Question

A square aluminium (shear modulus is $25 \times 10^9 Nm ^{-2}$ ) slab of side $60 \,cm$ and thickness $15\, cm$ is subjected to a shearing force (on its narrow face) of $18.0 \times 10^4 \,N$. The lower edge is riveted to the floor. The displacement of the upper edge is ______$\mu m$.

Answer 1

$ \frac{ F }{ A }=\eta \frac{ x }{\ell} \Rightarrow \frac{ F \ell}{ A \eta}= x$
$ \Rightarrow x =\frac{18 \times 10^4 \times 60 \times 10^{-2}}{60 \times 10^{-2} \times 15 \times 10^{-2} \times 25 \times 10^9}$
$ =48 \times 10^{-6} m =48\, \mu m$