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Q. A square lead slab of side 1 metre and thickness 1 metre is subjected to a shearing force on its narrow edge of $15 \times 10^{4} N$. The lower edge is riveted to the floor. How much will the upper edge be displaced?
$\left(G=5.6 \times 10^{9} N / m ^{2}\right)$

Mechanical Properties of Solids

Solution:

$\therefore G=\frac{F \times L}{A \times \Delta L}$
$\therefore \Delta L=\frac{F \times L}{A \times G}$
$=\frac{15 \times 10^{4} \times 1}{1 \times 5.6 \times 10^{9}}$
$=2.6 \times 10^{-5} m$