Answer
$(a)\space 710\space N/m^{2}$
$(b)\space 3.6\times10^{-8}$
$(c)\space 3.6\times10^{-10}m$
Work Step by Step
(a) According to equation 4.7 $f_{s}^{MAX}=\mu_{s}F_{N}$ We can get the shear force.
$F=f_{s}^{MAX}=\mu_{s}F_{N}=\mu_{s}mg$
Maximum possible = $\frac{F}{A}=\frac{\mu_{s}mg}{A}=\frac{(0.91)(7.2\times10^{-2}kg)(9.8\space m/s^{2})}{(3\times10^{-2}m)^{2}}=710\space N/m^{2}$
shear stress
(b) Maximum possible = $\frac{\Delta X}{L_{0}}=\frac{710\space N/m^{2}}{S}=\frac{710\space N/m^{2}}{2\times10^{10}N/m^{2}}=3.6\times10^{-8}$
shear strain
(c) We can write,
$\Delta X=Maximum\space possible\space shear\space strain\times L_{0}$
$\Delta X=(3.6\times10^{-8})(1\times10^{-2}m)=3.6\times10^{-10}m$
So, Maximum possible amount of shear deformation = $3.6\times10^{-10}m$
