A Girl on a High Heel

Q. A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. Taking g as 9.8 m/s², what is the pressure exerted by the heel on the horizontal floor, in pascals? expects 6.24 × 10⁶ Pa

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Q. A 50 kg $m_{g} = 50$ kg girl wearing high heel shoes balances on a single heel $F_{h} = m_{g} \cdot g$ . The heel is circular with a diameter 1.0 cm $d_{h} = 1.0$ cm. Taking g as 9.8 m/s² $g = 9.8$ m/s², what is the pressure exerted by the heel on the horizontal floor, in pascals?

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Inputs

mass

m_{g}

50 kg

diameter

d_{h}

1 cm

gravity

g

9.8 m/s²

Outputs

pressure

P_{h}

The solution

Solved with PascalsLaw

Given

$m_{g} = 50$ kg
$d_{h} = 1.0$ cm
$g = 9.8$ m/s²
$F_{h} = m_{g} \cdot g$ from “balances on a single heel”

Find $P_{h}$ .

Working

$d_{h} = 1.0$ cm $= 0.01$ m
balances on a single heel $F_{h} = m_{g} \cdot g$
$F_{h} = 50 \cdot 9.8$
$F_{h} = 490$ N
a round piston s area(heel) $A_{h} = \frac{π \cdot {d_{h}}^{2}}{4}$
$A_{h} = \frac{3.14159265 \cdot {0.01}^{2}}{4}$
$A_{h} = 7.85398163 \times 10^{−5}$ m²
pressure is force over area(heel) $P_{h} = \frac{F_{h}}{A_{h}}$
$P_{h} = \frac{490}{7.85398163 \times 10^{−5}}$
$P_{h} = 6.239 \times 10^{6}$ Pa
$P_{h} \approx 6.24 \times 10^{6}$ Pato 3 significant figures

Conclusion

Her whole weight rests on the tiny circle of one heel, so the pressure on the floor is enormous, far more than her flat feet would press.

The Ruby

1require_relative "../physics"23question :a_girl_on_a_high_heel do4  kind :numerical5  describes "A %{mass} girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter " \6            "%{diameter}. Taking g as %{gravity}, %{pressure}, in pascals?"7  given mass: "50 kg"8  given diameter: "1.0 cm"9  given gravity: "9.8 m/s2"10  asks pressure: "what is the pressure exerted by the heel on the horizontal floor"11end

1require_relative "../questions/a_girl_on_a_high_heel"2require_relative "../science/mechanical_properties_of_fluids/pascals_law"34solve :a_girl_on_a_high_heel, using: PascalsLaw do5  objects :girl, :heel67  take mass: girl.mass8  take diameter: heel.piston_diameter9  take gravity: g1011  deduce_from "balances on a single heel", heel.thrust => girl.mass * g1213  pressure = heel.pressure14  answer pressure:, in: "Pa", figures: 31516  conclusion "Her whole weight rests on the tiny circle of one heel, so the pressure on the floor is " \17             "enormous, far more than her flat feet would press."18end

1require_relative "../solutions/a_girl_on_a_high_heel"23play_with :a_girl_on_a_high_heel do4  title "A Girl on a High Heel"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 girl	$P_{g}$	Pa	0 … ∞
thrust on girl	$F_{g}$	N	—
area of girl	$A_{g}$	m²	—
pressure of heel	$P_{h}$	Pa	0 … ∞
thrust on heel	$F_{h}$	N	—
area of heel	$A_{h}$	m²	—
piston diameter of ground	$d_{g}$	m	0 … ∞
piston radius of ground	$r_{g}$	m	0 … ∞
piston diameter of girl	$d_{g}$	m	0 … ∞
piston radius of girl	$r_{g}$	m	0 … ∞
piston diameter of heel	$d_{h}$	m	0 … ∞
piston radius of heel	$r_{h}$	m	0 … ∞
volume pushed of ground	$V_{g}$	m³	—
stroke of ground	$s_{g}$	m	—
volume pushed of girl	$V_{g}$	m³	—
stroke of girl	$s_{g}$	m	—
volume pushed of heel	$V_{h}$	m³	—
stroke of heel	$s_{h}$	m	—
mass of girl	$m_{g}$	kg	0 … ∞
mass of heel	$m_{h}$	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_{g} = \frac{F_{g}}{A_{g}}

pressure is force over area(girl)

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

pressure is force over area(heel)

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

diameter is twice the radius(ground)

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

diameter is twice the radius(girl)

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

diameter is twice the radius(heel)

A_{g} = \frac{π \cdot {d_{g}}^{2}}{4}

a round piston s area(ground)

A_{g} = \frac{π \cdot {d_{g}}^{2}}{4}

a round piston s area(girl)

A_{h} = \frac{π \cdot {d_{h}}^{2}}{4}

a round piston s area(heel)

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

volume swept by a piston(ground)

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

volume swept by a piston(girl)

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

volume swept by a piston(heel)

F_{h} = m_{g} \cdot g

balances on a single heel