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Four sets of flexible cables, spaced at 120° intervals, are used to stabilize a 400-ft communications tower. The tower and one cable from each set is shown in Fig. The weight of each tower is 40 lb/ft, and the communica- tions equipment at the top weighs 2000 lb. Determine the axial forces transmitted by transverse cross sections at points A, B, C, and D of the tower. 50 ft 30 1500 lb 100 ft B 40° 1250 lb 100 ft 50° 1000 lb 100 ft 50 ft 60 750 Ib

Four sets of flexible cables, spaced at 120° intervals, are used to stabilize a 400-ft communications tower. The tower and one cable from each set is shown in Fig. The weight of each tower is 40 lb/ft, and the communica- tions equipment at the top weighs 2000 lb. Determine the axial forces transmitted by transverse cross sections at points A, B, C, and D of the tower. 50 ft 30 1500 lb 100 ft B 40° 1250 lb 100 ft 50° 1000 lb 100 ft 50 ft 60 750 Ib

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Four sets of flexible cables, spaced at 120° intervals, are used to stabilize a 400-ft communications tower. The tower and one cable from each set is shown in Fig. The weight of each tower is 40 lb/ft, and the communica- tions equipment at the top weighs 2000 lb. Determine the axial forces transmitted by transverse cross sections at points A, B, C, and D of the tower. 50 ft 30 1500 lb 100 ft B 40° 1250 lb 100 ft 50° 1000 lb 100 ft 50 ft 60 750 Ib
Transcribed Image Text:Four sets of flexible cables, spaced at 120° intervals, are used to stabilize a 400-ft communications tower. The tower and one cable from each set is shown in Fig. The weight of each tower is 40 lb/ft, and the communica- tions equipment at the top weighs 2000 lb. Determine the axial forces transmitted by transverse cross sections at points A, B, C, and D of the tower. 50 ft 30 1500 lb 100 ft B 40° 1250 lb 100 ft 50° 1000 lb 100 ft 50 ft 60 750 Ib
Expert Solution
Step 1

Draw the free body diagram of the section containing point A:

Mechanical Engineering homework question answer, step 1, image 1

The axial force transmitted through section at point A is the sum of axial forces acting on this section:

PA=2000 lb+1500cos 30° lb+40 lb/ft50 ftPA=5299.04 lb

 

 

 

Step 2

Draw the free body diagram of the section containing point B:

Mechanical Engineering homework question answer, step 2, image 1

The axial force transmitted through the section at point B is the sum of axial forces acting on this section:

PB=PA+1250cos 40° lb+40 lb/ft100 ftPB=5299.04 lb+1250cos 40° lb+40 lb/ft100 ftPB=10256.59 lb

 

 

 

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