The Pressure on the Thighbones

Q. The two thigh bones (femurs), each of cross-sectional area 10 cm², support the upper part of a human body of mass 40 kg. Taking g as 10 m/s², what is the average pressure sustained by the femurs, in N/m²? expects 2.0 × 10⁵ N/m²

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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?

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Inputs

area

A_{f}

10 cm²

mass

m_{u}

40 kg

gravity

g

10 m/s²

Outputs

pressure

P_{f2}

The solution

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 Ruby

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

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

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

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