6. Two independent forces F1~N(12,0.52) kg and F2~N(18,0.52) kg act at points B and C of cable ABCD.
(1) Determine the tension in the cable CD. (2) If the maximum load of the cable is 210 N, what is the
probability the cable CD might fail?
A
D
4m
4m
B
C
F2
F1
3m
5m
1m
Solution
3m
5m
A

TCD
4m
75.96°
F1
F2
(1) From FBD, we have:
M
A
 0; TCD cos75.96 (4)  TCD sin 75.96 (8)  F2 (9.81)(8)  F1 (9.81)(3)  0
Then, we can obtain the distribution of TCD:
T 
CD
 F (9.81)(8)   F (9.81)(3)
2
1
cos75.96 (4)  sin 75.96 (8)
T 
CD
 202.23 N
(9.81)  2 F2 (64)   2 F1 (9)
cos75.96 (4)  sin 75.96 (8)
 4.80
Thus, the distribution of cable CD is:
TCD ~ N (202.23, 4.802 ) N .
Ans.
(2) If the maximum load of the cable CD is 210N, then the probability that the system might fail is:
P(TCD  210)  1  P(TCD  210)  1  (
210  202.23
)  0.0529
4.8
Ans.