Q.A metal cube of side $10\ \text{cm}$ has its lower face fixed to a rigid table. A tangential force of $4000\ \text{N}$ is applied uniformly across its upper face, whose area is $0.01\ \text{m}^2$. If the modulus of rigidity of the material is $\eta = 8\times10^{10}\ \text{Pa}$, find

Given: cube side (height) $L = 0.10\ \text{m}$, tangential force $F = 4000\ \text{N}$ on the top face of area $A = 0.01\ \text{m}^2$, $\eta = 8\times10^{10}\ \text{Pa}$.

1. Shearing stress:

   $\text{Shear stress} = \frac{F}{A} = \frac{4000}{0.01} = 4.0\times10^5\ \text{Pa}$

2. Shearing strain, from $\eta = \text{shear stress}/\text{shear strain}$:

   $\text{Shear strain} = \frac{\text{Shear stress}}{\eta} = \frac{4.0\times10^5}{8\times10^{10}} = 5.0\times10^{-6}$

3. Lateral displacement, using shear strain $= \Delta x/L$:

   $\Delta x = \text{Shear strain} \times L = 5.0\times10^{-6} \times 0.10 = 5.0\times10^{-7}\ \text{m}$

   This very small displacement (half a micrometre) illustrates why solids, with their large shear moduli, deform only imperceptibly under ordinary shearing forces.
   ✓Final answer

   Shear stress $= 4.0\times10^5\ \text{Pa}$; shear strain $= 5.0\times10^{-6}$; lateral displacement $= 5.0\times10^{-7}\ \text{m}$.

Divide the tangential force by the face area to get shear stress; divide shear stress by the given modulus of rigidity $\eta$ to get shear strain; multiply shear strain by the cube's height $L$ to get the lateral displacement $\Delta x$.

• Using the wrong area (the face the force acts on, not a side face) when computing shear stress.
• Forgetting that shear strain is dimensionless (radians), so multiplying it by $L$, not dividing, gives the displacement $\Delta x$.