The Larger Piston Moves Out

solved PascalsLaw

Q. Two syringes of different cross-sections (without needles) filled with water $V_{l} = V_{s}$ are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1.0 cm $d_{s} = 1.0$ cm and 3.0 cm $d_{l} = 3.0$ cm respectively. If the smaller piston is pushed in through 6.0 cm $s_{s} = 6.0$ cm, how much does the larger piston move out, in cm?

Inputs

small diameter

d_{s}

1 cm

large diameter

d_{l}

3 cm

pushed

s_{s}

6 cm

Outputs

moved

s_{l}

How it's solved

Solved with PascalsLaw

Given

$d_{s} = 1.0$ cm
$d_{l} = 3.0$ cm
$s_{s} = 6.0$ cm
$V_{l} = V_{s}$ from “filled with water”

Find $s_{l}$ .

Working

$d_{s} = 1.0$ cm $= 0.01$ m
$d_{l} = 3.0$ cm $= 0.03$ m
$s_{s} = 6.0$ cm $= 0.06$ m
a round piston s area(small) $A_{s} = \frac{π \cdot {d_{s}}^{2}}{4}$
$A_{s} = \frac{3.14159265 \cdot {0.01}^{2}}{4}$
$A_{s} = 7.85398163 \times 10^{−5}$ m²
volume swept by a piston(small) $V_{s} = A_{s} \cdot s_{s}$
$V_{s} = 7.85398163 \times 10^{−5} \cdot 0.06$
$V_{s} = 4.71238898 \times 10^{−6}$ m³
filled with water $V_{l} = V_{s}$
$V_{l} = 4.71238898 \times 10^{−6}$
$V_{l} = 4.71238898 \times 10^{−6}$ m³
a round piston s area(large) $A_{l} = \frac{π \cdot {d_{l}}^{2}}{4}$
$A_{l} = \frac{3.14159265 \cdot {0.03}^{2}}{4}$
$A_{l} = 7.06858347 \times 10^{−4}$ m²
volume swept by a piston(large) $V_{l} = A_{l} \cdot s_{l}$
$4.71238898 \times 10^{−6} = 7.06858347 \times 10^{−4} \cdot s_{l}$
$s_{l} = 0.6667$ cm
$s_{l} = 0.00666666667$ m $= 0.666666667$ cm
$s_{l} \approx 0.67$ cmto 2 significant figures

Conclusion

Water does not squeeze, so the water the small piston pushes in fills the large syringe; spread over nine times the area, it moves the large piston a ninth as far.

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 :the_larger_piston_moves_out do4  kind :numerical5  describes "Two syringes of different cross-sections (without needles) filled with water are connected with a " \6            "tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are " \7            "%{small_diameter} and %{large_diameter} respectively. If the smaller piston is pushed in through " \8            "%{pushed}, %{moved}, in cm?"9  given small_diameter: "1.0 cm"10  given large_diameter: "3.0 cm"11  given pushed: "6.0 cm"12  asks moved: "how much does the larger piston move out"13end

solution.rb

1require_relative "../questions/the_larger_piston_moves_out"2require_relative "../science/mechanical_properties_of_fluids/pascals_law"34solve :the_larger_piston_moves_out, using: PascalsLaw do5  objects :small, :large67  take small_diameter: small.piston_diameter8  take large_diameter: large.piston_diameter9  take pushed: small.stroke1011  deduce_from "filled with water", large.volume_pushed => small.volume_pushed1213  moved = large.stroke14  answer moved:, in: "cm", figures: 21516  conclusion "Water does not squeeze, so the water the small piston pushes in fills the large syringe; " \17             "spread over nine times the area, it moves the large piston a ninth as far."18end

playground.rb

1require_relative "../solutions/the_larger_piston_moves_out"23play_with :the_larger_piston_moves_out do4  title "The Larger Piston Moves Out"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	—
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 … ∞
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	—
mass of small	$m_{s}$	kg	0 … ∞
mass of large	$m_{l}$	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)

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)

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)

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_{l} = V_{s}

filled with water