To determine the shear stress in \( \text{N/m}^2 \), which is equivalent to Pascals (Pa), we have already performed the necessary calculations using consistent SI units. Since 1 Pa is equal to 1 N/m², the shear stress we calculated is directly in N/m².

Here are the steps once again, confirming that the final result is in N/m²:

1. **Velocity Distribution Equation**:
   \[ u = \frac{3}{2} y - y^{3/2} \]

2. **Velocity Gradient Calculation**:
   \[ \frac{du}{dy} = \frac{3}{2} - \frac{3}{2} y^{1/2} \]
   At \( y = 0.09 \) m:
   \[ \frac{du}{dy} \bigg|_{y=0.09} = \frac{3}{2} - \frac{3}{2} \sqrt{0.09} \]
   \[ \frac{du}{dy} \bigg|_{y=0.09} = \frac{3}{2} - \frac{3}{2} \times 0.3 \]
   \[ \frac{du}{dy} \bigg|_{y=0.09} = 1.5 - 0.45 \]
   \[ \frac{du}{dy} \bigg|_{y=0.09} = 1.05 \, \text{m/s} \]

3. **Shear Stress Calculation**:
   Dynamic viscosity \( \mu = 8 \) poise = \( 8 \times 0.1 \) Pa·s = 0.8 Pa·s.
   \[ \tau = \mu \frac{du}{dy} \]
   \[ \tau = 0.8 \times 1.05 \]
   \[ \tau = 0.84 \, \text{Pa} \]

Therefore, the shear stress at \( y = 9 \) cm is \( 0.84 \) Pa, which is equivalent to \( 0.84 \) N/m².

Would you like more details or have any questions?

Here are 5 numbered questions you might want to ask:

1. How do you convert dynamic viscosity from poise to Pascal-seconds?
2. Can you explain the concept of shear stress in fluid mechanics?
3. What is the significance of the velocity gradient in calculating shear stress?
4. How would the shear stress change if the dynamic viscosity were different?
5. Can you provide a step-by-step explanation of taking the derivative of the velocity equation?

Tip: Always ensure units are consistent when performing calculations involving physical quantities.

Learn how to calculate shear stress in fluid mechanics using the velocity distribution equation, velocity gradient, and dynamic viscosity. This problem demonstrates the application of these concepts in determining shear stress at a specific point within a fluid flow.

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