Question

A steel plate of face area 4 cm² 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 × 10¹⁰ N m⁻².

Answer 1

Given:
Face area of steel plate A = 4 cm² = 4 × 10⁻⁴ m²
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 × 10¹⁰ N m⁻²
Let θ be the angular displacement.
Rigidity modulus $m=\frac{F}{A\theta }$
$$\begin{gathered} \Rightarrow m=\left(\frac{10}{4\times {10}^{-4}\theta }\right) \\ \Rightarrow \theta =\frac{10}{4\times {10}^{-4}\times 8.4\times {10}^{10}} \\ =0.297\times {10}^{-6} \end{gathered}$$

∴ 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.