q2. suppose the object in figure (b) is the brass base plate of an outdoor sculpture that experiences shear…

Explanation:

Step1: Recall the shear - modulus formula

The shear - modulus formula is $S=\frac{F/A}{x/h}$, where $S$ is the shear modulus, $F$ is the force applied, $A$ is the area over which the force is applied, $x$ is the displacement, and $h$ is the height (or thickness) of the object. We want to solve for $F$. First, we can re - arrange the formula for $F$: $F = S\times\frac{A\times x}{h}$.

Step2: Calculate the area $A$

The plate is square with side length $L = 0.80\ m$. The area $A$ of a square is $A = L^{2}$. So, $A=(0.80\ m)^{2}=0.64\ m^{2}$.

Step3: Convert the given values to SI units

The thickness $h = 0.50\ cm=0.50\times10^{- 2}\ m$ and the displacement $x = 0.16\ mm = 0.16\times10^{-3}\ m$. The shear modulus $S = 3.5\times10^{10}\ Pa$.

Step4: Substitute the values into the formula for $F$

$F=S\times\frac{A\times x}{h}$. Substitute $S = 3.5\times10^{10}\ Pa$, $A = 0.64\ m^{2}$, $x = 0.16\times10^{-3}\ m$, and $h = 0.50\times10^{-2}\ m$ into the formula: [ \begin{align*} F&=(3.5\times10^{10}\ Pa)\times\frac{0.64\ m^{2}\times0.16\times10^{-3}\ m}{0.50\times10^{-2}\ m}\ &=(3.5\times10^{10})\times\frac{0.64\times0.16\times10^{-3}}{0.50\times10^{-2}}\ N\ &=(3.5\times10^{10})\times\frac{0.1024\times10^{-3}}{0.50\times10^{-2}}\ N\ &=(3.5\times10^{10})\times\frac{1.024\times10^{-4}}{0.50\times10^{-2}}\ N\ &=(3.5\times10^{10})\times(2.048\times10^{-3})\ N\ & = 7.168\times10^{7}\ N \end{align*} ]

Answer:

$7.168\times10^{7}\ N$