1. What must be the pull in a L(10) meters tape in order to remove correction, given the following? a = M x10“ m² Weight of tape, 11 N E= 2x10" N/m² Po = G kg The tape lenoth is supported at both ends
1. What must be the pull in a L(10) meters tape in order to remove correction, given the following? a = M x10“ m² Weight of tape, 11 N E= 2x10" N/m² Po = G kg The tape lenoth is supported at both ends
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Equivalents
G = 6
M = 4
L = 8
![1. What must be the pull in a L(10) meters tape in order to remove the sag
correction, given the following?
a = M x10“ m?
E = 2x10" N/m?
Po = G kg
Weight of tape, 11 N
The tape length is supported at both ends.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F51ddf589-3be9-457c-ae42-c986899862a4%2Fd2b28bdc-eaa5-437e-a550-391db254480e%2F0q37hqs_processed.png&w=3840&q=75)
Transcribed Image Text:1. What must be the pull in a L(10) meters tape in order to remove the sag
correction, given the following?
a = M x10“ m?
E = 2x10" N/m?
Po = G kg
Weight of tape, 11 N
The tape length is supported at both ends.
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