The Pressure on the Thighbones

solved Pressure

Q. The two thigh bones $A_{f2} = 2 \cdot A_{f}$ (femurs), each of cross-sectional area 10 cm² $A_{f} = 10$ cm², support the upper part $F_{f2} = m_{u} \cdot g$ of a human body of mass 40 kg $m_{u} = 40$ kg. Taking g as 10 m/s² $g = 10$ m/s², what is the average pressure sustained by the femurs, in N/m²?

Inputs

area

A_{f}

10 cm²

mass

m_{u}

40 kg

gravity

g

10 m/s²

Outputs

pressure

P_{f2}

How it's solved

Solved with Pressure

Given

$A_{f} = 10$ cm²
$m_{u} = 40$ kg
$g = 10$ m/s²
$A_{f2} = 2 \cdot A_{f}$ from “The two thigh bones”
$F_{f2} = m_{u} \cdot g$ from “support the upper part”

Find $P_{f2}$ .

Working

$A_{f} = 10$ cm² $= 0.001$ m²
support the upper part $F_{f2} = m_{u} \cdot g$
$F_{f2} = 40 \cdot 10$
$F_{f2} = 400$ N
The two thigh bones $A_{f2} = 2 \cdot A_{f}$
$A_{f2} = 2 \cdot 0.001$
$A_{f2} = 0.002$ m²
pressure is force over area(both) $P_{f2} = \frac{F_{f2}}{A_{f2}}$
$P_{f2} = \frac{400}{0.002}$
$P_{f2} = 200000$ N/m²
$P_{f2} = 200000$ Pa $= 200000$ N/m²
$P_{f2} \approx 2.0 \times 10^{5}$ N/m²to 2 significant figures

Conclusion

The weight of the upper body is shared between both bones, so it is spread over the two cross-sections together.

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_pressure_on_the_thighbones do4  kind :numerical5  describes "The two thigh bones (femurs), each of cross-sectional area %{area}, support the upper part of a " \6            "human body of mass %{mass}. Taking g as %{gravity}, %{pressure}, in N/m²?"7  given area: "10 cm2"8  given mass: "40 kg"9  given gravity: "10 m/s2"10  asks pressure: "what is the average pressure sustained by the femurs"11end

solution.rb

1require_relative "../questions/the_pressure_on_the_thighbones"2require_relative "../science/mechanical_properties_of_fluids/pressure"34solve :the_pressure_on_the_thighbones, using: Pressure do5  objects :femur, :upper_body, both: :f267  take area: femur.area8  take mass: upper_body.mass9  take gravity: g1011  deduce_from "The two thigh bones", both.area => 2 * femur.area12  deduce_from "support the upper part", both.thrust => upper_body.mass * g1314  pressure = both.pressure15  answer pressure:, in: "N/m2", figures: 21617  conclusion "The weight of the upper body is shared between both bones, so it is spread over the two " \18             "cross-sections together."19end

playground.rb

1require_relative "../solutions/the_pressure_on_the_thighbones"23play_with :the_pressure_on_the_thighbones do4  title "The Pressure on the Thighbones"5end

The laws it drew on

quantity	written	units	within	from
mass	$m$	kg	0 … ∞	Pressure
gravity	$g$	m/s²	—	Pressure
thrust	$F$	N	—	Pressure
area	$A$	m²	—	Pressure
pressure	$P$	Pa	0 … ∞	Pressure
pressure of ground	$P_{g}$	Pa	0 … ∞
thrust on ground	$F_{g}$	N	—
area of ground	$A_{g}$	m²	—
pressure of femur	$P_{f}$	Pa	0 … ∞
thrust on femur	$F_{f}$	N	—
area of femur	$A_{f}$	m²	—
pressure of upper body	$P_{u}$	Pa	0 … ∞
thrust on upper body	$F_{u}$	N	—
area of upper body	$A_{u}$	m²	—
pressure of both	$P_{f2}$	Pa	0 … ∞
thrust on both	$F_{f2}$	N	—
area of both	$A_{f2}$	m²	—
mass of femur	$m_{f}$	kg	0 … ∞
mass of upper body	$m_{u}$	kg	0 … ∞
mass of both	$m_{f2}$	kg	0 … ∞

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

pressure is force over area

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

pressure is force over area(ground)

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

pressure is force over area(femur)

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

pressure is force over area(upper body)

P_{f2} = \frac{F_{f2}}{A_{f2}}

pressure is force over area(both)

A_{f2} = 2 \cdot A_{f}

The two thigh bones

F_{f2} = m_{u} \cdot g

support the upper part