A Car Lift — phryzby

solved PascalsLaw

Q. In a car lift compressed air exerts a force on a small piston having a radius of 5.0 cm $r_{s} = 5.0$ cm. This pressure is transmitted $P_{l} = P_{s}$ to a second piston of radius 15 cm $r_{l} = 15$ cm. The mass of the car to be lifted $F_{l} = m_{c} \cdot g$ is 1350 kg $m_{c} = 1350$ kg. Taking g as 9.8 m/s² $g = 9.8$ m/s², what force must the air exert on the small piston, in newtons?

Inputs

small radius

r_{s}

5 cm

large radius

r_{l}

15 cm

mass

m_{c}

1350 kg

gravity

g

9.8 m/s²

Outputs

force

F_{s}

How it's solved

Solved with PascalsLaw

Given

$r_{s} = 5.0$ cm
$r_{l} = 15$ cm
$m_{c} = 1350$ kg
$g = 9.8$ m/s²
$P_{l} = P_{s}$ from “This pressure is transmitted”
$F_{l} = m_{c} \cdot g$ from “The mass of the car to be lifted”

Find $F_{s}$ .

Working

$r_{s} = 5.0$ cm $= 0.05$ m
$r_{l} = 15$ cm $= 0.15$ m
The mass of the car to be lifted $F_{l} = m_{c} \cdot g$
$F_{l} = 1350 \cdot 9.8$
$F_{l} = 13230$ N
diameter is twice the radius(large) $d_{l} = 2 \cdot r_{l}$
$d_{l} = 2 \cdot 0.15$
$d_{l} = 0.3$ m
a round piston s area(large) $A_{l} = \frac{π \cdot {d_{l}}^{2}}{4}$
$A_{l} = \frac{3.14159265 \cdot {0.3}^{2}}{4}$
$A_{l} = 0.0706858347$ m²
pressure is force over area(large) $P_{l} = \frac{F_{l}}{A_{l}}$
$P_{l} = \frac{13230}{0.0706858347}$
$P_{l} = 187166.213$ Pa
This pressure is transmitted $P_{l} = P_{s}$
$187166.213 = P_{s}$
$P_{s} = 187166.213$ Pa
diameter is twice the radius(small) $d_{s} = 2 \cdot r_{s}$
$d_{s} = 2 \cdot 0.05$
$d_{s} = 0.1$ m
a round piston s area(small) $A_{s} = \frac{π \cdot {d_{s}}^{2}}{4}$
$A_{s} = \frac{3.14159265 \cdot {0.1}^{2}}{4}$
$A_{s} = 0.00785398163$ m²
pressure is force over area(small) $P_{s} = \frac{F_{s}}{A_{s}}$
$187166.213 = \frac{F_{s}}{0.00785398163}$
$F_{s} = 1470$ N
$F_{s} \approx 1.5 \times 10^{3}$ Nto 2 significant figures

Conclusion

The large piston has nine times the area of the small one, so the air needs only a ninth of the car's weight to hold it up.

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_car_lift do4  kind :numerical5  describes "In a car lift compressed air exerts a force on a small piston having a radius of %{small_radius}. This " \6            "pressure is transmitted to a second piston of radius %{large_radius}. The mass of the car to be lifted " \7            "is %{mass}. Taking g as %{gravity}, %{force}, in newtons?"8  given small_radius: "5.0 cm"9  given large_radius: "15 cm"10  given mass: "1350 kg"11  given gravity: "9.8 m/s2"12  asks force: "what force must the air exert on the small piston"13end

solution.rb

1require_relative "../questions/a_car_lift"2require_relative "../science/mechanical_properties_of_fluids/pascals_law"34solve :a_car_lift, using: PascalsLaw do5  objects :small, :large, :car67  take small_radius: small.piston_radius8  take large_radius: large.piston_radius9  take mass: car.mass10  take gravity: g1112  deduce_from "This pressure is transmitted", large.pressure => small.pressure13  deduce_from "The mass of the car to be lifted", large.thrust => car.mass * g1415  force = small.thrust16  answer force:, in: "N", figures: 21718  conclusion "The large piston has nine times the area of the small one, so the air needs only a ninth " \19             "of the car's weight to hold it up."20end

playground.rb

1require_relative "../solutions/a_car_lift"23play_with :a_car_lift do4  title "A Car Lift"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	PascalsLaw
mass	$m$	kg	0 … ∞	Pressure
gravity	$g$	m/s²	—	Pressure
thrust	$F$	N	—	Pressure
area	$A$	m²	—	Pressure
pressure	$P$	Pa	0 … ∞	Pressure
piston diameter	$d$	m	0 … ∞	PascalsLaw
piston radius	$r$	m	0 … ∞	PascalsLaw
stroke	$s$	m	—	PascalsLaw
volume pushed	$V$	m³	—	PascalsLaw
pressure of ground	$P_{g}$	Pa	0 … ∞
thrust on ground	$F_{g}$	N	—
area of ground	$A_{g}$	m²	—
pressure of small	$P_{s}$	Pa	0 … ∞
thrust on small	$F_{s}$	N	—
area of small	$A_{s}$	m²	—
pressure of large	$P_{l}$	Pa	0 … ∞
thrust on large	$F_{l}$	N	—
area of large	$A_{l}$	m²	—
pressure of car	$P_{c}$	Pa	0 … ∞
thrust on car	$F_{c}$	N	—
area of car	$A_{c}$	m²	—
piston diameter of ground	$d_{g}$	m	0 … ∞
piston radius of ground	$r_{g}$	m	0 … ∞
piston diameter of small	$d_{s}$	m	0 … ∞
piston radius of small	$r_{s}$	m	0 … ∞
piston diameter of large	$d_{l}$	m	0 … ∞
piston radius of large	$r_{l}$	m	0 … ∞
piston diameter of car	$d_{c}$	m	0 … ∞
piston radius of car	$r_{c}$	m	0 … ∞
volume pushed of ground	$V_{g}$	m³	—
stroke of ground	$s_{g}$	m	—
volume pushed of small	$V_{s}$	m³	—
stroke of small	$s_{s}$	m	—
volume pushed of large	$V_{l}$	m³	—
stroke of large	$s_{l}$	m	—
volume pushed of car	$V_{c}$	m³	—
stroke of car	$s_{c}$	m	—
mass of small	$m_{s}$	kg	0 … ∞
mass of large	$m_{l}$	kg	0 … ∞
mass of car	$m_{c}$	kg	0 … ∞

P_{s} = \frac{F_{s}}{A_{s}}

pressure is force over area

d_{p} = 2 \cdot r_{p}

diameter is twice the radius

A_{p} = \frac{π \cdot {d_{p}}^{2}}{4}

a round piston s area

V_{p} = A_{p} \cdot s_{p}

volume swept by a piston

P_{g} = \frac{F_{g}}{A_{g}}

pressure is force over area(ground)

P_{s} = \frac{F_{s}}{A_{s}}

pressure is force over area(small)

P_{l} = \frac{F_{l}}{A_{l}}

pressure is force over area(large)

P_{c} = \frac{F_{c}}{A_{c}}

pressure is force over area(car)

d_{g} = 2 \cdot r_{g}

diameter is twice the radius(ground)

d_{s} = 2 \cdot r_{s}

diameter is twice the radius(small)

d_{l} = 2 \cdot r_{l}

diameter is twice the radius(large)

d_{c} = 2 \cdot r_{c}

diameter is twice the radius(car)

A_{g} = \frac{π \cdot {d_{g}}^{2}}{4}

a round piston s area(ground)

A_{s} = \frac{π \cdot {d_{s}}^{2}}{4}

a round piston s area(small)

A_{l} = \frac{π \cdot {d_{l}}^{2}}{4}

a round piston s area(large)

A_{c} = \frac{π \cdot {d_{c}}^{2}}{4}

a round piston s area(car)

V_{g} = A_{g} \cdot s_{g}

volume swept by a piston(ground)

V_{s} = A_{s} \cdot s_{s}

volume swept by a piston(small)

V_{l} = A_{l} \cdot s_{l}

volume swept by a piston(large)

V_{c} = A_{c} \cdot s_{c}

volume swept by a piston(car)

P_{l} = P_{s}

This pressure is transmitted

F_{l} = m_{c} \cdot g

The mass of the car to be lifted