28. Two independently normally distributed weight m~ N(10, 0.52) kg and force P ~ N(100, 1.52) N are
shown in the system. If the cable AB is in the position shown below, determine the probability that the
system may fail if the allowable tension of AB is 85 N.
C
45°
P
B
θ
W
5 ft
A
Solution
W
P
45°
θ
T
F
x
 0; P  mg cos 45  T sin   0;
F
y
Solve the above equations, we have
 0; mg sin 45  T cos  0;
 P  mg cos 45 
  23.83
 mg sin 45 
  arctan 
T  mg sin 45 / cos
T  m g sin 45 / cos  75.83 N
 T   m g sin 45 / cos  3.79
Therefore, we have
T ~ N (75.83, 3.792 ) N
The maximum tension of cable AB is 80 N, thus the probability that the system might fail is P(Y  0) ,
where Y  T  TMax
Finally, the probability is
P(Y  0)  1  P(Y  0)  1  (
Y
Y
)  0.78%
Ans.