A Chairlift Cable

solved YoungsModulus

Q. A steel cable with a radius of 1.5 cm $r = 1.5$ cm supports a chairlift at a ski area. If the maximum stress is not to exceed 1 × 10⁸ N/m² $σ = 1 \times 10^{8}$ N/m², what is the maximum load the cable can support, in newtons?

Inputs

radius

r

1.5 cm

stress

σ

1e+08 N/m²

Outputs

load

F

How it's solved

Solved with YoungsModulus

Given

$r = 1.5$ cm
$σ = 1 \times 10^{8}$ N/m²

Find $F$ .

Working

$r = 1.5$ cm $= 0.015$ m
$σ = 1 \times 10^{8}$ N/m² $= 1 \times 10^{8}$ Pa
a round rod s cross section(cable) $A = π \cdot r^{2}$
$A = 3.14159265 \cdot {0.015}^{2}$
$A = 7.06858347 \times 10^{−4}$ m²
stress is load over area(cable) $σ = \frac{F}{A}$
$1 \times 10^{8} = \frac{F}{7.06858347 \times 10^{−4}}$
$F = 70690$ N
$F \approx 7.07 \times 10^{4}$ Nto 3 significant figures

Conclusion

Stress is load over cross-section, so the largest load is the largest stress the steel may bear spread over the cable's round cross-section.

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 :a_chairlift_cable do4  kind :numerical5  describes "A steel cable with a radius of %{radius} supports a chairlift at a ski area. If the maximum stress is " \6            "not to exceed %{stress}, %{load}, in newtons?"7  given radius: "1.5 cm"8  given stress: "1e8 N/m2"9  asks load: "what is the maximum load the cable can support"10end

solution.rb

1require_relative "../questions/a_chairlift_cable"2require_relative "../science/mechanical_properties_of_solids/youngs_modulus"34solve :a_chairlift_cable, using: YoungsModulus do5  objects :cable67  take radius: cable.thickness8  take stress: cable.stress910  load = cable.load11  answer load:, in: "N", figures: 31213  conclusion "Stress is load over cross-section, so the largest load is the largest stress the steel may " \14             "bear spread over the cable's round cross-section."15end

playground.rb

1require_relative "../solutions/a_chairlift_cable"23play_with :a_chairlift_cable do4  title "A Chairlift Cable"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	YoungsModulus
mass	$m$	kg	0 … ∞	YoungsModulus
gravity	$g$	m/s²	—	YoungsModulus
load	$F$	N	—	YoungsModulus
thickness	$r$	m	0 … ∞	YoungsModulus
diameter	$D$	m	—	YoungsModulus
cross section	$A$	m²	—	YoungsModulus
original length	$L$	m	—	YoungsModulus
extension	$Δ L$	m	—	YoungsModulus
stress	$σ$	Pa	—	YoungsModulus
strain	$ε$	—	—	YoungsModulus
youngs modulus	$Y$	Pa	—	YoungsModulus
thickness of ground	$r_{g}$	m	0 … ∞
diameter of ground	$D_{g}$	m	—
thickness of cable	$r$	m	0 … ∞
diameter of cable	$D$	m	—
cross section of ground	$A_{g}$	m²	—
cross section of cable	$A$	m²	—
stress of ground	$σ_{g}$	Pa	—
load on ground	$F_{g}$	N	—
stress of cable	$σ$	Pa	—
load on cable	$F$	N	—
strain of ground	$ε_{g}$	—	—
extension of ground	$Δ L_{g}$	m	—
original length of ground	$L_{g}$	m	—
strain of cable	$ε$	—	—
extension of cable	$Δ L$	m	—
original length of cable	$L$	m	—
youngs modulus of ground	$Y_{g}$	Pa	—
youngs modulus of cable	$Y$	Pa	—
mass of cable	$m$	kg	0 … ∞

r_{r} = \frac{D_{r}}{2}

radius is half the diameter

A_{r} = π \cdot {r_{r}}^{2}

a round rod s cross section

σ_{r} = \frac{F_{r}}{A_{r}}

stress is load over area

ε_{r} = \frac{Δ L_{r}}{L_{r}}

strain is stretch over length

Y_{r} = \frac{σ_{r}}{ε_{r}}

hookes law

r_{g} = \frac{D_{g}}{2}

radius is half the diameter(ground)

r = \frac{D}{2}

radius is half the diameter(cable)

A_{g} = π \cdot {r_{g}}^{2}

a round rod s cross section(ground)

A = π \cdot r^{2}

a round rod s cross section(cable)

σ_{g} = \frac{F_{g}}{A_{g}}

stress is load over area(ground)

σ = \frac{F}{A}

stress is load over area(cable)

ε_{g} = \frac{Δ L_{g}}{L_{g}}

strain is stretch over length(ground)

ε = \frac{Δ L}{L}

strain is stretch over length(cable)

Y_{g} = \frac{σ_{g}}{ε_{g}}

hookes law(ground)

Y = \frac{σ}{ε}

hookes law(cable)