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^o to the cube causes the cube to deform through an angle of 1.24^o.

The force of magnitude $F=1375N$

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

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

Shear Modulus is defined as the ratio of shear stress and shear strain.

$\mathit{S}\mathbf{=}\frac{\mathbf{Shear}\mathbf{}\mathbf{stress}}{\mathbf{Shear}\mathbf{}\mathbf{strain}}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{II}}\mathbf{/}\mathbf{A}}{\mathbf{x}\mathbf{/}\mathbf{h}}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{II}}\mathbf{h}}{\mathbf{Ax}}$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦.(1)

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

Use equation (1) such that,

$S=\frac{F_{âˆ\yen }}{A\mathrm{φ}}$

Where F_II 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.

$\begin{array}{rcl}{F}_{âˆ\yen }& =& 1375N\cos \left(8.5°\right)\\ & =& 1360N\end{array}$

and

$\begin{array}{rcl}\mathrm{φ}& =& 1.24°\left(\frac{\mathrm{π}rad}{180°}\right)\\ & =& 0.0216rad\end{array}$

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

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

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