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

Given two springs of identical size and shape, one made of steel and the other made of aluminum, which has the higher spring constant? Why? Does the difference depend more on the shear modulus or the bulk modulus of the material?

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?

You have two identical silver spheres and two unknown fluids, \(A\) and \(B\). You place one sphere in fluid \(A\), and it sinks; you place the other sphere in fluid \(\mathrm{B}\), and it floats. What can you conclude about the buoyant force of fluid \(\mathrm{A}\) versus that of fluid \(\mathrm{B} ?\)

If you turn on the faucet in the bathroom sink, you will observe that the stream seems to narrow from the point at which it leaves the spigot to the point at which it hits the bottom of the sink. Why does this occur?

In what direction does a force due to water flowing from a showerhead act on a shower curtain, inward toward the shower or outward? Explain.
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Engineering Physics

Read Explanation

Fluids

Read Explanation

Work Energy and Power

Read Explanation

Quantum Physics

Read Explanation

Magnetism

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