---
title: "Force to lift plate in liquid gap"
subject: "Fluid Mechanics"
chapter: 1
key: "semester_3/fluid_mechanics:plate-lifted-in-gap"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/plate-lifted-in-gap
site: BCT Engineering
---

# Force to lift plate in liquid gap

Fluid Mechanics · Chapter 1: Fluid and its physical properties

- Asked in: 2076 Asoj, 5 marks
- Video: [Example-3 | Plate Lifted in Oil | Newton’s Law of Viscosity | Fluid Mechanics Solved Problem](https://www.youtube.com/watch?v=A18-D_unBvQ) — EASY SCIENCE PREP
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/plate-lifted-in-gap
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-1

## Question

A 2.2 cm wide gap between two vertical plane surfaces is filled with liquid of specific gravity 0.9 and dynamic viscosity 1.75 N s/m². A metal plate 1.5 m $\times$ 1.5 m $\times$ 0.2 cm thick and weighing 40 N is placed midway in the gap. Find the force required if the plate is to be lifted with constant velocity of 0.15 m/s.

## Short answer

Clearance each side t = (2.2−0.2)/2 = 1 cm = 0.01 m. Viscous drag F_v = 2 μ A V/t = 2×1.75×2.25×0.15/0.01 = 118.13 N. Buoyancy F_b = ρ g V_p = 39.73 N. Force needed F = W − F_b + F_v = 40 − 39.73 + 118.13 = 118.4 N upward.

## Answer

**Data:** gap $=2.2$ cm, plate thickness $=0.2$ cm, so the clearance on each side is $\dfrac{2.2-0.2}{2}=1.0$ cm $=0.01$ m. $A=1.5\times1.5=2.25\ \text{m}^2$, $V=0.15$ m/s, $\mu=1.75$ N·s/m², $W=40$ N, $\rho=0.9\times1000=900\ \text{kg/m}^3$.

```text
  wall |  1 cm | plate | 1 cm |  wall
       |  oil  | 0.2cm | oil  |
                  ^ lifted at 0.15 m/s
```

**Viscous drag on both faces** (linear profile):

$$
F_v=2\mu A\frac{V}{t}=2\times1.75\times2.25\times\frac{0.15}{0.01}=118.13\ \text{N}
$$

**Buoyancy** (plate volume $=1.5\times1.5\times0.002=0.0045\ \text{m}^3$):

$$
F_b=\rho g V_p=900\times9.81\times0.0045=39.73\ \text{N}
$$

**Force balance** for upward motion at constant velocity:

$$
F+F_b=W+F_v\ \Rightarrow\ F=40-39.73+118.13=118.39\ \text{N}
$$

**Answer: $F\approx118.4$ N upward** (about 118 N; if buoyancy is ignored, $F=158.1$ N).

## Check yourself

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

1. Gap 2.2 cm, plate thickness 0.2 cm, plate midway. What is the oil clearance on each side?
   A. 2.2 cm
   B. 1.0 cm
   C. 1.1 cm
   D. 0.2 cm
   Answer: B. (2.2 - 0.2)/2 = 1.0 cm = 0.01 m. (Key point: Clearance each side)

2. Which expression gives the viscous drag on the plate?
   A. mu A V / t (one face)
   B. 2 mu A V / t (both faces)
   C. mu A t / V
   D. 2 mu A t / V
   Answer: B. Oil acts on both faces, so F_v = 2 mu A V/t = 118.1 N. (Key point: Viscous drag)

3. For upward motion at constant velocity, which equation holds?
   A. F = W + F_v + F_b
   B. F + F_b = W + F_v
   C. F = W - F_v - F_b
   D. F + W = F_v + F_b
   Answer: B. Upward force plus buoyancy equals weight plus downward viscous drag. (Key point: Direction of forces)

4. How does including buoyancy change the required force (118.4 N versus 158.1 N)?
   A. Buoyancy supports part of the weight, lowering F
   B. Buoyancy adds to the weight, raising F
   C. Buoyancy has no effect on F
   D. Buoyancy replaces the viscous drag
   Answer: A. The upward buoyant force of about 39.7 N reduces the required lifting force. (Key point: Buoyancy effect)

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