Two Syringes Joined

solved PascalsLaw

Q. Two syringes of different cross-sections (without needles) filled with water are connected with a tightly fitted rubber tube filled with water $P_{l} = P_{s}$ . 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. What is the force exerted on the larger piston when a force of 10 N $F_{s} = 10$ N is applied to the smaller piston, in newtons?

Inputs

small diameter

d_{s}

1 cm

large diameter

d_{l}

3 cm

push

F_{s}

10 N

Outputs

force

F_{l}

How it's solved

Solved with PascalsLaw

Given

$d_{s} = 1.0$ cm
$d_{l} = 3.0$ cm
$F_{s} = 10$ N
$P_{l} = P_{s}$ from “connected with a tightly fitted rubber tube filled with water”

Find $F_{l}$ .

Working

$d_{s} = 1.0$ cm $= 0.01$ m
$d_{l} = 3.0$ cm $= 0.03$ 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
pressure is force over area(small) $P_{s} = \frac{F_{s}}{A_{s}}$
$P_{s} = \frac{10}{7.85398163 \times 10^{−5}}$
$P_{s} = 127323.954$ Pa
connected with a tightly fitted rubber tube filled with water $P_{l} = P_{s}$
$P_{l} = 127323.954$
$P_{l} = 127323.954$ Pa
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
pressure is force over area(large) $P_{l} = \frac{F_{l}}{A_{l}}$
$127323.954 = \frac{F_{l}}{7.06858347 \times 10^{−4}}$
$F_{l} = 90$ N
$F_{l} \approx 90$ Nto 2 significant figures

Conclusion

The water carries the pressure across undiminished, so each piston feels it over its own area: three times the diameter is nine times the area and nine times the force.

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 :two_syringes_joined 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. %{force} when a force of %{push} is applied to " \8            "the smaller piston, in newtons?"9  given small_diameter: "1.0 cm"10  given large_diameter: "3.0 cm"11  given push: "10 N"12  asks force: "What is the force exerted on the larger piston"13end

solution.rb

1require_relative "../questions/two_syringes_joined"2require_relative "../science/mechanical_properties_of_fluids/pascals_law"34solve :two_syringes_joined, using: PascalsLaw do5  objects :small, :large67  take small_diameter: small.piston_diameter8  take large_diameter: large.piston_diameter9  take push: small.thrust1011  deduce_from "connected with a tightly fitted rubber tube filled with water", large.pressure => small.pressure1213  force = large.thrust14  answer force:, in: "N", figures: 21516  conclusion "The water carries the pressure across undiminished, so each piston feels it over its own " \17             "area: three times the diameter is nine times the area and nine times the force."18end

playground.rb

1require_relative "../solutions/two_syringes_joined"23play_with :two_syringes_joined do4  title "Two Syringes Joined"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)

P_{l} = P_{s}

connected with a tightly fitted rubber tube filled with water