1. For the concurrent forces shown, solve for its resultant and it'sdirection.188kN169kN214KN175kN143kN25197kN42°0.51552399kN15°213kN102kN210kN155kN304kN 1:23< NotesNote: in 102 kn the slope is 2/3December 23, 2123Adqwe:)> Ortyui орasdfg hjklbnmZ X CVDonespacexXreturn

Answered: Image /qna-images/answer/6f26f2a7-0746-436d-8d49-7d4fc41e8c5f.jpg

Answered: 1. For the concurrent forces shown,…

1. For the concurrent forces shown, solve for its resultant and it's direction. 188kN 169kN 214KN 175kN 143kN 25 10.5 197kN 42° 15 23° 99kN 15° 210kN 213kN 102KN 155kN 304KN

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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Concept explainers

Laws of Equilibrium

A body is said to be in equilibrium if the system of forces acting on the body tends to keep the body at rest, in other words, there is no translation motion or net torque on the body. Under such conditions, the equilibrium conditions are to be applied to analyze the condition. The branch of mechanics that deals with this condition is known as 'statics'. Under equilibrium, there are no net forces on the body.

Question

1. For the concurrent forces shown, solve for its resultant and it's
direction.
188kN
169kN
214KN
175kN
143kN
25
197kN
42°
0.5
155
23
99kN
15°
213kN
102kN
210kN
155kN
304kN

1:23
< Notes
Note: in 102 kn the slope is 2/3
December 23, 2
123
Ad
qwe
:)
> O
rtyui ор
asdfg hjkl
bnm
Z X CV
Done
space
x
X
return

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