Hello I am try to solve this loads and reactive forces problem 1.7 problem as I am having difficult time understanding how the distributed load is 7 ft. Where did the 7ft come from and why is it 7ft and and not 3ft? what am I doing wrong? How I got 3 ft . I added 3+3 = 6ft which means that the distributed load is 3ft
Hello I am try to solve this loads and reactive forces problem 1.7 problem as I am having difficult time understanding how the distributed load is 7 ft. Where did the 7ft come from and why is it 7ft and and not 3ft? what am I doing wrong? How I got 3 ft . I added 3+3 = 6ft which means that the distributed load is 3ft
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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Hello I am try to solve this loads and reactive forces problem 1.7 problem as I am having difficult time understanding how the distributed load is 7 ft. Where did the 7ft come from and why is it 7ft and and not 3ft? what am I doing wrong? How I got 3 ft . I added 3+3 = 6ft which means that the distributed load is 3ft
![Given information:
The total length of the beam is 14 ft (4.2 m).
The downward load P, acting at 3 ft from the right support is 4,000
Ib (17.8 kN).
The uniformly distributed load wi is 6,000 lb (26.7 kN) acting
throughout the span.
The uniformly distributed load wz is 2,000 lb (8.90 kN) acting along
8 ft from the right support.
Calculation:
Show the given beam as in Figure (1).
2,000 lb
4,000 lb (17.8 kN)
6,000 lb
(8.90 kN)
(26.7 kN)
6 ft (1.8 m)
5 ft (1.5 m)
3 ft (0.9 m)
L= 14 ft (4.2 m)
R1
R2
Figure 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffeb710f9-5b43-4ced-a801-b89c99bfcf7c%2F03b8f76d-57fe-4d92-b86b-accc4e1fce47%2Fitewvkr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Given information:
The total length of the beam is 14 ft (4.2 m).
The downward load P, acting at 3 ft from the right support is 4,000
Ib (17.8 kN).
The uniformly distributed load wi is 6,000 lb (26.7 kN) acting
throughout the span.
The uniformly distributed load wz is 2,000 lb (8.90 kN) acting along
8 ft from the right support.
Calculation:
Show the given beam as in Figure (1).
2,000 lb
4,000 lb (17.8 kN)
6,000 lb
(8.90 kN)
(26.7 kN)
6 ft (1.8 m)
5 ft (1.5 m)
3 ft (0.9 m)
L= 14 ft (4.2 m)
R1
R2
Figure 1
![W2 = 2,000 lb
(8.90 kN)
W = 6,000 lb
(26.7 kN)
P = 4,000 lb (17.8 kN)
1 =7 ft (2.1 m)
2= 3 ft
4 = 3 ft
(0.9 m) =1 ft (0.9 m)
R2
R1
(0.3 m)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffeb710f9-5b43-4ced-a801-b89c99bfcf7c%2F03b8f76d-57fe-4d92-b86b-accc4e1fce47%2Fklsfot_processed.jpeg&w=3840&q=75)
Transcribed Image Text:W2 = 2,000 lb
(8.90 kN)
W = 6,000 lb
(26.7 kN)
P = 4,000 lb (17.8 kN)
1 =7 ft (2.1 m)
2= 3 ft
4 = 3 ft
(0.9 m) =1 ft (0.9 m)
R2
R1
(0.3 m)
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