A mine elevator is supported by a 2.2 cm diameter steel cable (Ultimate Strength
ID: 1699654 • Letter: A
Question
A mine elevator is supported by a 2.2 cm diameter steel cable (Ultimate Strength = 480 MN/m2). The mass of the elevator and its contents is 710 kg. The modulus of elasticity of the cable is 570x103 MN/m2.
A) If the elevator is stationary, by how much is the cable stretched when the elevator is 248 m below Earth's surface?
A: 0.7972 (cm)
B) If the elevator is accelerating down at 3.0 m/s2, by how much is the cable stretched when the elevator is 248 m below Earth's surface? (cm)
A: 0.55340 (cm)
C) If the elevator is accelerating up at 3.0 m/s2, by how much is the cable stretched when the elevator is 248 m below Earth's surface?
A: 1.040996 (cm)
D) For the three cases examined in parts A-C, what is the minimum factor of safety?
Explanation / Answer
The radius of the cable r = 1.1cm or 0.011m
The mass of the elevator m = 710kg
The modulus of elasticity E = 570000*10^6 N/m^2
The length of the cable l = 248m
(a) If the elevator is stationary
then the stretch of the cable
l = (1/E)(F/A)l
= (1/E)(mg/r^2)l
=( 1/570000*10^6)(710*9.8/(0.011)^2)(248)
= 79679*10^-6 m
= 0.7972cm
(b) if the elevator is moving down at a =3.0m/s^2
Then the stretch of the cable is
l = (1/E) [m(g-a)/ r^2]l
= (1/570000*10^6)[ (710)(9.8-3)/(0.011)^2] 248
= 5528*10^-6 m
= 0.553 cm
(c) If the elevator is moving up at a = 3.0m/s^2
Then the stretch of the cable is
l = (1/E)[ (710)(9.8+3)/(0.011)^2] 248
= (1/570000*10^6)[ (710)(12.8)/(0.011)^2] (248)
= 10409.96*10^-6 m
= 1.040996 cm
(d)
If the elevator is stationary
then the stretch of the cable
l = (1/E)(F/A)l
= (1/E)(mg/r^2)l
=( 1/570000*10^6)(480*10^6)(248)
= 0.2088m or 20.88cm
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