A Block on an Oil Film

Q. A metal block of area 0.10 m² is connected to a 0.010 kg mass via a string that passes over an ideal pulley. A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085 m/s. Taking g as 9.8 m/s², what is the coefficient of viscosity of the liquid, in Pa s? expects 0.00346 Pa s

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

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

The solution

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 Ruby

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

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

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

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