Q11-37E - University Physics with Modern Physics

Question

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. What is the shear modulus of material?

Step-by-Step Solution

Answer

Shear modulus is $7.36\times {10}^6$ Pa

Step 1: The given data

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 $F=1375 N$

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

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

Step 2: Formula used

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

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

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

Step 3: Find Shear modulus

Use equation (1) such that,

$S=\frac{F_{\parallel }}{A\phi }$

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 $\phi$  is the shear strain.

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

and

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

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

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

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