A Bingham plastic (plastic viscosity μ = 10 Pa·s, yield stress τ0 = 10 kPa) is sheared between parallel plates separated by 1×10^-3 m. The top plate moves at 1 m/s. What shear stress acts on the moving plate?

Introduction / Context:
Bingham plastic fluids (e.g., some slurries, paints) exhibit a yield stress plus a linear viscous term once yielded. Computing shear stress in simple shear between plates tests understanding of constitutive laws and basic kinematics of shear flow.

Given Data / Assumptions:

• Plastic viscosity μ = 10 Pa·s.
• Yield stress τ0 = 10 kPa = 10,000 Pa.
• Gap h = 1×10^-3 m; top plate speed U = 1 m/s; bottom plate stationary.
• Fully yielded uniform shear assumed (no plug zone with these numbers).

Concept / Approach:

Bingham law: τ = τ0 + μ * (du/dy) for |τ| ≥ τ0. For simple Couette flow, shear rate du/dy ≈ U/h when fully yielded. First verify τ exceeds τ0 to ensure yielding; then compute τ directly from the linear relation.

Step-by-Step Solution:

Verification / Alternative check:

Check yield: τ (20 kPa) > τ0 (10 kPa), so the fluid is indeed yielded across the gap. Hence the Couette assumption is consistent and the calculated τ is valid.

Why Other Options Are Wrong:

10 kPa: Counts only yield, ignores viscous term. 30 kPa or 40 kPa: Overestimates by adding extra multiples of the viscous term not supported by the data.

Common Pitfalls:

Using Newtonian formula τ = μ*(du/dy) and forgetting τ0; mis-converting kPa to Pa; using the wrong gap or speed units.

Final Answer:

20 kPa