not sure how to get to the given solution from my last step shown 10:49 PM Sun Feb 6全40%< 88 Qengineeringpaper (1) -•..engineeringpaper (1)EGG 211-700THelveticaNeue24 O95t - 1207 + 7S û N|-145,6867 - 82.5T + 23.3683 R82.5J-50.686 Î - 37.59+ 103.368 ū r)* Next Px ?(- So.6867- 37.59+ 103.36る50.6867 -37.59+ 103.3682)x(.37 tli65p thik)Solution:M= 212î + 190 j + 34.9 k N mlili 10:49 PM Sun Feb 6全40%< 88 Qengineeringpaper (1) -•..engineeringpaper (1)EGG 211-700HelveticaNeue24 OFebruary 7, 2022Assignment Week 3Alfonso SernaProblem 1:Find the moment about point O.Moment of Force:F,F1 = 95 i- 120 j +75 k NF2 = 165 N, a = 152°, ß = 60°, y = 80.1°X1 = 1.3 m, Y1 = 1.65 m, Z1 = 1.1 mF = 9S;- 120j+as k N1.3i+1.65; + LIK=ノ165(cosait cosp Ì + cos y k)(65 ( cos IS2°it cos 60°î t cos 80.1k)CoS165( - 0.8829 + 0.Sĵ t 0.171929 KlI65(-0.88297) +l65(05;) +i65(0.1719)F2145.68635? + 82.5Î t 28.3683 K-* Next F,+ F295€-1207 + 7S û N-145,686782 SJ *+ 28,3683 ñ-50.686 î - 37.5ît 103.368 K37,5li

Answered: Image /qna-images/answer/c42a7391-54d6-4bcc-bccd-6e9e029045df.jpg

Answered: * Next Px ? |(-50.6867-37.59+ 103.368…

Engineering

Civil Engineering

* Next Px ? |(-50.6867-37.59+ 103.368 2)× (.37+li65p t1.lk)

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

not sure how to get to the given solution from my last step shown

10:49 PM Sun Feb 6
全40%
< 88 Q
engineeringpaper (1) -
•..
engineeringpaper (1)
EGG 211-700
T
HelveticaNeue
24 O
95t - 1207 + 7S û N
|
-145,6867 - 82.5T + 23.3683 R
82.5J
-50.686 Î - 37.59+ 103.368 ū r)
* Next Px ?
(- So.6867- 37.59+ 103.36る
50.6867 -37.59+ 103.3682)x(.37 tli65p thik)
Solution:
M= 212î + 190 j + 34.9 k N m
lili

10:49 PM Sun Feb 6
全40%
< 88 Q
engineeringpaper (1) -
•..
engineeringpaper (1)
EGG 211-700
HelveticaNeue
24 O
February 7, 2022
Assignment Week 3
Alfonso Serna
Problem 1:
Find the moment about point O.
Moment of Force:
F,
F1 = 95 i- 120 j +75 k N
F2 = 165 N, a = 152°, ß = 60°, y = 80.1°
X1 = 1.3 m, Y1 = 1.65 m, Z1 = 1.1 m
F = 9S;- 120j+as k N
1.3i+1.65; + LIK
=ノ
165(cosait cosp Ì + cos y k)
(65 ( cos IS2°it cos 60°î t cos 80.1k)
CoS
165
( - 0.8829 + 0.Sĵ t 0.171929 Kl
I65(-0.88297) +l65(05;) +i65(0.1719)
F2
145.68635? + 82.5Î t 28.3683 K
-
* Next F,+ F2
95€-1207 + 7S û N
-145,6867
82 SJ *
+ 28,3683 ñ
-50.686 î - 37.5ît 103.368 K
37,5
li

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