A solid steel column is 4.0 m long and 0.20 m in diameter. Young's modulus for this steel is $2.0 \times$ $10 ^ { 11 } \mathrm {~N} / \mathrm { m } ^ { 2 }$
By what distance does the column shrink when a 5000-\mathrm{kg} truck is supported by it?
A) $3.2 \times 10 ^ { - 6 } \mathrm {~m}$
B) $7.8 \times 10 ^ { - 6 } \mathrm {~m}$
C) $3.1 \times 10 ^ { - 5 } \mathrm {~m}$
D) $8.0 \times 10 ^ { - 7 } \mathrm {~m}$

Question

A solid steel column is  4.0 m long and  0.20 m in diameter. Young's modulus for this steel is $2.0 \times$ $10 ^ { 11 } \mathrm {~N} / \mathrm { m } ^ { 2 }$
By what distance does the column shrink when a  5000-\mathrm{kg}  truck is supported by it?
A) $3.2 \times 10 ^ { - 6 } \mathrm {~m}$
B) $7.8 \times 10 ^ { - 6 } \mathrm {~m}$
C) $3.1 \times 10 ^ { - 5 } \mathrm {~m}$
D) $8.0 \times 10 ^ { - 7 } \mathrm {~m}$

Quizplus · Accepted Answer

The shrinkage can be calculated using the formula $\Delta L = \frac{F \cdot L}{A \cdot Y}$, where $F = mg$, $A = \pi r^2$, and $Y$ is Young's modulus. Substituting the given values, $\Delta L = \frac{(5000 \times 9.8) \times 4}{\pi \times (0.1)^2 \times 2.0 \times 10^{11}}$, which results in approximately $3.1 \times 10^{-5} \mathrm{~m}$.

A Solid Steel Column Is 4 2.0×2.0 \times2.0× 1011 N/m210 ^ { 11 } \mathrm {~N} / \mathrm { m } ^ { 2 }1011 N/m2

A Solid Steel Column Is 4 $2.0 \times$ $10 ^ { 11 } \mathrm {~N} / \mathrm { m } ^ { 2 }$