A shear force of 400 N is applied to one face of an aluminum cube having sides of 30 cm each. What is the resulting displacement of this face of the cube if the opposite face is held fixed? The shear modulus for aluminum is $2.5 \times 10 ^ { 10 } \mathrm {~N} / \mathrm { m } ^ { 2 }$
A) $4.4 \times 10 ^ { - 8 } \mathrm {~m}$
B) $5.3 \times 10 ^ { - 8 } \mathrm {~m}$
C) $1.9 \times 10 ^ { - 8 } \mathrm {~m}$
D) $8.2 \times 10 ^ { - 8 } \mathrm {~m}$

Question

A shear force of  400 N  is applied to one face of an aluminum cube having sides of  30 cm each. What is the resulting displacement of this face of the cube if the opposite face is held fixed? The shear modulus for aluminum is $2.5 \times 10 ^ { 10 } \mathrm {~N} / \mathrm { m } ^ { 2 }$
A) $4.4 \times 10 ^ { - 8 } \mathrm {~m}$
B) $5.3 \times 10 ^ { - 8 } \mathrm {~m}$
C) $1.9 \times 10 ^ { - 8 } \mathrm {~m}$
D) $8.2 \times 10 ^ { - 8 } \mathrm {~m}$

Quizplus · Accepted Answer

The displacement can be calculated using the formula for shear deformation: $\delta = \frac{F}{A} \cdot \frac{L}{G}$, where $F$ is the force applied, $A$ is the area, $L$ is the initial length (height of the cube in this case), and $G$ is the shear modulus. Given $F = 400 \, \text{N}$, $L = 0.3 \, \text{m}$ (since the cube's side is 30 cm), $G = 2.5 \times 10^{10} \, \text{N/m}^2$, and $A = 0.3 \, \text{m} \times 0.3 \, \text{m}$, the displacement $\delta$ can be calculated, resulting in $5.3 \times 10^{-8} \, \text{m}$.

A Shear Force of 400 N Is Applied 2.5×1010 N/m22.5 \times 10 ^ { 10 } \mathrm {~N} / \mathrm { m } ^ { 2 }2.5×1010 N/m2

A Shear Force of 400 N Is Applied $2.5 \times 10 ^ { 10 } \mathrm {~N} / \mathrm { m } ^ { 2 }$