steel wire hangs from a high-altitude balloon. The steel has unit weight of 77 kN/m $77 \mathrm { kN } / \mathrm { m } ^ { 3 }$ and yield stress of 280 MPa. The required factor of safety against yield is 2.0. The maximum permissible length of the wire is
Approximately:

Question

Quizplus · Accepted Answer

The maximum permissible length of the wire can be calculated using the stress due to the self-weight of the wire and the yield stress with the factor of safety applied. The stress ($\sigma$) due to the self-weight is given by $\sigma = \rho g h$, where $\rho$ is the density (unit weight in this case), $g$ is the acceleration due to gravity (approximately $9.81 \, \text{m/s}^2$), and $h$ is the height or length of the wire. The yield stress is given as 280 MPa, and with a factor of safety of 2, the allowable stress is $280 \, \text{MPa} / 2 = 140 \, \text{MPa}$.Given:- $\rho = 77 \, \text{kN/m}^3 = 77 \times 10^3 \, \text{N/m}^3$- $g = 9.81 \, \text{m/s}^2$- Allowable stress = $140 \, \text{MPa} = 140 \times 10^6 \, \text{N/m}^2$Solving for $h$:$$140 \times 10^6 = 77 \times 10^3 \times 9.81 \times h$$$$h = \frac{140 \times 10^6}{77 \times 10^3 \times 9.81}$$$$h \approx 1842 \, \text{m}$$Therefore, the maximum permissible length of the wire is approximately 1800 m, which is closest to option A.

Steel Wire Hangs from a High-Altitude Balloon 77kN/m377 \mathrm { kN } / \mathrm { m } ^ { 3 }77kN/m3

Steel Wire Hangs from a High-Altitude Balloon $77 \mathrm { kN } / \mathrm { m } ^ { 3 }$