Question

A square shaped slab of lead with side $50\text{cm}$ and thickness $10.0\text{cm}$ is subjected to a shearing force (on its narrow face) of magnitude $9.0\times {10}^4\text{N}$. The lower edge is riveted to the floor as shown in figure. How much is the upper edge displaced, if the shear modulus of lead is $5.6\times {10}^9\text{Pa}$?

• A. $4.8\times {10}^{-4}\text{m}$
• B. $3.2\times {10}^{-4}\text{m}$
• C. $1.6\times {10}^{-4}\text{m}$
• D. $0.8\times {10}^{-4}\text{m}$

Answer 1

The correct option is C $1.6\times {10}^{-4}\text{m}$

Given,
$L=50\text{cm}=50\times {10}^{-2}\text{m}$
$t=10\text{cm}=10\times {10}^{-2}\text{m}$
$G=5.6\times {10}^9\text{Pa}$
$F=9.0\times {10}^4\text{N}$
Area of the face on which force is applied,
$A=50\times 10=500{\text{cm}}^2=0.05{\text{m}}^2$

Let $\Delta L$ be the displacement of the upper edge of the slab, due to tangential force $F$ applied.
Then, $G=\frac{\left(\frac{F}{A}\right)}{\left(\frac{\Delta L}{L}\right)}$
$\Rightarrow \Delta L=\frac{FL}{GA}=\frac{9\times {10}^4\times 50\times {10}^{-2}}{5.6\times {10}^9\times 0.05}$
$\therefore \Delta L=1.6\times {10}^{-4}\text{m}$