Question

A cube has side length \(5 \text{ cm}\) and modulus of rigidity \(10^5 \text{ N/m}^2\). The displacement produced by a force of \(10 \text{ N}\) in the upper face of the cube is ____ \text{mm}.

Answer 1

Correct Answer: 2

Step 1: Understanding the Question:
The question asks for the lateral (shear) displacement (\(\Delta x\)) of the top face of a cube when a tangential force is applied. This involves the concept of the modulus of rigidity (\(\eta\)), which is the ratio of shear stress to shear strain.
Step 2: Key Formula or Approach:
The formula for the modulus of rigidity is:
\[ \eta = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{F / A}{\Delta x / L} \]
Rearranging for displacement (\(\Delta x\)):
\[ \Delta x = \frac{F \cdot L}{A \cdot \eta} \]
Step 3: Detailed Explanation:
Given data:
Side length of the cube, \(L = 5 \text{ cm} = 0.05 \text{ m}\).
Area of the upper face, \(A = L^2 = (0.05 \text{ m})^2 = 0.0025 \text{ m}^2 = 25 \times 10^{-4} \text{ m}^2\).
Applied force, \(F = 10 \text{ N}\).
Modulus of rigidity, \(\eta = 10^5 \text{ N/m}^2\).
Substitute the values into the rearranged formula:
\[ \Delta x = \frac{10 \times 0.05}{(25 \times 10^{-4}) \times 10^5} \]
\[ \Delta x = \frac{0.5}{25 \times 10} \]
\[ \Delta x = \frac{0.5}{250} \text{ m} \]
\[ \Delta x = 0.002 \text{ m} \]
Convert the result from meters to millimeters:
\[ \Delta x = 0.002 \times 1000 \text{ mm} = 2 \text{ mm} \]
Step 4: Final Answer:
The displacement produced is \(2 \text{ mm}\).