Question

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

• A. 0.16 mm
• B. 1.6 mm
• C. 0.16 cm
• D. 1.6 cm

Answer 1

Answer: (A)

0.16 mm

Answer 2

: The lead slab is fixed and force is applied parallel to the narrow face as shown in the figure.

Area of the face parallel to which this force is applied is

$A = 50cm \times 10cm = 0.5m \times 0.1m = 0.05m^{2}$

If $\Delta L$is the displacement of the upper edge of the slab due to tangential force F, then

$\eta = \frac{F/A}{\Delta L/L}or\Delta L = \frac{FL}{\eta A}$

Substituting the given values, we get

$\Delta L = \frac{(9 \times 10^{4}N)(0.5m)}{5.6 \times 10^{9}Nm^{- 2} \times 0.05m^{2}}$

$= 1.6 \times 10^{- 4}m = 0.16mm$