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. Eз4-193 GPa E F G Take: A= 1.5 m meter B= 2.5 m C= 2500 N B meter Fo.5 m- 0.9 m B 1.5 m -0.5 m Solution: 0.5m Im FDE CF 0.5 m 1 m 1.5m 0.5m F, BG AH Page 1 1.5m 0.5 m Internal Forces in the wires : EM,= 0; 2F,G – C(1.5)=0 EF, = 0; F,+F,6-C =0 1875 N F 625 N AH EM, =0; 1.5F, -0.5F, =0 208.33 N CF AH EF = 0; F +Fg-FH = F, N 416.66 DE
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. Eз4-193 GPa E F G Take: A= 1.5 m meter B= 2.5 m C= 2500 N B meter Fo.5 m- 0.9 m B 1.5 m -0.5 m Solution: 0.5m Im FDE CF 0.5 m 1 m 1.5m 0.5m F, BG AH Page 1 1.5m 0.5 m Internal Forces in the wires : EM,= 0; 2F,G – C(1.5)=0 EF, = 0; F,+F,6-C =0 1875 N F 625 N AH EM, =0; 1.5F, -0.5F, =0 208.33 N CF AH EF = 0; F +Fg-FH = F, N 416.66 DE
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
answer missing box
![Displacement :
FOLDE
0.21588
mm
DE
mm
0.108
Ac E
0.108
56
mm
1
1.5
8g = 6g + ốc =
mm
0.18
tan α
0.108000
Ans: a =
1500
mm
0.1943
AH
mm
0.26
mm
1.62
tan 6
1.619
Ans: B =
0.03
2000](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F36041524-0a65-49c4-957a-f2a779038cf9%2F90187959-06d7-436c-adf8-54e5e870d723%2F69swrts_processed.png&w=3840&q=75)
Transcribed Image Text:Displacement :
FOLDE
0.21588
mm
DE
mm
0.108
Ac E
0.108
56
mm
1
1.5
8g = 6g + ốc =
mm
0.18
tan α
0.108000
Ans: a =
1500
mm
0.1943
AH
mm
0.26
mm
1.62
tan 6
1.619
Ans: B =
0.03
2000
![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
Take:
A=
1.5
m
A meter
B=
2.5
m
C=
2500
N
B meter
1 1o.5 m
1 m
0.9 m
1.5 m
0.5 m
Solution:
0. 5m
Im
F
DE
F
CF
0.5 m
1 m
1.5m
F
0.5m
Page 1
1.5m
0.5 m
Internal Forces in the wires :
EM, = 0; 2Fac – C(1.5)= 0
EF, = 0; F+F -C = 0
F
F
1875
N
BG
N
625
AH
AH
Хм, - 0; 1.5F, -0.5F, -0
EF, = 0; F +F -FH
F
N
=
208.33
AH
= 0
416.66
N
DE
E E Z](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F36041524-0a65-49c4-957a-f2a779038cf9%2F90187959-06d7-436c-adf8-54e5e870d723%2F2qzwu9_processed.png&w=3840&q=75)
Transcribed Image Text: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
Take:
A=
1.5
m
A meter
B=
2.5
m
C=
2500
N
B meter
1 1o.5 m
1 m
0.9 m
1.5 m
0.5 m
Solution:
0. 5m
Im
F
DE
F
CF
0.5 m
1 m
1.5m
F
0.5m
Page 1
1.5m
0.5 m
Internal Forces in the wires :
EM, = 0; 2Fac – C(1.5)= 0
EF, = 0; F+F -C = 0
F
F
1875
N
BG
N
625
AH
AH
Хм, - 0; 1.5F, -0.5F, -0
EF, = 0; F +F -FH
F
N
=
208.33
AH
= 0
416.66
N
DE
E E Z
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