The maximum length of a steel cable that can hang vertically supported from one end can be determined using the concepts of tensile strength and Young’s modulus of the material.
Given that:
- Young’s Modulus (E) of steel = 2.0 x 1011 N/m2
- Tensile Strength (σmax) of steel = 8.0 x 108 N/m2
- Density (ρ) of steel = 7.9 x 103 kg/m3
The maximum tensile stress that the cable can withstand before failing is equal to its tensile strength:
σmax = F/A
Where F is the force acting on the cable and A is its cross-sectional area. The weight of the cable itself will create tension in the cable:
F = m * g = ρ * V * g = ρ * A * L * g
Where:
- V is the volume of the cable,
- L is the length of the cable,
- g is the acceleration due to gravity (approximately 9.81 m/s2).
Now, substituting this into our initial tensile strength equation, we have:
σmax = (ρ * A * L * g) / A
Which simplifies to:
σmax = ρ * L * g
Now we can solve for L:
L = σmax / (ρ * g)
Substituting in the values:
L = (8.0 x 108 N/m2) / ((7.9 x 103 kg/m3) * 9.81 m/s2)
Calculating the above:
L ≈ 1.01 x 105 m
Therefore, the maximum length of the steel cable that can hang vertically supported from one end is approximately 101,000 meters or 101 kilometers.