An aluminum wire 2.0 m long and 2.0 mm in diameter supports a 10.0-kg fixture. What is the stress in the wire? Young's modulus for aluminum is $7.0 \times 10 ^ { 10 } \mathrm {~N} / \mathrm { m } ^ { 2 }$
A) $6.2 \times 107 \mathrm {~N} / \mathrm { m } _ { 2 }$
B) $1.2 \times 108 \mathrm {~N} / \mathrm { m } 2$
C) $9.3 \times 107 \mathrm {~N} / \mathrm { m } ^ { 2 }$
D) $3.1 \times 107 \mathrm {~N} / \mathrm { m } ^ { 2 }$

Question

An aluminum wire  2.0 m long and  2.0 mm in diameter supports a  10.0-kg  fixture. What is the stress in the wire? Young's modulus for aluminum is $7.0 \times 10 ^ { 10 } \mathrm {~N} / \mathrm { m } ^ { 2 }$
A) $6.2 \times 107 \mathrm {~N} / \mathrm { m } _ { 2 }$
B) $1.2 \times 108 \mathrm {~N} / \mathrm { m } 2$
C) $9.3 \times 107 \mathrm {~N} / \mathrm { m } ^ { 2 }$
D) $3.1 \times 107 \mathrm {~N} / \mathrm { m } ^ { 2 }$

Quizplus · Accepted Answer

Stress is calculated as force divided by area. The force is the weight of the fixture, $10.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 98 \, \text{N}$. The cross-sectional area of the wire is $\pi \times (0.001 \, \text{m})^2$. Stress = $98 \, \text{N} / (\pi \times (0.001 \, \text{m})^2) = 3.1 \times 10^7 \, \text{N/m}^2$.

An Aluminum Wire 2 7.0×1010 N/m27.0 \times 10 ^ { 10 } \mathrm {~N} / \mathrm { m } ^ { 2 }7.0×1010 N/m2

An Aluminum Wire 2 $7.0 \times 10 ^ { 10 } \mathrm {~N} / \mathrm { m } ^ { 2 }$