A Chairlift Cable

Q. A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 1 × 10⁸ N/m², what is the maximum load the cable can support, in newtons? expects 7.07 × 10⁴ N

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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?

The playground

Inputs

radius

r

1.5 cm

stress

σ

1e+08 N/m²

Outputs

load

F

The solution

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 Ruby

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

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

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

The laws

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)