Answer
(a) $\Delta x = 0.28~cm$
(b) The shear modulus is $2.04\times 10^4~N/m^2$
Work Step by Step
(a) We can find $\Delta x$:
$\frac{\Delta x}{L} = tan~\gamma$
$\Delta x = L~tan~\gamma$
$\Delta x = (2.0~cm)~tan~8.0^{\circ}$
$\Delta x = 0.28~cm$
(b) $G = \frac{F/A}{\Delta x/L}$
$G$ is shear modulus
$F$ is the force
$A$ is the cross-sectional area parallel to the force
$\Delta x$ is the distance the surface is displaced
$L$ is the perpendicular length
We can find the shear modulus:
$G = \frac{F/A}{\Delta x/L}$
$G = \frac{F~L}{A~\Delta x}$
$G = \frac{(12~N)(2.0~cm)}{(42\times 10^{-4}~m^2)(0.28~cm)}$
$G = 2.04\times 10^4~N/m^2$
The shear modulus is $2.04\times 10^4~N/m^2$
