A Block on an Oil Film

solved Viscosity

Q. A metal block of area 0.10 m² $A_{b} = 0.10$ m² is connected to a 0.010 kg $m_{h} = 0.010$ kg mass via a string that passes over an ideal pulley. A liquid with a film thickness of 0.30 mm $l_{b} = 0.30$ mm is placed between the block and the table. When released the block moves to the right with a constant speed $F_{b} = m_{h} \cdot g$ of 0.085 m/s $v_{b} = 0.085$ m/s. Taking g as 9.8 m/s² $g = 9.8$ m/s², what is the coefficient of viscosity of the liquid, in Pa s?

Inputs

area

A_{b}

0.1 m²

mass

m_{h}

0.01 kg

thickness

l_{b}

0.3 mm

speed

v_{b}

0.1 m/s

gravity

g

9.8 m/s²

Outputs

viscosity

η

How it's solved

Solved with Viscosity

Given

$A_{b} = 0.10$ m²
$m_{h} = 0.010$ kg
$l_{b} = 0.30$ mm
$v_{b} = 0.085$ m/s
$g = 9.8$ m/s²
$F_{b} = m_{h} \cdot g$ from “a constant speed”

Find $η$ .

Working

$l_{b} = 0.30$ mm $= 3 \times 10^{−4}$ m
a constant speed $F_{b} = m_{h} \cdot g$
$F_{b} = 0.01 \cdot 9.8$
$F_{b} = 0.098$ N
drag of a sheared film(block) $F_{b} = \frac{η \cdot A_{b} \cdot v_{b}}{l_{b}}$
$0.098 = \frac{η \cdot 0.1 \cdot 0.085}{3 \times 10^{−4}}$
$η = 0.003459$ Pa s
$η \approx 0.00346$ Pa sto 3 significant figures

Conclusion

At a steady speed the film's drag just balances the hanging weight, and the drag of a sheared film is the viscosity times the area times the speed over the film's thickness.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_block_on_an_oil_film do4  kind :numerical5  describes "A metal block of area %{area} is connected to a %{mass} mass via a string that passes over an ideal " \6            "pulley. A liquid with a film thickness of %{thickness} is placed between the block and the table. When " \7            "released the block moves to the right with a constant speed of %{speed}. Taking g as %{gravity}, " \8            "%{viscosity}, in Pa s?"9  given area: "0.10 m2"10  given mass: "0.010 kg"11  given thickness: "0.30 mm"12  given speed: "0.085 m/s"13  given gravity: "9.8 m/s2"14  asks viscosity: "what is the coefficient of viscosity of the liquid"15end

solution.rb

1require_relative "../questions/a_block_on_an_oil_film"2require_relative "../science/mechanical_properties_of_fluids/viscosity"34solve :a_block_on_an_oil_film, using: Viscosity do5  objects :block, :hanging_mass67  take area: block.area8  take mass: hanging_mass.mass9  take thickness: block.film_thickness10  take speed: block.speed11  take gravity: g1213  deduce_from "a constant speed", block.viscous_drag => hanging_mass.mass * g1415  viscosity = eta16  answer viscosity:, in: "Pa s", figures: 31718  conclusion "At a steady speed the film's drag just balances the hanging weight, and the drag of a " \19             "sheared film is the viscosity times the area times the speed over the film's thickness."20end

playground.rb

1require_relative "../solutions/a_block_on_an_oil_film"23play_with :a_block_on_an_oil_film do4  title "A Block on an Oil Film"5end

The laws it drew on

quantity	written	units	within	from
mass	$m$	kg	0 … ∞	Viscosity
gravity	$g$	m/s²	—	Viscosity
area	$A$	m²	—	Viscosity
film thickness	$l$	m	—	Viscosity
speed	$v$	m/s	—	Viscosity
viscous drag	$F$	N	—	Viscosity
coefficient of viscosity	$η$	Pa s	—	Viscosity
viscous drag on ground	$F_{g}$	N	—
area of ground	$A_{g}$	m²	—
speed of ground	$v_{g}$	m/s	—
film thickness of ground	$l_{g}$	m	—
viscous drag on block	$F_{b}$	N	—
area of block	$A_{b}$	m²	—
speed of block	$v_{b}$	m/s	—
film thickness of block	$l_{b}$	m	—
viscous drag on hanging mass	$F_{h}$	N	—
area of hanging mass	$A_{h}$	m²	—
speed of hanging mass	$v_{h}$	m/s	—
film thickness of hanging mass	$l_{h}$	m	—
mass of block	$m_{b}$	kg	0 … ∞
mass of hanging mass	$m_{h}$	kg	0 … ∞

F_{p} = \frac{η \cdot A_{p} \cdot v_{p}}{l_{p}}

drag of a sheared film

F_{g} = \frac{η \cdot A_{g} \cdot v_{g}}{l_{g}}

drag of a sheared film(ground)

F_{b} = \frac{η \cdot A_{b} \cdot v_{b}}{l_{b}}

drag of a sheared film(block)

F_{h} = \frac{η \cdot A_{h} \cdot v_{h}}{l_{h}}

drag of a sheared film(hanging mass)

F_{b} = m_{h} \cdot g

a constant speed