/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 A 63 -kg tightrope walker stands... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 63 -kg tightrope walker stands in the middle of a rope. Her weight makes the rope sag, with each half at a \(9.5^{\circ}\) angle to the horizontal. (a) Make a force diagram for the walker. (b) Find the rope tension.

Short Answer

Expert verified
The tension in each side of the rope is approximately 1874.2 N.

Step by step solution

01

Identify the Forces

Draw a force diagram showing the forces acting on the tightrope walker. The forces include the gravitational force (weight) acting downward and the tension forces from the rope, which act along the rope at an angle to the horizontal.
02

Weight Calculation

Calculate the gravitational force (weight) acting on the tightrope walker using the formula: \[ W = mg \]where \( m = 63 \text{ kg} \) is the mass of the walker and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. The result is \( W = 63 \times 9.8 = 617.4 \text{ N} \).
03

Resolve Tension Forces

Resolve the tension forces into horizontal and vertical components. For each side, the vertical component is \( T \sin(9.5^{ ext{o}}) \) and the horizontal component is \( T \cos(9.5^{ ext{o}}) \). Since the walker is at equilibrium, the vertical components of the tension must sum to her weight, and the horizontal components must cancel out.
04

Vertical Force Balance Equation

Write the equation for the balance of vertical forces:\[ 2T \sin(9.5^{ ext{o}}) = 617.4 \text{ N} \] This equation accounts for both sides of the rope, each contributing a vertical tension component of \( T \sin(9.5^{ ext{o}}) \).
05

Solve for Rope Tension

To find the tension \( T \), solve the vertical force equation:\[ T = \frac{617.4}{2\sin(9.5^{ ext{o}})} \]Calculate \( \sin(9.5^{ ext{o}}) \) and substitute to find:\[ T = \frac{617.4}{2 \times 0.165} \approx 1874.2 \text{ N} \].
06

Conclusion

Each side of the rope exerts a tension force of approximately 1874.2 N to balance the weight of the tightrope walker.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Tension Calculation
When dealing with the equilibrium of forces, tension calculation is crucial, specifically in scenarios such as a tightrope walker standing mid-rope. Tension is the force conducted along the rope that helps to support the walker’s weight.
To solve for tension, you first need to understand the conditions of equilibrium:
  • The sum of vertical forces should be zero.
  • The sum of horizontal forces should also be zero.
In the given exercise, the tightrope creates two tension forces due to the sag from the weight of the walker. Each side of the rope pulls upwards and inwards, balancing the downward gravitational force.
Set up the mathematical equation to find the correct tension value using trigonometry, as the angle of sag causes the force to split into components. Solving these equations gives you an accurate measure of rope tension required to maintain equilibrium.
Creating a Force Diagram
A force diagram is a simple visual representation that identifies and illustrates all the forces acting on an object. In our case, depicting the forces acting on the tightrope walker is pivotal.

Here's how you create a useful force diagram for this scenario:
  • Draw the tightrope walker as a point or small box, centered on the diagram.
  • Include arrows to represent forces:
    • A downward arrow for gravitational force (weight).
    • Two arrows at an angle (in this case, each at a hyperbolic-looking angle of 9.5° from the horizontal) to represent rope tension on both sides.
These graphical elements help visualize how the components of tension balance the weight, enabling a clear understanding of all involved forces.
Balancing Vertical Forces
Vertical force balance is about ensuring that there is equilibrium in the vertical direction. Equilibrium occurs when the sum of all vertical forces equals zero, meaning there's no unbalanced force causing the walker to move vertically.
The total weight of the walker, calculated by multiplying mass and gravitational acceleration (\[ W = mg = 63 \times 9.8 = 617.4 \text{ N} \]), acts downward. The vertical components of the tension from each half of the rope act upwards.

In equilibrium, these components must add up to balance the weight:\[ 2T \sin(9.5^{\circ}) = 617.4 \text{ N} \].
This equation represents the vertical force balance, emphasizing the necessity for the tension’s components to counteract the gravitational pull.
Using Trigonometric Components in Physics
Trigonometric components in physics are useful for breaking down forces that are not aligned with a standard axis, as seen in tension forces causing a sagged rope. When dealing with angles, using trigonometry allows us to find horizontal and vertical components of these forces.
For any angle \(\theta\), a force \(T\) can be resolved into:
  • Horizontal component: \(T \cos(9.5^{\circ})\)
  • Vertical component: \(T \sin(9.5^{\circ})\)
This separation is essential because it lets us work with components that align with the standard x-y axis. Armed with these components, we solve each axis separately and then integrate to get the full picture, as done in tension calculation and balancing forces in this exercise. Such analysis forms a cornerstone in understanding more complex physics problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Describe how each of the following changes over time from the moment a skydiver opens her parachute to just before she reaches the ground: gravitational force, drag force, net force, speed, and acceleration.

A \(1.0-\mathrm{kg}\) cannonball and a \(5.0-\mathrm{kg}\) cannonball are both in free fall. Compared with the \(1-\mathrm{kg}\) ball, the gravitational force on the \(5-\mathrm{kg}\) ball is (a) one-fifth as great; (b) the same; (c) five times as great; (d) 25 times as great.

A wooden block slides down a \(28^{\circ}\) incline with acceleration \(3.85 \mathrm{~m} / \mathrm{s}^{2}\). (a) Make a force diagram for the block. (b) Find the coefficient of kinetic friction.

Whiplash injuries. Struck from behind, a \(950-\mathrm{kg}\) stopped car accelerates from rest to \(32 \mathrm{~km} / \mathrm{h}\) in \(75 \mathrm{~ms}\). Typically, a person's head makes up \(6.0 \%\) of their body weight. (a) Draw a force diagram for a \(65.0-\mathrm{kg}\) person's head during the collision if there's no headrest in the car. (b) What horizontal force on the head would accelerate it along with the rest of the body? Express your answer in newtons and as a multiple of the head's weight. What exerts this force on the head? (c) Would the head in fact accelerate along with the rest of the body? Why? What would it actually do? Explain why this can lead to neck in juries.

In a popular amusement park ride, people stand on a platform around the inside of a large cylinder. The cylinder spins, the floor drops away, and the people are left pressed with their backs against the cylinder wall. (a) Why don't they fall? Draw a force diagram for a person inside the spinning cylinder to aid your explanation. (b) Take a cylinder with inner radius \(3.0 \mathrm{~m}\), rotating with period \(3.3 \mathrm{~s}\). Find the minimum coefficient of static friction required between the person and the cylinder wall.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.