A steel plate of face area 4 cm^{2} and thickness 0.5 cm is fixed rigidly at the lower surface. A tangential force of 10 N is applied on the upper surface. Find the lateral displacement of the upper surface with respect to the lower surface. Rigidity modulus of steel = 8.4 × 10^{10} N m^{−2}.

#### Solution

Given:

Face area of steel plate *A *= 4 cm^{2} = 4 × 10^{−4} m^{2}

Thickness of steel plate *d* = 0.5 cm = 0.5 × 10^{−2} m

Applied force on the upper surface *F* = 10 N

Rigidity modulus of steel = 8.4 × 10^{10} N m^{−2}

Let *θ* be the angular displacement.

Rigidity modulus \[\text{ m } = \frac{F}{A\theta}\]

\[\Rightarrow m = \left( \frac{10}{4 \times {10}^{- 4} \theta} \right)\]

\[ \Rightarrow \theta = \frac{10}{4 \times {10}^{- 4} \times 8 . 4 \times {10}^{10}}\]

\[ = 0 . 297 \times {10}^{- 6}\]

∴ Lateral displacement of the upper surface with respect to the lower surface = *θ* × *d*

⇒ (0.297) × 10^{−6} × (0.5) × 10^{−2}

⇒ 1.5 × 10^{−9} m

Hence, the required lateral displacement of the steel plate is 1.5 × 10^{−9} m.