Answer
(a) $7.3KN$
(b) reduced by a factor of 2
Work Step by Step
(a) We know that
$F=S(\frac{\Delta x}{L_{\circ}})A$
We plug in the known values to obtain:
$F=(8.1\times 10^{10}N/m^2)(\frac{1.1\times 10^{-4}m}{0.28m})(2.3\times 10^{-4}m^2)$
$F=7.3KN$
(b) We know that $S\propto A$. This relation shows that the cross-sectional area and shear deformation are mutually inversely proportional. Thus, if the cross sectional area is doubled then the shear deformation will decrease by a factor of 2.
