A student foolishly attempts to stop a steel bar, of length L = 1 m and at a temperature of 20ºC, from thermally expanding by attaching it to a wooden support with a nail at each end. Steel's Young's modulus is Y = 1.1 × 1011 N/m2 and it's linear thermal expansion coefficient is α = 13 × 10-6 1/C. Randomized VariablesY = 1.1 × 1011 N/m2α = 13 × 10-6 1/CWhat is the volue of the stress, in pascals, that develops due to a rise of temprature to 21 C? Assuming the nails have a cross- sectional area of A= 10^-5 m^2 all of which is perpendicular to the stress force from the bar, what is the force acting on each due to that temperature rise?

Answered: Image /qna-images/answer/d973eb88-d224-41f8-823a-1b264520dab8.jpg

Answered: A student foolishly attempts to stop a…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Concept explainers

Question

A student foolishly attempts to stop a steel bar, of length L = 1 m and at a temperature of 20ºC, from thermally expanding by attaching it to a wooden support with a nail at each end. Steel's Young's modulus is Y = 1.1 × 10¹¹ N/m² and it's linear thermal expansion coefficient is α = 13 × 10^-6 1/C.

Randomized Variables

Y = 1.1 × 10¹¹ N/m²
α = 13 × 10^-6 1/C

What is the volue of the stress, in pascals, that develops due to a rise of temprature to 21 C?

Assuming the nails have a cross- sectional area of A= 10^-5 m^2 all of which is perpendicular to the stress force from the bar, what is the force acting on each due to that temperature rise?

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions