Displacement of Cube Upper Face Under Shearing Force

To find the displacement produced in the upper face of the cube, we use the definition of the modulus of rigidity ($\eta$):

$\eta = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{\left(\dfrac{F}{A}\right)}{\left(\dfrac{\Delta x}{L}\right)}$

where:

• $F = 10 \text{ N}$ is the tangential force applied to the upper face,
• $L = 5 \text{ cm} = 5 \times 10^{-2} \text{ m}$ is the side length (height) of the cube,
• $A = L^2 = (5 \times 10^{-2} \text{ m})^2 = 25 \times 10^{-4} \text{ m}^2$ is the area of the face,
• $\eta = 10^5 \text{ N/m}^2$ is the modulus of rigidity,
• $\Delta x$ is the lateral displacement of the upper face.

Rearranging the formula to solve for the displacement $\Delta x$:

$\Delta x = \frac{F \cdot L}{A \cdot \eta} = \frac{F}{L \cdot \eta}$

Substituting the given values into the expression:

$\Delta x = \frac{10}{5 \times 10^{-2} \times 10^5}$

$\Delta x = \frac{10}{5000} = \frac{1}{500} \text{ m} = 0.002 \text{ m}$

Converting the displacement to millimeters ($\text{mm}$):

$\Delta x = 0.002 \times 10^3 \text{ mm} = 2 \text{ mm}$

Thus, the displacement produced in the upper face of the cube is $2$ $\text{mm}$.