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

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

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

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

∴ 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.