To solve the problem step by step, we will follow these calculations:
### Step 1: Determine the Area of the Top Surface
The dimensions of the rubber eraser are given as 3 cm x 1 cm. The area \( A \) of the top surface can be calculated as:
\[
A = \text{length} \times \text{width} = 3 \, \text{cm} \times 1 \, \text{cm} = 3 \, \text{cm}^2
\]
To convert this area into square meters:
\[
A = 3 \, \text{cm}^2 = 3 \times 10^{-4} \, \text{m}^2
\]
### Step 2: Calculate Shear Stress
Shear stress \( \sigma \) is defined as the force \( F \) applied divided by the area \( A \):
\[
\sigma = \frac{F}{A}
\]
Given \( F = 2.1 \, \text{N} \) and \( A = 3 \times 10^{-4} \, \text{m}^2 \):
\[
\sigma = \frac{2.1 \, \text{N}}{3 \times 10^{-4} \, \text{m}^2} = 0.7 \times 10^{4} \, \text{N/m}^2 = 7000 \, \text{N/m}^2
\]
### Step 3: Calculate Shear Strain
The shear strain \( \theta \) can be calculated using the relationship between shear stress and shear modulus \( G \):
\[
\theta = \frac{\sigma}{G}
\]
Given \( G = 1.4 \times 10^{5} \, \text{N/m}^2 \):
\[
\theta = \frac{7000 \, \text{N/m}^2}{1.4 \times 10^{5} \, \text{N/m}^2} = 0.05 \, \text{radians}
\]
### Step 4: Relate Shear Strain to Horizontal Displacement
The horizontal displacement \( x \) at the top face can be related to the angle \( \theta \) and the height \( h \) (which is 8 cm or 0.08 m):
\[
\tan(\theta) = \frac{x}{h}
\]
For small angles, \( \tan(\theta) \approx \theta \):
\[
\theta \approx \frac{x}{h}
\]
Thus, we can rearrange to find \( x \):
\[
x = \theta \cdot h
\]
Substituting \( \theta = 0.05 \, \text{radians} \) and \( h = 0.08 \, \text{m} \):
\[
x = 0.05 \cdot 0.08 = 0.004 \, \text{m}
\]
### Step 5: Convert Displacement to Millimeters
To convert \( x \) from meters to millimeters:
\[
x = 0.004 \, \text{m} = 4 \, \text{mm}
\]
### Final Answer
The horizontal displacement \( x \) of the top face is:
\[
\boxed{4 \, \text{mm}}
\]
