A Brass Wire Cooled Between Supports

Q. A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. Co-efficient of linear expansion of brass = 2.0 × 10⁻⁵ K⁻¹; Young's modulus of brass = 0.91 × 10¹¹ Pa. If the wire is cooled to a temperature of -39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm, in newtons? expects 3.8 × 10² N

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Q. A brass wire 1.8 m $L_{1} = 1.8$ m long at 27 °C $T_{1} = 27$ °C is held taut $L = L_{1}$ with little tension between two rigid supports $Δ L = - Δ l$ . Co-efficient of linear expansion of brass = 2.0 × 10⁻⁵ K⁻¹ $α_{l} = 2.0 \times 10^{−5}$ K⁻¹; Young's modulus of brass = 0.91 × 10¹¹ Pa $Y = 0.91 \times 10^{11}$ Pa. If the wire is cooled to a temperature of -39 °C $T_{2} = -39$ °C, what is the tension developed in the wire, if its diameter is 2.0 mm $r = 1.0$ mm, in newtons?

The playground

Inputs

length

L_{1}

1.80 m

warm

T_{1}

27 °C

expansion

α_{l}

2e-05 K⁻¹

modulus

Y

9.1e+10 Pa

cold

T_{2}

-39 °C

Outputs

tension

F

The solution

Solved with ThermalExpansion, YoungsModulus

Given

$L_{1} = 1.8$ m
$T_{1} = 27$ °C
$α_{l} = 2.0 \times 10^{−5}$ K⁻¹
$Y = 0.91 \times 10^{11}$ Pa
$T_{2} = -39$ °C
$r = 1.0$ mm from “its diameter is 2.0 mm”
$L = L_{1}$ from “held taut”
$Δ L = - Δ l$ from “between two rigid supports”

Find $F$ .

Working

$T_{1} = 27$ °C $= 300.15$ K
$T_{2} = -39$ °C $= 234.15$ K
$r = 1.0$ mm $= 0.001$ m
linear expansion(wire) $L_{2} - L_{1} = α_{l} \cdot L_{1} \cdot (T_{2} - T_{1})$
$L_{2} - 1.8 = 2 \times 10^{−5} \cdot 1.8 \cdot (234.15 - 300.15)$
$L_{2} = 1.797624$ m
the change is the length after less the length at first(wire) $Δ l = L_{2} - L_{1}$
$Δ l = 1.797624 - 1.8$
$Δ l = −0.002376$ m
between two rigid supports $Δ L = - Δ l$
$Δ L = - (−0.002376)$
$Δ L = 0.002376$ m
held taut $L = L_{1}$
$L = 1.8$
$L = 1.8$ m
strain is stretch over length(wire) $ε = \frac{Δ L}{L}$
$ε = \frac{0.002376}{1.8}$
$ε = 0.00132$
hookes law(wire) $Y = \frac{σ}{ε}$
$9.1 \times 10^{10} = \frac{σ}{0.00132}$
$σ = 1.2012 \times 10^{8}$ Pa
a round rod s cross section(wire) $A = π \cdot r^{2}$
$A = 3.14159265 \cdot {0.001}^{2}$
$A = 3.14159265 \times 10^{−6}$ m²
stress is load over area(wire) $σ = \frac{F}{A}$
$1.2012 \times 10^{8} = \frac{F}{3.14159265 \times 10^{−6}}$
$F = 377.4$ N
$F \approx 3.8 \times 10^{2}$ Nto 2 significant figures

Conclusion

Cooled, the wire would shrink, but the supports hold it at its old length; it is stretched by just what it would have shrunk, and that strain sets the tension.

The Ruby

1require_relative "../physics"23question :a_brass_wire_cooled_between_supports do4  kind :numerical5  describes "A brass wire %{length} long at %{warm} is held taut with little tension between two rigid supports. " \6            "Co-efficient of linear expansion of brass = %{expansion}; Young's modulus of brass = %{modulus}. If the " \7            "wire is cooled to a temperature of %{cold}, %{tension}, if its diameter is 2.0 mm, in newtons?"8  given length: "1.8 m"9  given warm: "27 °C"10  given expansion: "2.0e-5 K-1"11  given modulus: "0.91e11 Pa"12  given cold: "-39 °C"13  asks tension: "what is the tension developed in the wire"14end

1require_relative "../questions/a_brass_wire_cooled_between_supports"2require_relative "../science/thermal_properties_of_matter/thermal_expansion"3require_relative "../science/mechanical_properties_of_solids/youngs_modulus"45solve :a_brass_wire_cooled_between_supports, using: [ ThermalExpansion, YoungsModulus ] do6  objects :wire78  take length: wire.length_at_first9  take warm: wire.temperature_at_first10  take expansion: wire.coefficient_of_linear_expansion11  take modulus: wire.youngs_modulus12  take cold: wire.temperature_after1314  deduce_from "its diameter is 2.0 mm", wire.thickness => "1.0 mm"15  deduce_from "held taut", wire.original_length => wire.length_at_first16  deduce_from "between two rigid supports", wire.extension => -wire.change_in_length1718  tension = wire.load19  answer tension:, in: "N", figures: 22021  conclusion "Cooled, the wire would shrink, but the supports hold it at its old length; it is stretched " \22             "by just what it would have shrunk, and that strain sets the tension."23end

1require_relative "../solutions/a_brass_wire_cooled_between_supports"23play_with :a_brass_wire_cooled_between_supports do4  title "A Brass Wire Cooled Between Supports"5end

The laws

quantity	written	units	within	from
length at first	$L_{1}$	m	—	ThermalExpansion
length after	$L_{2}$	m	—	ThermalExpansion
change in length	$Δ l$	m	—	ThermalExpansion
temperature at first	$T_{1}$	K	—	ThermalExpansion
temperature after	$T_{2}$	K	—	ThermalExpansion
coefficient of linear expansion	$α_{l}$	K⁻¹	—	ThermalExpansion
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
length after of ground	$L_{2,g}$	m	—
length at first of ground	$L_{1,g}$	m	—
coefficient of linear expansion of ground	$α_{l,g}$	K⁻¹	—
temperature after of ground	$T_{2,g}$	K	—
temperature at first of ground	$T_{1,g}$	K	—
length after of wire	$L_{2}$	m	—
length at first of wire	$L_{1}$	m	—
coefficient of linear expansion of wire	$α_{l}$	K⁻¹	—
temperature after of wire	$T_{2}$	K	—
temperature at first of wire	$T_{1}$	K	—
change in length of ground	$Δ l_{g}$	m	—
change in length of wire	$Δ l$	m	—
thickness of ground	$r_{g}$	m	0 … ∞
diameter of ground	$D_{g}$	m	—
thickness of wire	$r$	m	0 … ∞
diameter of wire	$D$	m	—
cross section of ground	$A_{g}$	m²	—
cross section of wire	$A$	m²	—
stress of ground	$σ_{g}$	Pa	—
load on ground	$F_{g}$	N	—
stress of wire	$σ$	Pa	—
load on wire	$F$	N	—
strain of ground	$ε_{g}$	—	—
extension of ground	$Δ L_{g}$	m	—
original length of ground	$L_{g}$	m	—
strain of wire	$ε$	—	—
extension of wire	$Δ L$	m	—
original length of wire	$L$	m	—
youngs modulus of ground	$Y_{g}$	Pa	—
youngs modulus of wire	$Y$	Pa	—
mass of wire	$m$	kg	0 … ∞

L_{2,b} - L_{1,b} = α_{l,b} \cdot L_{1,b} \cdot (T_{2,b} - T_{1,b})

linear expansion

Δ l_{b} = L_{2,b} - L_{1,b}

the change is the length after less the length at first

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

L_{2,g} - L_{1,g} = α_{l,g} \cdot L_{1,g} \cdot (T_{2,g} - T_{1,g})

linear expansion(ground)

L_{2} - L_{1} = α_{l} \cdot L_{1} \cdot (T_{2} - T_{1})

linear expansion(wire)

Δ l_{g} = L_{2,g} - L_{1,g}

the change is the length after less the length at first(ground)

Δ l = L_{2} - L_{1}

the change is the length after less the length at first(wire)

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

radius is half the diameter(ground)

r = \frac{D}{2}

radius is half the diameter(wire)

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(wire)

σ_{g} = \frac{F_{g}}{A_{g}}

stress is load over area(ground)

σ = \frac{F}{A}

stress is load over area(wire)

ε_{g} = \frac{Δ L_{g}}{L_{g}}

strain is stretch over length(ground)

ε = \frac{Δ L}{L}

strain is stretch over length(wire)

Y_{g} = \frac{σ_{g}}{ε_{g}}

hookes law(ground)

Y = \frac{σ}{ε}

hookes law(wire)

L = L_{1}

held taut

Δ L = - Δ l

between two rigid supports