Answer
$4.4\times10^{-6}m$
Work Step by Step
Let's apply equation 10.18 $F=S(\frac{\Delta X}{L_{0}})A$ Where $S$ - Shear modulus of brass, $L_{0}$ -Distance between the upper and lower surfaces, $A$ - Cross sectional area of surface. So we can write,
$\Delta X=\frac{FL_{0}}{SA}$ ; Let's plug known values into this equation.
$\Delta X=\frac{(770\space N)(0.04\space m)}{(3.5\times10^{10}N/m^{2})(0.01\space m)(0.02\space m)}\approx4.4\times10^{-6}m$
