---
title: "Plate towed between two parallel walls"
subject: "Fluid Mechanics"
chapter: 1
key: "semester_3/fluid_mechanics:plate-between-walls-force"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/plate-between-walls-force
site: BCT Engineering
---

# Plate towed between two parallel walls

Fluid Mechanics · Chapter 1: Fluid and its physical properties

- Asked in: 2078 Bhadra, 8 marks
- Video: [Fluid Mechanics Exam Problem: Shear Stress Between Parallel Plates](https://www.youtube.com/watch?v=H00cycc0VNA) — Fluid Matters
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/plate-between-walls-force
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-1

## Question

The space between two large flat and parallel walls 25 mm apart is filled with a liquid of dynamic viscosity 0.7 Pa.s. Within this space a thin flat plate 250 mm $\times$ 250 mm is towed at a velocity of 150 mm/s at a distance of 6 mm from one wall, the plate and its movement being parallel to the walls. Assuming linear variations of velocity between the plate and the walls, determine the force exerted by the liquid on the plate.

## Short answer

Shear stresses on both sides: τ₁ = μ V / y₁ = 17.5 Pa, τ₂ = 5.526 Pa. Forces: F₁ = τ₁ A = 1.094 N, F₂ = 0.345 N. Total force opposing motion is F = 1.439 N ≈ 1.44 N.

## Answer

**Data:** $\mu=0.7$ Pa·s, $A=0.25\times0.25=0.0625\ \text{m}^2$, $V=0.15$ m/s. Total gap $=25$ mm; the plate is 6 mm from one wall, so its distance from the other wall is $25-6=19$ mm (plate thickness neglected).

```text
 wall |<-- 6 mm -->[plate]<---- 19 mm ---->| wall
```

With linear velocity profiles on both sides, the shear stresses are

$$
\tau_1=\mu\frac{V}{y_1}=0.7\times\frac{0.15}{0.006}=17.5\ \text{N/m}^2
$$

$$
\tau_2=\mu\frac{V}{y_2}=0.7\times\frac{0.15}{0.019}=5.526\ \text{N/m}^2
$$

Forces on the two faces (both oppose the motion):

$$
F_1=\tau_1A=17.5\times0.0625=1.094\ \text{N},\qquad F_2=\tau_2A=0.345\ \text{N}
$$

$$
F=F_1+F_2=1.094+0.345=1.439\ \text{N}
$$

**Answer: the force exerted by the liquid on the plate is about 1.44 N, opposing the motion.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How do the shear forces on the two faces of the towed plate act?
   A. Both oppose the motion and add up
   B. They cancel each other
   C. Only the narrow-gap side matters
   D. The wide-gap side assists the motion
   Answer: A. The liquid on both sides resists the plate's motion, so F = F1 + F2. (Key point: Shear on both faces)

2. What is the shear stress on the 6 mm side (mu = 0.7 Pa·s, V = 0.15 m/s)?
   A. 17.5 N/m^2
   B. 5.53 N/m^2
   C. 0.7 N/m^2
   D. 4.2 N/m^2
   Answer: A. tau = mu V / y = 0.7 x 0.15/0.006 = 17.5 N/m^2. (Key point: Shear stress on narrow side)

3. What is the total force exerted by the liquid on the plate?
   A. 1.09 N
   B. 2.88 N
   C. 0.345 N
   D. 1.44 N
   Answer: D. F = 17.5 x 0.0625 + 5.526 x 0.0625 = 1.094 + 0.345 = 1.44 N. (Key point: Total force)

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