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Telephone and electrical lines are allowed to sag between poles so that the tension will not be too great when something hits or sits on the line. (a) Is it possible to have the lines perfectly horizontal? Why or why not? (b) Suppose that a line were stretched almost perfectly horizontally between two poles that are \(30 \mathrm{~m}\) apart. If a \(0.25-\mathrm{kg}\) bird perches on the wire midway between the poles and the wire sags \(1.0 \mathrm{~cm},\) what would be the tension in the wire? (Neglect the mass of the wire.)

Short Answer

Expert verified
(a) No, perfectly horizontal lines require infinite tension. (b) The tension is approximately 3675 N.

Step by step solution

01

Understanding Tension and Angle

For perfectly horizontal lines, the angle of sag would be zero. If the line cannot be horizontal, it suggests that no finite tension could support any load at zero angle, as it would require an infinite tension force to counteract any additional weight or force applied onto the line.
02

Analyzing Forces on the Wire (Part B)

We assume the center of the wire sags by 1 cm due to the weight of a bird. This creates a right angle triangle with the sag and half the distance between poles (30/2 = 15 m) as legs. Applying trigonometry, the horizontal tension forces balance the bird's downward force.
03

Calculating Sag Component

We use the vertical component of the tension (T_y) and equate it to the bird's weight. If T is the tension in the wire, then T_y can be expressed by trigonometric relations as T sin(θ). The angle θ can be calculated using sag and half the distance (0.01 m/15 m = tan(θ)).
04

Calculating Bird's Weight

The bird's weight (W) is calculated using W = mg, where m = 0.25 kg and g = 9.81 m/s². Therefore, W = 0.25 kg * 9.81 m/s² = 2.4525 N.
05

Using Trigonometric Relations

With tan(θ) = opposite/adjacent = 0.01 m / 15 m, we find sin(θ) ≈ tan(θ) ≈ 0.01/15. We estimate sin(θ) = 0.000667 since the sag is small.
06

Solving for Tension in the Wire

Equating T_y to the bird's weight: T · 0.000667 = 2.4525 N. Solving for T, we get T ≈ 2.4525 / 0.000667 ≈ 3675 N.

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Key Concepts

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

Sag tension problem
In the context of physics, when we talk about tension in wires or ropes, it's crucial to consider the concept of sag tension. This tension is particularly important in situations where wires or ropes are used to span distances, such as between telephone poles or electricity towers. Sag happens because gravity pulls down on the line, creating a natural curve. Not allowing a line to sag would mean it needs to be perfectly straight, requiring infinite tension to support any weight on it. This is why you never see perfectly horizontal wires; they couldn't physically handle the added weight without breaking.
Trigonometrical analysis in physics
Trigonometry is a powerful tool in physics used to analyze forces in cases like the sag tension problem. When a bird perches on a wire, the sag creates a right triangle where:
  • One leg is the sag amount (in this case, 1 cm or 0.01 m),
  • The other is half the distance between the poles, 15 m.
By understanding this right-angle triangle, we can use trigonometric ratios like tangent to find angles or components of force. In this exercise, tan(θ) is used as the ratio of the sag (the opposite side) over half the wire's span (the adjacent side), providing a way to calculate angles and tensions accurately.
Physics problem-solving steps
Solving physics problems involves a systematic approach to understanding and calculating relevant quantities. In this exercise, we followed several key steps:
  • First, comprehend the limitation of perfectly horizontal lines and why some sag is necessary.
  • Next, relate the forces acting on the system, like how the bird contributes to the sag by providing a downward force.
  • Then, break down these forces using trigonometry to find the relationship between the wire's tension and the bird's weight.
  • Finally, calculate the necessary quantities, like the angle and the tension, to fully understand the physics of the problem.
This structured method helps to tackle even the most complex scenarios you’ll encounter in physics, ensuring nothing is overlooked.
Bird's weight calculation
To understand the effect a bird has on a wire, you must first calculate its weight. Weight is determined using the formula: \[ W = mg \]where \( m \) is the mass (0.25 kg for our bird) and \( g \) is the acceleration due to gravity (9.81 m/s² on Earth). Put these together, and you calculate the bird's weight:\[ W = 0.25 \times 9.81 = 2.4525 \text{ N} \]This number helps us quantify the force the bird applies downwards on the wire, a necessary step to figure out how much tension the wire will experience. Knowing the bird's weight allows you to equate it to the vertical component of tension (\( T_y \)) and use trigonometry to solve for the total tension in the wire, as we did in this exercise.

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Most popular questions from this chapter

A uniform sphere of mass \(2.50 \mathrm{~kg}\) and radius \(15.0 \mathrm{~cm}\) is released from rest at the top of an incline that is \(5.25 \mathrm{~m}\) long and makes an angle of \(35^{\circ}\) with the horizontal. Assuming it rolls without slipping, (a) determine its total kinetic energy at the bottom of the incline. (b) Determine its rotational kinetic energy at the bottom of the incline. (c) What type of friction, static or kinetic, is acting on the surface of the sphere? Explain. (d) Determine the force of friction in part (d).

In a tumbling clothes dryer, the cylindrical drum (radius \(50.0 \mathrm{~cm}\) and mass \(35.0 \mathrm{~kg}\) ) rotates once every second. (a) Determine the rotational kinetic energy about its central axis. (b) If it started from rest and reached that speed in \(2.50 \mathrm{~s}\), determine the average net torque on the dryer drum.

(a) When a disk rolls without slipping, should the product \(r \omega\) be (1) greater than, (2) equal to, or (3) less than \(v_{\mathrm{CM}}\) ? (b) A disk with a radius of \(0.15 \mathrm{~m}\) rotates through \(270^{\circ}\) as it travels \(0.71 \mathrm{~m}\). Does the disk roll without slipping? Prove your answer.

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A kitten stands on the edge of a lazy Susan (a turntable). Assume that the lazy Susan has frictionless bearings and is initially at rest. (a) If the kitten starts to walk around the edge of the lazy Susan, the lazy Susan will (1) remain lazy and stationary, (2) rotate in the direction opposite that in which the kitten is walking, or (3) rotate in the direction the kitten is walking. Explain. (b) The mass of the kitten is \(0.50 \mathrm{~kg},\) and the lazy Susan has a mass of \(1.5 \mathrm{~kg}\) and a radius of \(0.30 \mathrm{~m}\). If the kitten walks at a speed of \(0.25 \mathrm{~m} / \mathrm{s},\) relative to the ground, what will be the angular speed of the lazy Susan? (c) When the kitten has walked completely around the edge and is back at its starting point, will that point be above the same point on the ground as it was at the start? If not, where is the kitten relative to the starting point? (Speculate on what might happen if everyone on the Earth suddenly started to run eastward. What effect might this have on the length of a day?)

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