A vertical 30-cm steel rod,1.0 cm in diameter,supports a 300-kg mass.What is the change in length of the rod caused by this mass? Young's modulus for steel is 2.0 × 10¹¹ N/m².

Question

Quizplus · Accepted Answer

The change in length ($\Delta L$) of the rod can be calculated using the formula $\Delta L = \frac{FL}{AE}$, where $F$ is the force applied, $L$ is the original length of the rod, $A$ is the cross-sectional area, and $E$ is Young's modulus. The force $F$ is the weight of the mass, $F = mg$, where $m = 300$ kg and $g \approx 9.8$ m/s$^2$. The cross-sectional area $A = \pi r^2$, where $r = 0.5$ cm = $0.005$ m. Plugging in the values: $F = 300 \times 9.8$, $A = \pi \times (0.005)^2$, $L = 0.3$ m, and $E = 2.0 \times 10^{11}$ N/m$^2$, we find $\Delta L = 5.6 \times 10^{-5}$ m.

A Vertical 30-Cm Steel Rod,1