A Hydraulic Automobile Lift

solved PascalsLaw

Q. A hydraulic $P_{s} = P_{l}$ automobile lift is designed to lift cars with a maximum mass of 3000 kg $m_{c} = 3000$ kg. The area of cross-section of the piston carrying the load $F_{l} = m_{c} \cdot g$ is 425 cm² $A_{l} = 425$ cm². Taking g as 9.8 m/s² $g = 9.8$ m/s², what maximum pressure would the smaller piston have to bear, in pascals?

Inputs

mass

m_{c}

3000 kg

area

A_{l}

425 cm²

gravity

g

9.8 m/s²

Outputs

pressure

P_{s}

How it's solved

Solved with PascalsLaw

Given

$m_{c} = 3000$ kg
$A_{l} = 425$ cm²
$g = 9.8$ m/s²
$F_{l} = m_{c} \cdot g$ from “the piston carrying the load”
$P_{s} = P_{l}$ from “hydraulic”

Find $P_{s}$ .

Working

$A_{l} = 425$ cm² $= 0.0425$ m²
the piston carrying the load $F_{l} = m_{c} \cdot g$
$F_{l} = 3000 \cdot 9.8$
$F_{l} = 29400$ N
pressure is force over area(large) $P_{l} = \frac{F_{l}}{A_{l}}$
$P_{l} = \frac{29400}{0.0425}$
$P_{l} = 691764.706$ Pa
hydraulic $P_{s} = P_{l}$
$P_{s} = 691764.706$
$P_{s} = 691800$ Pa
$P_{s} \approx 6.92 \times 10^{5}$ Pato 3 significant figures

Conclusion

The liquid passes the pressure on undiminished, so the small piston bears the same pressure as the large one: the car's weight over the large piston's area.

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_hydraulic_automobile_lift do4  kind :numerical5  describes "A hydraulic automobile lift is designed to lift cars with a maximum mass of %{mass}. The area of " \6            "cross-section of the piston carrying the load is %{area}. Taking g as %{gravity}, %{pressure}, in " \7            "pascals?"8  given mass: "3000 kg"9  given area: "425 cm2"10  given gravity: "9.8 m/s2"11  asks pressure: "what maximum pressure would the smaller piston have to bear"12end

solution.rb

1require_relative "../questions/a_hydraulic_automobile_lift"2require_relative "../science/mechanical_properties_of_fluids/pascals_law"34solve :a_hydraulic_automobile_lift, using: PascalsLaw do5  objects :car, :large, :small67  take mass: car.mass8  take area: large.area9  take gravity: g1011  deduce_from "the piston carrying the load", large.thrust => car.mass * g12  deduce_from "hydraulic", small.pressure => large.pressure1314  pressure = small.pressure15  answer pressure:, in: "Pa", figures: 31617  conclusion "The liquid passes the pressure on undiminished, so the small piston bears the same pressure " \18             "as the large one: the car's weight over the large piston's area."19end

playground.rb

1require_relative "../solutions/a_hydraulic_automobile_lift"23play_with :a_hydraulic_automobile_lift do4  title "A Hydraulic Automobile 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 car	$P_{c}$	Pa	0 … ∞
thrust on car	$F_{c}$	N	—
area of car	$A_{c}$	m²	—
pressure of large	$P_{l}$	Pa	0 … ∞
thrust on large	$F_{l}$	N	—
area of large	$A_{l}$	m²	—
pressure of small	$P_{s}$	Pa	0 … ∞
thrust on small	$F_{s}$	N	—
area of small	$A_{s}$	m²	—
piston diameter of ground	$d_{g}$	m	0 … ∞
piston radius of ground	$r_{g}$	m	0 … ∞
piston diameter of car	$d_{c}$	m	0 … ∞
piston radius of car	$r_{c}$	m	0 … ∞
piston diameter of large	$d_{l}$	m	0 … ∞
piston radius of large	$r_{l}$	m	0 … ∞
piston diameter of small	$d_{s}$	m	0 … ∞
piston radius of small	$r_{s}$	m	0 … ∞
volume pushed of ground	$V_{g}$	m³	—
stroke of ground	$s_{g}$	m	—
volume pushed of car	$V_{c}$	m³	—
stroke of car	$s_{c}$	m	—
volume pushed of large	$V_{l}$	m³	—
stroke of large	$s_{l}$	m	—
volume pushed of small	$V_{s}$	m³	—
stroke of small	$s_{s}$	m	—
mass of car	$m_{c}$	kg	0 … ∞
mass of large	$m_{l}$	kg	0 … ∞
mass of small	$m_{s}$	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_{c} = \frac{F_{c}}{A_{c}}

pressure is force over area(car)

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

pressure is force over area(large)

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

pressure is force over area(small)

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

diameter is twice the radius(ground)

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

diameter is twice the radius(car)

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

diameter is twice the radius(large)

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

diameter is twice the radius(small)

A_{g} = \frac{π \cdot {d_{g}}^{2}}{4}

a round piston s area(ground)

A_{c} = \frac{π \cdot {d_{c}}^{2}}{4}

a round piston s area(car)

A_{l} = \frac{π \cdot {d_{l}}^{2}}{4}

a round piston s area(large)

A_{s} = \frac{π \cdot {d_{s}}^{2}}{4}

a round piston s area(small)

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

volume swept by a piston(ground)

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

volume swept by a piston(car)

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

volume swept by a piston(large)

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

volume swept by a piston(small)

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

the piston carrying the load

P_{s} = P_{l}

hydraulic