8.38 The equation of the catenary shown is y = 100 cosh(x/100) where x and y are measured in feet (the catenary is the shape of a cable suspended between two points). Locate the y-coordinate of the centroid of the catenary by numerical integration using Ax = 25 ft. 25 ft - 100 ft-100 t- Fig. P8.38 Answer: 119.7ft 154.3 ft 154.3 ft
8.38 The equation of the catenary shown is y = 100 cosh(x/100) where x and y are measured in feet (the catenary is the shape of a cable suspended between two points). Locate the y-coordinate of the centroid of the catenary by numerical integration using Ax = 25 ft. 25 ft - 100 ft-100 t- Fig. P8.38 Answer: 119.7ft 154.3 ft 154.3 ft
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
![8.38 The equation of the catenary shown is y = 100 cosh(x/100) where x and y
are measured in feet (the catenary is the shape of a cable suspended between
two points). Locate the y-coordinate of the centroid of the catenary by numerical
integration using Ax = 25 ft.
25 ft
- 100 ft-
- 100 ft-
Fig. P8.38
Answer: 119.7ft
154.3 ft
U00L
154.3 ft](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0172dfe6-fb6c-4afc-9805-7d7f28fc3db7%2F2d135be5-0f2c-4fb1-b1d6-147d50cf25dd%2Fwxnr74j_processed.png&w=3840&q=75)
Transcribed Image Text:8.38 The equation of the catenary shown is y = 100 cosh(x/100) where x and y
are measured in feet (the catenary is the shape of a cable suspended between
two points). Locate the y-coordinate of the centroid of the catenary by numerical
integration using Ax = 25 ft.
25 ft
- 100 ft-
- 100 ft-
Fig. P8.38
Answer: 119.7ft
154.3 ft
U00L
154.3 ft
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