q1 please answer quickly Internal Forces in the wires:EM, = 0; 2F3G -C(1.5)= 0EF, = 0; FAH +F30 -C = 0F301877.25NFNAHEM, = 0; 1.5FCF -0.5F = 0EF, = 0; FDE +Fg - FAH = 0F =AHFDENDisplacement :FOELDE-mmApgEFeg LeFAcgE%3Dmmmm%3D1.58g = 8g. + dc =%3Dmmtan a =Ans: a =1500FLAE =mmAEHF.8, = 84 +84H =mmBGmmBGtan B =Ans: B=2000I||| Question 1:The load is supported by the four 304 stainless stell wires that are connected to the rigid membersAB and DC. Determine the angle of tilt of each member after the C (Newton) load is applied.The members were originally horizontal, and each wire has a cross-sectional area of 15 mm.E304=193 GPaEFTake:A=1.2mA meterB=2.2mB meterC=2497N10.5 m1 m0.9 m0.5 mt1.5 mSolution:1mFDE0.5m0.5 m1mFBGFAH1.5m0.5m1.5m0.5 mC

Answered: Image /qna-images/answer/a1fa06f9-2bc8-4ce9-a681-9153bd57a81a.jpg

Answered: uestion 1: ne load is supported by the…

uestion 1: ne load is supported by the four 304 sta B and DC. Determine the angle of tilt o he members were originally horizontal, 20=193 GPa

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter6: Beams And Cables

Section: Chapter Questions

Problem 6.76P: The 50-ft measuring tape weighs 2.4 lb. Compute the span L of the tape to four significant figures.

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Question

q1 please answer quickly

Internal Forces in the wires:
EM, = 0; 2F3G -C(1.5)= 0
EF, = 0; FAH +F30 -C = 0
F30
1877.25
N
F
N
AH
EM, = 0; 1.5FCF -0.5F = 0
EF, = 0; FDE +Fg - FAH = 0
F =
AH
FDE
N
Displacement :
FOELDE-
mm
ApgE
Feg LeF
AcgE
%3D
mm
mm
%3D
1.5
8g = 8g. + dc =
%3D
mm
tan a =
Ans: a =
1500
FLAE =
mm
AE
HF.
8, = 84 +84H =
mm
BG
mm
BG
tan B =
Ans: B=
2000
I|||

Question 1:
The load is supported by the four 304 stainless stell wires that are connected to the rigid members
AB and DC. Determine the angle of tilt of each member after the C (Newton) load is applied.
The members were originally horizontal, and each wire has a cross-sectional area of 15 mm.
E304=193 GPa
E
F
Take:
A=
1.2
m
A meter
B=
2.2
m
B meter
C=
2497
N
10.5 m
1 m
0.9 m
0.5 mt
1.5 m
Solution:
1m
FDE
0.5m
0.5 m
1m
FBG
FAH
1.5m
0.5m
1.5m
0.5 m
C

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