A Car Lift — phryzby

Q. In a car lift compressed air exerts a force on a small piston having a radius of 5.0 cm. This pressure is transmitted to a second piston of radius 15 cm. The mass of the car to be lifted is 1350 kg. Taking g as 9.8 m/s², what force must the air exert on the small piston, in newtons? expects 1.5 × 10³ N

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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?

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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}

The solution

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 Ruby

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

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

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