$A$ square lead slab of side $50\, cm$ and thickness $10\, cm$ is subject to a shearing force (on its narrow face) of $9.0 \times 10^{4} \, N$. The lower edge is riveted to the floor. How much will the upper edge be displaced? (Given: Shear modulus of lead $G = 5.6 \times 10^{9} \, N/m^2$)

(N/A) The lead slab is fixed at the bottom,and a shearing force $F = 9.0 \times 10^{4} \, N$ is applied parallel to the top narrow face.
The area $A$ of the face on which the force is applied is:
$A = 50 \, cm \times 10 \, cm = 0.5 \, m \times 0.1 \, m = 0.05 \, m^2$
The shearing stress is given by:
$\text{Stress} = \frac{F}{A} = \frac{9.0 \times 10^{4} \, N}{0.05 \, m^2} = 1.8 \times 10^{6} \, N/m^2$
We know that the shearing strain is defined as:
$\text{Strain} = \frac{\Delta x}{L} = \frac{\text{Stress}}{G}$
Where $L = 50 \, cm = 0.5 \, m$ is the height of the slab and $G = 5.6 \times 10^{9} \, N/m^2$ is the shear modulus of lead.
Therefore,the displacement $\Delta x$ is:
$\Delta x = \frac{\text{Stress} \times L}{G} = \frac{1.8 \times 10^{6} \, N/m^2 \times 0.5 \, m}{5.6 \times 10^{9} \, N/m^2}$
$\Delta x = \frac{0.9 \times 10^{6}}{5.6 \times 10^{9}} \, m \approx 0.1607 \times 10^{-3} \, m$
$\Delta x \approx 1.6 \times 10^{-4} \, m = 0.16 \, mm$