Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 26

Find the minimum diameter of a \(50.0-\mathrm{m}\) -long nylon string that will stretch no more than \(1.00 \mathrm{~cm}\) when a load of \(70.0 \mathrm{~kg}\) is suspended from its lower end. Assume that \(Y_{\text {nylon }}=3.51 \cdot 10^{8} \mathrm{~N} / \mathrm{m}^{2}\)

Short Answer

Expert verified
Answer: The minimum diameter of the nylon string that will stretch no more than 1.00 cm when a load of 70.0 kg is suspended from its lower end is 3.52 mm.

Step by step solution

01

Convert the given data to the proper units

First, we need to convert the given data to the proper units. The load is given in kg, so we need to convert it to Newtons (N) using the equation \(F = mg\), where \(m\) is the mass and \(g\) is the acceleration due to gravity (9.8 m/s²). The elongation is given in cm, so we need to convert it to meters. \(F = (70.0 ~kg)(9.8 ~m/s^2) = 686 ~N\) \(\Delta L = 1.00 ~cm = 0.0100 ~m\) Now we have: - \(F = 686 ~N\) - \(L = 50.0 ~m\) - \(\Delta L = 0.0100 ~m\) - \(Y = 3.51 × 10^8 ~N/m^{2}\)
02

Find the cross-sectional area (A)

Now we can solve for A by rearranging the elongation equation and plugging in our known values: \(A = \frac{FL}{Y \Delta L} = \frac{(686 ~N)(50.0 ~m)}{(3.51 × 10^8 ~N/m^2)(0.0100 ~m)} = 9.785 × 10^{-6} ~m^2\)
03

Find the diameter of the string

Since the cross-sectional area of the string is circular, we can use the formula \(A = \pi r^2\), where \(r\) is the radius of the string. To find the diameter, we simply need to find the radius and multiply by 2: \(r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{9.785 × 10^{-6} ~m^2}{\pi}} = 0.00176 ~m\) The diameter is twice the radius: \(diameter = 2r = 2(0.00176 ~m) = 0.00352 ~m\)
04

Convert the diameter to the proper units

Finally, let's convert the diameter from meters to millimeters for the final answer: \(diameter = 0.00352 ~m = 3.52 ~mm\) The minimum diameter of the nylon string that will stretch no more than 1.00 cm when a load of 70.0 kg is suspended from its lower end is 3.52 mm.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A tourist of mass \(60.0 \mathrm{~kg}\) notices a chest with a short chain attached to it at the bottom of the ocean. Imagining the riches it could contain, he decides to dive for the chest. He inhales fully, thus setting his average body density to \(945 \mathrm{~kg} / \mathrm{m}^{3}\), jumps into the ocean (with saltwater density = \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) ), grabs the chain, and tries to pull the chest to the surface. Unfortunately, the chest is too heavy and will not move. Assume that the man does not touch the bottom. a) Draw the man's free-body diagram, and determine the tension on the chain. b) What mass (in kg) has a weight that is equivalent to the tension force in part (a)? c) After realizing he cannot free the chest, the tourist releases the chain. What is his upward acceleration (assuming that he simply allows the buoyant force to lift him up to the surface)?

Which of the following assumptions is not made in the derivation of Bernoulli's Equation? a) Streamlines do not cross. c) There is negligible friction. b) There is negligible d) There is no turbulence. viscosity. e) There is negligible gravity.

A wooden block floating in seawater has two thirds of its volume submerged. When the block is placed in mineral oil, \(80.0 \%\) of its volume is submerged. Find the density of the (a) wooden block, and (b) the mineral oil.

A water pipe narrows from a radius of \(r_{1}=5.00 \mathrm{~cm}\) to a radius of \(r_{2}=2.00 \mathrm{~cm} .\) If the speed of the water in the wider part of the pipe is \(2.00 \mathrm{~m} / \mathrm{s}\), what is the speed of the water in the narrower part?

A beaker is filled with water to the rim. Gently placing a plastic toy duck in the beaker causes some of the water to spill out. The weight of the beaker with the duck floating in it is a) greater than the weight before adding the duck. b) less than the weight before adding the duck. c) the same as the weight before adding the duck. d) greater or less than the weight before the duck was added, depending on the weight of the duck.
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Solid State Physics

Read Explanation

Quantum Physics

Read Explanation

Particle Model of Matter

Read Explanation

Magnetism

Read Explanation

Translational Dynamics

Read Explanation
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.

Sign-up for free