Answer
The magnitude of the tangential force is $0.301~N$
Work Step by Step
$G = \frac{F/A}{\Delta L/L}$
$G$ is shear modulus
$F$ is the force
$A$ is the cross-sectional area parallel to the force
$\Delta L$ is the distance the surface is displaced
$L$ is the perpendicular length
We can find the required force:
$F = \frac{G~A~\Delta L}{L}$
$F = \frac{(940~N/m^2)(5.0\times 10^{-2}~m)^2(0.64\times 10^{-2}~m)}{5.0\times 10^{-2}~m}$
$F = 0.301~N$
The magnitude of the tangential force is $0.301~N$.
