/*! 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 43 A swing is made from a rope that... [FREE SOLUTION] | 91Ó°ÊÓ

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A swing is made from a rope that will tolerate a maximum tension of \(8.00 \times 10^{2} \mathrm{~N}\) without breaking. Initially, the swing hangs vertically. The swing is then pulled back at an angle of \(60.0^{\circ}\) with respect to the vertical and released from rest. What is the mass of the heaviest person who can ride the swing?

Short Answer

Expert verified
The maximum mass that can ride the swing is approximately 40.76 kg.

Step by step solution

01

Determine the Forces Involved

When the swing is pulled back and released, gravitational force acts downward, and the tension in the rope acts along the rope. At the bottom of the swing, when the swing is vertical, these forces combine to create a net centripetal force needed for circular motion. The formula for tension at the lowest point is given by \( T = mg + \frac{mv^2}{r} \), where \( T \) is the tension, \( m \) is the mass, \( g \) is gravitational acceleration, \( v \) is the velocity at the lowest point, and \( r \) is the length of the rope.
02

Calculate Velocity at Lowest Point

To find the velocity \( v \), use energy conservation principles. Initially, the potential energy is converted into kinetic energy at the lowest point. The height \( h \) from which it falls can be found using trigonometry: \( h = r(1 - \cos \theta) \). Therefore, kinetic energy \( \frac{1}{2}mv^2 \) equals potential energy \( mgh \). Thus, \( v = \sqrt{2gh} = \sqrt{2g(r(1 - \cos \theta))} \).
03

Substitute Values and Solve for Mass

Substitute \( v \) back into the tension equation: \[ T = mg + \frac{m(\sqrt{2g(r(1 - \cos \theta))})^2}{r} \]Simplifying gives \[ T = mg + 2mg(1 - \cos \theta) \]Now solve for \( m \) given that \( T = 8.00 \times 10^{2} \text{ N} \), \( g = 9.81 \text{ m/s}^2 \), and \( \cos 60^{\circ} = 0.5 \): \[ 8.00 \times 10^{2} = mg + 2mg(1-0.5) \] \[ 8.00 \times 10^{2} = mg + mg \] \[ 8.00 \times 10^{2} = 2mg \] \[ m = \frac{8.00 \times 10^{2}}{2 \times 9.81} \]
04

Calculate the Mass

Solve for \( m \): \[ m \approx \frac{800}{19.62} \approx 40.76 \text{ kg} \] Thus, the maximum mass is approximately \( 40.76 \) kg.

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Key Concepts

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

Centripetal Force
When dealing with circular motion, understanding centripetal force is key. Centripetal force is the inward force that keeps an object moving in a circular path. Without it, the object would continue in a straight line due to inertia. Centripetal force is calculated using the formula:
  • \( F_c = \frac{mv^2}{r} \)
Here, \( m \) is the mass of the object, \( v \) is its velocity, and \( r \) is the radius of the circular path. In the context of a swing, at the lowest point of its path, the tension in the rope contributes to this force along with gravitational force. Understanding this concept is essential when calculating the tension in the rope to ensure it does not exceed its maximum strength.
Energy Conservation
The principle of energy conservation is crucial in solving many physics problems. It states that energy cannot be created or destroyed. It can only change forms. In the swing problem, initially, all of the system's energy is potential energy when the swing is at its highest point, pulled back to an angle. As it is released and descends, this potential energy converts into kinetic energy at the lowest point of the swing. The energy conservation formula can be expressed as:
  • Potential Energy: \( PE = mgh \), where \( h \) is the height.
  • Kinetic Energy: \( KE = \frac{1}{2}mv^2 \)
Equating both at the lowest most point where all potential energy has been converted gives us the basis for calculating the velocity, which is critical for understanding the swing's dynamics.
Tension in Rope
Tension is a force experienced by a string, cable, or similar object when it is pulled tight. In the swing problem, tension plays a crucial role. As the swing moves, the rope experiences different amounts of tension. At the lowest point of the swing, the tension is at its maximum because it must support the gravitational force and provide the centripetal force needed for circular motion. The tension formula used in these situations is:
  • \( T = mg + \frac{mv^2}{r} \)
Where \( T \) is the tension, and other symbols represent their usual meanings. Calculating the tension helps to ensure that the rope can support the weight without breaking.
Trigonometry in Physics
Trigonometry is widely used in physics to solve problems related to angles and heights. In the swing problem, trigonometry helps determine the vertical height from which the swing descends. When the swing is pulled to a certain angle, the change in height can be calculated using:
  • \( h = r(1 - \cos \theta) \)
Where \( r \) is the length of the rope, and \( \theta \) is the angle with the vertical. This height is part of calculating the potential energy at the starting point. Trigonometry, hence, bridges the geometry of the swing's path and energy concepts in this scenario.
Maximum Mass Calculation
Determining the maximum mass the swing can handle involves understanding the interplay of forces and energy. The rope of the swing has a maximum tension limit, and this dictates the heaviest person it can support. Using the tension equation and substituting known values, you solve for the mass (
  • \( T = 2mg \)
Where \( T \) is the maximum tension the rope can handle, and \( g \) is the acceleration due to gravity. Rearranging the formula allows you to solve for the mass:
  • \( m = \frac{T}{2g} \)
For example, with a tension limit of \(8.00 \times 10^2 \) Newtons and \( g = 9.81 \text{ m/s}^2\), the maximum mass comes out to approximately 40.76 kg. This calculated mass ensures safety while maximizing the use of the swing's capacity.

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Most popular questions from this chapter

"Rocket man" has a propulsion unit strapped to his back. He starts from rest on the ground, fires the unit, and is propelled straight upward. At a height of \(16 \mathrm{~m}\), his speed is \(5.0 \mathrm{~m} / \mathrm{s} .\) His mass, including the propulsion unit, has the approximately constant value of \(136 \mathrm{~kg} .\) Find the work done by the force generated by the propulsion unit.

The brakes of a truck cause it to slow down by applying a retarding force of \(3.0 \times 10^{3} \mathrm{~N}\) to the truck over a distance of \(850 \mathrm{~m}\). What is the work done by this force on the truck? Is the work positive or negative? Why?

An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. (a) Is the direction of the force the same as or opposite to the direction of the displacement of the asteroid? Why? (b) Does the force do positive, negative, or zero work? Justify your answer. (c) What type of energy is changing as the object slows down? (d) What is the relationship between the work done by this force and the change in the object's energy? The asteroid has a mass of \(4.5 \times 10^{4} \mathrm{~kg}\), and the force causes its speed to change from 7100 to \(5500 \mathrm{~m} / \mathrm{s}\). (a) What is the work done by the force? (b) If the asteroid slows down over a distance of \(1.8 \times 10^{6} \mathrm{~m}\), determine the magnitude of the force. Verify that your answers are consistent with the answers to the Concept Questions.

A \(1200-\mathrm{kg}\) car is being driven up a \(5.0^{\circ}\) hill. The frictional force is directed opposite to the motion of the car and has a magnitude of \(f=524 \mathrm{~N}\). A force \(\overrightarrow{\mathrm{F}}\) is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight \(\overrightarrow{\mathrm{W}}\) and the normal force \(\overrightarrow{\mathrm{F}}_{\mathrm{N}}\) directed perpendicular to the road surface. The length of the road up the hill is \(290 \mathrm{~m}\). What should be the magnitude of \(\overrightarrow{\mathbf{F}}\), so that the net work done by all the forces acting on the car is \(+150 \mathrm{~kJ}\) ?

A \(5.00 \times 10^{2}-\) kg hot-air balloon takes off from rest at the surface of the earth. The nonconservative wind and lift forces take the balloon up, doing \(+9.70 \times 10^{4} \mathrm{~J}\) of work on the balloon in the process. At what height above the surface of the earth does the balloon have a speed of \(8.00 \mathrm{~m} / \mathrm{s} ?\)

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