A mine elevator is supported by a 2.2 cm diameter steel cable (Ultimate Strength

Question

A mine elevator is supported by a 2.2 cm diameter steel cable (Ultimate Strength = 600 MN/m2). The mass of the elevator and its contents is 800 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 274 m below Earth's surface? (include units with answer)

B.) If the elevator is accelerating down at 3.1 m/s2, by how much is the cable stretched when the elevator is 274 m below Earth's surface? (include units with answer)

C.) If the elevator is accelerating up at 3.1 m/s2, by how much is the cable stretched when the elevator is 274 m below Earth's surface? (include units with answer)

D.) For the three cases examined in parts A-C, what is the minimum factor of safety?

I will rate, please include answers with work so i can understand though.

Explanation / Answer

L = 274 m

mg = 800*9.8 = 7840 N

A = pi*r^2 = 3.14*0.011*0.011 = 3.7994e-4 m^2

Young's modulus Y = stess/strain

stress = T/A

strain = e/L

Y = T*L/Ae

part A)

at stationary

T = m*g

stress = T/A = 7840/3.7994e-4 = 20.6 MN/m^2

e = stress*L/Y

e = 9.92*10^-3 m = 9.92 mm

part B)

T - mg = -ma

T = m*(g-a) = 800*(9.8-3.1) = 5360 N

stress = T/A = 5360/3.7994e-4 = 14.1 MN/m^2

e = stress*L/Y = 14.1e6*274/570e9

e = 6.78*10^-3 m = 6.78 mm

part C)

T - mg = ma

T = m*(g+a) = 800*(9.8+3.1) = 10320 N

stress = T/A = 10320/3.7994e-4 = 27.1 MN/m^2

e = stress*L/Y = 27.1e6*274/570e9

e = 13*10^-3 m = 13 mm

part D)

part C

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