A Hydraulic Automobile Lift

Q. A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm². Taking g as 9.8 m/s², what maximum pressure would the smaller piston have to bear, in pascals? expects 6.92 × 10⁵ Pa

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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?

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Inputs

mass

m_{c}

3000 kg

area

A_{l}

425 cm²

gravity

g

9.8 m/s²

Outputs

pressure

P_{s}

The solution

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 Ruby

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

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

1require_relative "../solutions/a_hydraulic_automobile_lift"23play_with :a_hydraulic_automobile_lift do4  title "A Hydraulic Automobile Lift"5end

The laws

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