91Ó°ÊÓ

Given that in lab tests on a 9.25 cm cube of a certain material, a force of 1375 N directed at $8.50°$to the cube causes the cube to deform through an angle of $1.24°$.

The force of magnitude F = 1375 N

The side of the cube is $\mathrm{a}=9.25\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.0925\mathrm{m}\mathrm{m}$m

Area of cube $\mathrm{A}={\mathrm{a}}^2={\left(0.0925\mathrm{m}\right)}^2$

Shear Modulus is defined as the ratio of shear stress and shear strain. …………….(1)

$\mathit{S}\mathbf{=}\frac{\mathbf{\text{Shear stress}}}{\mathbf{\text{Shear strain}}}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{a}}\mathbf{/}\mathbf{A}}{\mathbf{φ}}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{◻}}}{\mathbf{A}\mathbf{φ}}$

Where F is the force applied tangent to the surface of the object, $\mathbf{φ}$ is shear strain, and A is the area over which force is exerted

Use equation (1) such that,

$\mathrm{S}=\frac{\mathrm{F}}{\mathrm{´¡Ï†}}$

Where F is the force applied tangent to the surface of the object, A is the area over which force is exerted, S is the shear modulus and$\mathrm{φ}$ is the shear strain.

$$\begin{gathered} \mathrm{F}=1375\mathrm{Ncos}(8.5°) \\ =1360\mathrm{N} \end{gathered}$$

and

$$\begin{gathered} \mathrm{φ}=1.24°\left(\frac{\mathrm{ττ}\mathrm{rad}}{180°}\right) \\ =0.0216\mathrm{rad} \end{gathered}$$

Therefore, by substituting the given information in equation (2), we get

$\begin{array}{r}\mathrm{S}=\frac{1360\mathrm{N}}{\left(0.0925\mathrm{m}{)}^2\right(0.0216\mathrm{rad})}\\ =7.36×{10}^6\mathrm{Pa}\end{array}$

Hence, the shear modulus is $7.36×{10}^6\mathrm{Pa}$