Tensile stress and strain: A 1000-kg object hangs from the lower end of a steel rod 5.0 m long that is suspended vertically. The diameter of the rod is 0.80 cm and Young's modulus for the rod is 210,000 MN/m². What is the elongation of the rod due to this object? A) 1.2 cm B) 0.46 cm C) 0.12 cm D) 1.8 cm E) 0.047 cm

Question

Quizplus · Accepted Answer

The elongation ($\Delta L$) of the rod can be calculated using the formula $\Delta L = \frac{FL}{AE}$, where $F$ is the force applied (weight of the object), $L$ is the original length of the rod, $A$ is the cross-sectional area of the rod, and $E$ is Young's modulus for the material of the rod. Given that the weight of the object is $1000 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 9800 \, \text{N}$, the length of the rod is $5.0 \, \text{m}$, the diameter of the rod is $0.80 \, \text{cm} = 0.008 \, \text{m}$, and Young's modulus is $210,000 \, \text{MN/m}^2 = 210 \times 10^9 \, \text{N/m}^2$, the cross-sectional area $A = \pi r^2 = \pi (0.004)^2 \, \text{m}^2$. Substituting these values into the formula gives the elongation, which is approximately $0.46 \, \text{cm}$.

Tensile Stress and Strain: a 1000-Kg Object Hangs from the Lower