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Chapter 9: Problem 75

A tightly stretched horizontal "high wire" is 36 m long. It sags vertically 2.1 m when a 60.0-kg tightrope walker stands at its center. What is the tension in the wire? Is it possible to increase the tension in the wire so that there is no sag?

Short Answer

Expert verified
Tension in the wire is approximately 2529 N. It is not possible to eliminate sag by only increasing tension.

Step by step solution

01

Understand the Problem Setup

We have a 36-meter long wire that sags under the weight of a 60.0 kg tightrope walker. The midpoint of the wire sags by 2.1 m. We need to calculate the tension in the wire and determine if increasing it can eliminate the sag.
02

Understand the Forces at Play

The tightrope walker is in equilibrium at the midpoint, meaning the vertical components of the tension in the wire have to balance the weight of the walker. The force due to the walker's weight is \( mg = 60 \times 9.8 \text{ m/s}^2 = 588 \text{ N} \).
03

Calculate the Tension in the Wire

At the sagging point in the middle, the wire forms an angle \( \theta \) with the horizontal on each side. There are two identical tension forces, one from each side. The vertical component of these tensions must add up to 588 N. If \( T \) is the tension in each side of the wire, then \( 2T \sin(\theta) = 588 \).
04

Find the Angle with Trigonometry

The wire geometry at sag mimics a right triangle from the midpoint to one of the supports. The horizontal part is half the wire, \( 18 \text{ m} \), and the vertical sag is 2.1 m. Using \( \tan(\theta) = \frac{2.1}{18} \), we find \( \theta \).
05

Solve for \( \theta \)

Calculate \( \tan(\theta) = \frac{2.1}{18} \approx 0.1167 \). Find \( \theta = \arctan(0.1167) \approx 6.65^\circ \).
06

Calculate the Tension using \( \sin(\theta) \)

From the right triangle, \( \sin(\theta) = \frac{2.1}{\sqrt{2.1^2 + 18^2}} \). Substitute into \( 2T \sin(\theta) = 588 \), solve for \( T \).
07

Solve for Tension \( T \)

\( \sin(\theta) \approx 0.116 \). Substitute to get \( 2T (0.116) = 588 \) which gives \( T \approx 2529 \, \text{N} \).
08

Consider the Possibility of No Sag

For no sagging, the angle \( \theta = 0 \) degrees such that \( \sin(\theta) = 0 \), requiring infinite tension. Thus, no sagging isn't practically attainable just by increasing tension.

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Key Concepts

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

Physics Equilibrium
In the tightrope walker problem, equilibrium refers to the state where all forces acting on the walker are perfectly balanced. This ensures that the walker does not accelerate or move in any direction.
For the walker standing still at the midpoint of the wire, the forces include the weight of the walker directed downward and the forces of tension in the wire directed upward at angles.
  • The walker's weight pulls downward due to gravity, measured as mass times the gravitational acceleration (\( mg = 60 imes 9.8 ext{ m/s}^2 \)).
  • The tension in the wire provides an upward force, divided into two components on each side.
Equilibrium here means the sum of upward forces from the wire's tension components equals the downward gravitational force, which is 588 N.
Force Components
When analyzing forces in a problem like this, it's crucial to break the tension in the wire down into components. These are split into horizontal and vertical forces.
  • The vertical component is responsible for balancing the weight of the walker.
  • The horizontal components ensure the tightrope does not move to the sides and remain taut.

In mathematical terms, at the center of the wire, the sum of the vertical components must counteract the gravitational pull.
This leads to the equation:
\[2T \, \sin(\theta) = 588 \, \text{N}\]
Here, \( T \) is the tension in each side of the wire and \( \theta \) is the angle the wire makes with the horizontal. Breaking up the forces this way helps simplify solving the tension required to maintain equilibrium.
Trigonometry in Physics
Trigonometry becomes an essential tool when working with angles and forces like in our tightrope scenario. It helps to determine how much of the tension force acts vertically and horizontally.
  • The vertical sag (2.1 m) and half the wire length (18 m) form a right triangle. Within this triangle, the angle (\( \theta \)) and the lengths can help find the necessary relations.
  • The tangent function relates the vertical and horizontal components: \( \tan(\theta) = \frac{2.1}{18} \).

This calculation gives us \( \theta \approx 6.65^\circ \).

To find the sine of the angle to use in tension calculations, use:
\[\sin(\theta) = \frac{2.1}{\sqrt{2.1^2 + 18^2}}\]
Understanding these relationships allows for precise solving of the tension force, making trigonometry a vital part of physics.
Tightrope Walker Problem
The tightrope walker problem is a common physics scenario that highlights the key principles of equilibrium, force distribution, and the limitations of structural tension.

This problem is about maintaining balance with proper force distribution to keep the walker safe on the wire.
  • As the walker balances, the wire sags, resulting in significant tension forces that support the walker's weight.
  • Mathematically, solving for the wire's tension demonstrates that fully eliminating sag isn't practical. Theoretical calculations show that complete removal of sag would require infinite tension.

The problem illustrates how, in reality, increasing the tension to eliminate sag creates additional structural challenges, such as increased tension needing unfeasibly strong materials.
Tightrope walkers, therefore, rely on the wire's stiffness and their own balancing skills to manage the natural sag safely.

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Most popular questions from this chapter

(II) The Leaning Tower of Pisa is 55 m tall and about 7.7 m in radius. The top is 4.5 m off center. Is the tower in stable equilibrium? If so, how much farther can it lean before it becomes unstable? Assume the tower is of uniform composition.

In a mountain-climbing technique called the "Tyrolean traverse," a rope is anchored on both ends (to rocks or strong trees) across a deep chasm, and then a climber traverses the rope while attached by a sling as in Fig. 9-91. This technique generates tremendous forces in the rope and anchors, so a basic understanding of physics is crucial for safety. A typical climbing rope can undergo a tension force of perhaps 29 kN before breaking, and a "safety factor" of 10 is usually recommended. The length of rope used in the Tyrolean traverse must allow for some "sag" to remain in the recommended safety range. Consider a 75-kg climber at the center of a Tyrolean traverse, spanning a 25-m chasm. (a) To be within its recommended safety range, what minimum distance x must the rope sag? (b) If the Tyrolean traverse is set up incorrectly so that the rope sags by only one-fourth the distance found in (a), determine the tension in the rope. Ignore stretching of the rope.Will the rope break?

A 2.0-m-high box with a 1.0-m-square base is moved across a rough floor as in Fig. 9-89. The uniform box weighs 250 N and has a coefficient of static friction with the floor of 0.60. What minimum force must be exerted on the box to make it slide? What is the maximum height \(h\) above the floor that this force can be applied without tipping the box over? Note that as the box tips, the normal force and the friction force will act at the lowest corner.

(II) A 172-cm-tall person lies on a light (massless) board which is supported by two scales, one under the top of her head and one beneath the bottom of her feet (Fig. 9-64). The two scales read, respectively, 35.1 and 31.6 kg. What distance is the center of gravity of this person from the bottom of her feet?

(III) A door 2.30 m high and 1.30 m wide has a mass of 13.0 kg. A hinge 0.40 m from the top and another hinge 0.40 m from the bottom each support half the door's weight (Fig. 9-69). Assume that the center of gravity is at the geometrical center of the door, and determine the horizontal and vertical force components exerted by each hinge on the door.
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