Please correct my understanding of this question. -In the equation of the moment at point A, there is a value, 36(11.5). I assume that is a force from the distributed load. -If you are finding the moment at point A, wouldn't you only need one value from the distributed load which is 27(10)? So my question is how do you get the value of 36 and 11.5, separately. And what is it for?
Please correct my understanding of this question. -In the equation of the moment at point A, there is a value, 36(11.5). I assume that is a force from the distributed load. -If you are finding the moment at point A, wouldn't you only need one value from the distributed load which is 27(10)? So my question is how do you get the value of 36 and 11.5, separately. And what is it for?
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
Please correct my understanding of this question.
-In the equation of the moment at point A, there is a value, 36(11.5). I assume that is a force from the distributed load.
-If you are finding the moment at point A, wouldn't you only need one value from the distributed load which is 27(10)?
So my question is how do you get the value of 36 and 11.5, separately. And what is it for?
Transcribed Image Text:50 kN
4 m
Radius of pulley = 1m
B
3 m
10 kN/m
9 m
4 kN/m
3 m
D
2 m
E
2 m
8 m
2m
2 m
Transcribed Image Text:ΣΜ
Az F
Ау
:0
1-
50
lom
-
50
11.5m
DG
= 0 => 50 (3) - 50 (1)-27 (10) - 36 (11.5) +DQ sin (45) (23)_ DG Cos145) (11)
A
=> DG = 104.18 KN 45°1
27 36
↓↓
内
[Fx = 0 => Ax + 50_104.18 sin (45) = 0 => A₂ = 23.67 KN →
ΣFy=0 => Ay -50 - 27 - 36 + 104. 18 Cos (45) = 0 => Ay = 39.33 KN ↑
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