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 `=8.4xx10^10Nm^-2`.

Explanation

To solve the problem, we need to find the lateral displacement (Δx) of the upper surface of the steel plate when a tangential force is applied. We will use the relationship between shear stress, shear strain, and the modulus of rigidity (rigidity modulus) to arrive at the solution.

### Step-by-Step Solution:

1. **Convert the given dimensions to SI units:**
- Face area \( A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2 \)
- Thickness \( d = 0.5 \, \text{cm} = 0.005 \, \text{m} \)

2. **Calculate the shear stress (σs):**
- The formula for shear stress is given by:
\[ \sigma_s = \frac{F}{A} \]
- Where \( F = 10 \, \text{N} \) (the applied force).
- Substituting the values:
\[ \sigma_s = \frac{10 \, \text{N}}{4 \times 10^{-4} \, \text{m}^2} = 25000 \, \text{N/m}^2 \]

3. **Calculate the shear strain (εs):**
- The formula for shear strain is given by:
\[ \epsilon_s = \frac{\Delta x}{d} \]
- Rearranging gives:
\[ \Delta x = \epsilon_s \cdot d \]

4. **Use the modulus of rigidity (G) to relate shear stress and shear strain:**
- The relationship is given by:
\[ G = \frac{\sigma_s}{\epsilon_s} \]
- Rearranging gives:
\[ \epsilon_s = \frac{\sigma_s}{G} \]

5. **Substituting the known values:**
- Given that the rigidity modulus \( G = 8.4 \times 10^{10} \, \text{N/m}^2 \):
\[ \epsilon_s = \frac{25000 \, \text{N/m}^2}{8.4 \times 10^{10} \, \text{N/m}^2} \approx 2.976 \times 10^{-7} \]

6. **Calculate the lateral displacement (Δx):**
- Now substituting back to find Δx:
\[ \Delta x = \epsilon_s \cdot d = (2.976 \times 10^{-7}) \cdot (0.005) \approx 1.488 \times 10^{-9} \, \text{m} \]

7. **Final result:**
- The lateral displacement of the upper surface with respect to the lower surface is approximately:
\[ \Delta x \approx 1.5 \times 10^{-9} \, \text{m} \]