A sign (mass 2000 kg) hangs from the end of avertical steel girder with a cross-sectional area of0.014 m². Ignore the mass of the girder itself.The elastic modulus for steel is2.0 x 10¹¹ N/m².What is the strain on the girder?Express your answer to two significant figures.strain 7.00x10-6SubmitPart CCorrectAl=If the girder is 9.70 m long, how much is it lengthened?Express your answer to two significant figures and include the appropriate units.SubmitPrevious AnswersCHHA 1ValueProvide FeedbackRequest AnswerP PearsonMacBook AirUnitsCourse Home?Copyright © 2022 Pearson Education Inc. All rights reserved. | Terms of Use | Privacy Policy | Permissions | Contact Us |Review I CNext

GivenMass, m=2000 kgArea, A=0.014 m2Elastic modulus for steel, E=2.0×1011 Nm2

A sign (mass 2000 kg ) hangs from the end of a vertical steel girder with a cross-sectional area of 0.014 m². Ignore the mass of the girder itself. The elastic modulus for steel is 2.0 x 10¹¹ N/m². What is the strain on the gin Express your answer to t strain = 7.00x10-6 Submit Previous An

A sign (mass 2000 kg ) hangs from the end of a vertical steel girder with a cross-sectional area of 0.014 m². Ignore the mass of the girder itself. The elastic modulus for steel is 2.0 x 10¹¹ N/m². What is the strain on the gin Express your answer to t strain = 7.00x10-6 Submit Previous An

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter14: Static Equilibrium, Elasticity, And Fracture

Section: Chapter Questions

Problem 38PQ: At a museum, a 1300-kg model aircraft is hung from a lightweight beam of length 12.0 m that is free...

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Concept explainers

Mechanical Properties of Solids

There are three types of matter- solid, liquid and gas. Out of them, solids have a rigid shaped and size. The intermolecular forces of attractions in solids are greatest. Mechanical properties are the physical properties of a material which it shows, when it is provided with certain amount of force or load. These properties are to be kept in mind for various purposes. When a force is applied to an object, the object may or may not get deformed depending upon its elastic property. But every time the object develops a restoring force against the force applied. This is commonly known as stress. If the object changes its dimensions due to the force applied it develops strain ( which is the ratio of change in dimension to the original dimension).

Question

A sign (mass 2000 kg) hangs from the end of a
vertical steel girder with a cross-sectional area of
0.014 m². Ignore the mass of the girder itself.
The elastic modulus for steel is
2.0 x 10¹¹ N/m².
What is the strain on the girder?
Express your answer to two significant figures.
strain 7.00x10-6
Submit
Part C
Correct
Al=
If the girder is 9.70 m long, how much is it lengthened?
Express your answer to two significant figures and include the appropriate units.
Submit
Previous Answers
CHHA 1
Value
Provide Feedback
Request Answer
P Pearson
MacBook Air
Units
Course Home
?
Copyright © 2022 Pearson Education Inc. All rights reserved. | Terms of Use | Privacy Policy | Permissions | Contact Us |
Review I C
Next

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At a museum, a 1300-kg model aircraft is hung from a lightweight beam of length 12.0 m that is free to pivot about its base and is supported by a massless cable (Fig. P14.38). Ignore the mass of the beam. a. What is the tension in the section of the cable between the beam and the wall? b. What are the horizontal and vertical forces that the pivot exerts on the beam? FIGURE P14.38 (a) From the free-body diagram, the angle that the string tension makes with the beam is = 55.0 + 18.0 = 73.0, and the perpendicular component of the string tension is FT sin73.0. Summing torques around the base of the rod gives (Eq. 14.2): =0:(12.0m)(1300kg)(9.81m/s2)cos55.0+FT(12.0m)sin73.0=0FT=(12.0m)(1300kg)(9.81m/s2)cos55.0(12.0m)sin73.0FT=7.65103N Figure P14.38ANS (b) Using force balance (Eq. 14.1): Fx=0:FHFTcos18.0=0FH=FTcos18.0=[(12.0m)(1300kg)(9.81m/s2)cos55.0(12.0m)sin73.0]cos18.0=7.27103NFy=0:FVFTsin18.0(1300kg)(9.81m/s2)=0 FV=FTsin18.0+(1300kg)gFV=[(12.0m)(1300kg)(9.81m/s2)cos55.0(12.0m)sin73.0]sin18.0+(1300kg)(9.81m/s2)FV=1.51104N
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