---
title: "Plate sliding on inclined oil film"
subject: "Fluid Mechanics"
chapter: 1
key: "semester_3/fluid_mechanics:incline-plate-oil-viscosity"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/incline-plate-oil-viscosity
site: BCT Engineering
---

# Plate sliding on inclined oil film

Fluid Mechanics · Chapter 1: Fluid and its physical properties

- Asked in: 2079 Baisakh, 4 marks
- Video: [Fluid Mechanics 1.4 - Viscosity Problem with Solution - Terminal Velocity on Inclined Plate](https://www.youtube.com/watch?v=0Y3VvPiF7hg) — Rasim Guldiken
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/incline-plate-oil-viscosity
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-1

## Question

A square plate of size 1 m $\times$ 1 m and weighing 350 N slides down an inclined plane with a uniform velocity of 1.5 m/s. The inclined plane is laid on a slope of 5 vertical to 12 horizontal and has an oil film of 1 mm thickness. Calculate the dynamic viscosity of oil.

## Short answer

Weight component along slope: F = W sinθ = 350 × 5/13 = 134.62 N. With linear velocity profile, τ = μ V/t = F/A, so μ = F t/(A V) = 134.62 × 0.001 / (1 × 1.5) = 0.0897 Pa·s.

## Answer

**Data:** $A=1\ \text{m}^2$, $W=350$ N, $V=1.5$ m/s, $t=1\ \text{mm}=0.001$ m, $\tan\theta=5/12$ so $\sin\theta=5/13$.

```text
        /|
       / |  5
      /  |
     /___|
       12
  plate slides down, oil film t = 1 mm
```

At uniform velocity there is no acceleration, so the shear force from the oil balances the component of weight along the plane:

$$
F=W\sin\theta=350\times\frac{5}{13}=134.62\ \text{N}
$$

The velocity profile across the thin film is linear:

$$
\tau=\mu\frac{V}{t}=\frac{F}{A}
$$

$$
\mu=\frac{F\,t}{A\,V}=\frac{134.62\times0.001}{1\times1.5}=0.0897\ \text{Pa·s}
$$

**Answer: $\mu\approx0.0897\ \text{N·s/m}^2$ (0.897 poise).**

## Check yourself

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

1. The plate slides down the incline at uniform velocity. What balances the oil's shear force?
   A. The component of weight along the plane, W sin(theta)
   B. The full weight W
   C. The component of weight normal to the plane, W cos(theta)
   D. Zero, since there is no acceleration
   Answer: A. With no acceleration the shear force on the plate equals W sin(theta) along the slope. (Key point: Force balance at uniform velocity)

2. For a slope of 5 vertical to 12 horizontal, what is sin(theta)?
   A. 5/12
   B. 12/13
   C. 5/13
   D. 12/5
   Answer: C. The hypotenuse is 13, so sin(theta) = 5/13, giving F = 350 x 5/13 = 134.6 N. (Key point: Slope angle)

3. Which relation gives the dynamic viscosity in this problem?
   A. mu = F t / (A V)
   B. mu = F A / (t V)
   C. mu = F V / (A t)
   D. mu = A V / (F t)
   Answer: A. tau = F/A = mu V/t, so mu = F t/(A V) = 0.0897 Pa·s. (Key point: Viscosity from shear)

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