A metallic system OA is subjected to two forces FAB and FAC. The force in cable AC is 5 KN. Given FAB = 0.912i + 1.713j - 2.286k (KN) 1.6 m A FAB FAC 2 m *--* 1.5 m 0.8 m B The force FAC (KN) as a vector is represented as The projection of FAB along AC in KN is The angle between AB and AC is Activate Windows Go to Settings to activate Windows. 1.5 m 0.8 m B The force FAC (KN) as a vector is represented as The projection of FAB along AC in KN is The angle between AB and AC is −1.078i + 1.01j – 1.346k 1.452i + 0.851j - 1.034k -1.452i + 0.851j - 1.034k | | -2.695i + 2.525j - 3.365k 2.474 1.355 1.355 1.911 2.314 50.4 65.81 71.02 44.12
A metallic system OA is subjected to two forces FAB and FAC. The force in cable AC is 5 KN. Given FAB = 0.912i + 1.713j - 2.286k (KN) 1.6 m A FAB FAC 2 m *--* 1.5 m 0.8 m B The force FAC (KN) as a vector is represented as The projection of FAB along AC in KN is The angle between AB and AC is Activate Windows Go to Settings to activate Windows. 1.5 m 0.8 m B The force FAC (KN) as a vector is represented as The projection of FAB along AC in KN is The angle between AB and AC is −1.078i + 1.01j – 1.346k 1.452i + 0.851j - 1.034k -1.452i + 0.851j - 1.034k | | -2.695i + 2.525j - 3.365k 2.474 1.355 1.355 1.911 2.314 50.4 65.81 71.02 44.12
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:Andrew Pytel And Jaan Kiusalaas
Chapter1: Introduction To Statics
Section: Chapter Questions
Problem 1.63P: Let A and B be two nonparallel vectors that lie in a common plane S. If C=A(AB), which of the...
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Transcribed Image Text:A metallic system OA is subjected to two forces FAB and FAC. The force in cable AC is 5 KN. Given FAB = 0.912i + 1.713j - 2.286k
(KN)
1.6 m
A
FAB
FAC
2 m
*--*
1.5 m
0.8 m
B
The force FAC (KN) as a vector is represented as
The projection of FAB along AC in KN is
The angle between AB and AC is
Activate Windows
Go to Settings to activate Windows.

Transcribed Image Text:1.5 m
0.8 m
B
The force FAC (KN) as a vector is represented as
The projection of FAB along AC in KN is
The angle between AB and AC is
−1.078i + 1.01j – 1.346k 1.452i + 0.851j - 1.034k -1.452i + 0.851j - 1.034k | | -2.695i + 2.525j - 3.365k
2.474 1.355
1.355 1.911 2.314
50.4 65.81 71.02 44.12
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