechanics is the branch of physics that deals with forces and motion, and it is fundamental to understanding how physical systems behave. In this scenario, we apply mechanics to understand the forces within the chain and on the wrecking ball. This branch involves understanding concepts such as:
  • The weight of objects, calculated using the formula for weight, which is the product of mass and acceleration due to gravity: \( W = mg \).
  • The distribution of this weight along the chain, which affects the tension in different segments.
  • How tension varies depending on the position along the chain, from the top to the bottom.
Mechanics allows us to predict how the system will react under various conditions, such as calculating when and where the maximum and minimum tensions occur. Through mechanics, we understand not just the 'what', but the 'why' and 'how' of physical phenomena. This understanding is crucial for solving real-world engineering problems as well.
Force Calculation
Calculating forces, especially tensions in chains or ropes, involves understanding the distribution of weight and the effects of gravity. Here, we calculate forces based on the gravitational pull on the wrecking ball and the chain. To do this:
  • First, compute the gravitational force (weight) of both the wrecking ball and the chain using \( W = mg \).
  • Identify how tensions within the chain vary, with max tension supporting both the wrecking ball and the chain, and min tension right above the wrecking ball.
  • For specific segments, such as three-fourths up the chain, calculate the weight of the chain below the segment and add it to the weight of the wrecking ball.
This straightforward approach to force calculation helps to determine the tension at any given point along the chain. Understanding this calculation process is essential for those studying physics and engineering, as it forms the foundation of analyzing system dynamics. With practice, solving such mechanics problems becomes intuitive and builds a deeper appreciation for the physics governing our world.

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Most popular questions from this chapter

You are riding in a school bus. As the bus rounds a flat curve at constant speed, a lunch box with mass 0.500 \(\mathrm{kg}\) , suspended from the ceiling of the bus by a string 1.80 \(\mathrm{m}\) long, is found to hang at rest relative to the bus when the string makes an angle of \(30.0^{\circ}\) with the vertical. In this position the lunch box is 50.0 \(\mathrm{m}\) from the center of curvature of the curve What is the speed \(v\) of the bus?

A bowling ball weighing 71.2 \(\mathrm{N}(16.0 \mathrm{lb})\) is attached to the ceiling by a \(3.80-\mathrm{m}\) rope. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.20 \(\mathrm{m} / \mathrm{s} .\) (a) What is the acceleration of the bowling ball, in magnitude and direction, at this instant? (b) What is the tension in the rope at this instant?

Maximum Safe Speed. As you travel every day to campus, the road makes a large turn that is approximately an are of a circle. You notice the warning sign at the start of the turm, asking for a maximum speed of 55 \(\mathrm{mi} / \mathrm{h}\) . You also notice that in the curved portion the road is level - that is, not banked at all. On a dry day with very little traffic, you enter the turn at a constant speed of 80 \(\mathrm{mi} / \mathrm{h}\) and feel that the car may skid if you do not slow down quickly. You conclude that your speed is at the limit of safety for this curve and you slow down. However, you remember reading that on dry pavement new tires have an average coefficient of static friction of about 0.76 .while under the worst winter driving conditions, you may encounter wet ice for which the coefficient of static friction can be as low as \(0.20 .\) Wet ice is not unheard of on this road, so you ask yourself whether the speed limit for the turn on the roadside warning sign is for the worst-case scenario. (a) Estimate the radius of the curve from your \(80-\) mi/h experience in the dry turn. (b) Use this estimate to find the maximum speed limit in the turn under the worst wet-ice conditions. How does this compare with the speed limit on the sign? Is the sign misleading drivers?(c) On a rainy day, the coefficient of static friction would be about \(0.37 .\) What is the maximum safe speed of for the turn when the road is wet? Does your answer help you understand the maximum-speed sign?

A hammer is hanging by a light rope from the ceiling of a bus. The ceiling of the bus is parallel to the roadway. The bus is traveling in a straight line on a horizontal street. You observe that the hammer hangs at rest with respect to the bus when the angle between the rope and the ceiling of the bus is \(74^{\circ} .\) What is the acceleration of the bus?

Stopping Distance. (a) If the coefficient of kinctic friction between tires and dry pavement is 0.80 , what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 28.7 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} ) ?\) (b) On wet pavement the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

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