Copper and Steel End to End

Q. A copper wire of length 2.2 m and a steel wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. What is the load applied, in newtons? expects 1.8 × 10² N

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Q. A copper wire of length 2.2 m $L_{c} = 2.2$ m and a steel wire of length 1.6 m $L_{s} = 1.6$ m, both of diameter $D_{s} = D_{c}$ 3.0 mm $D_{c} = 3.0$ mm, are connected end to end $F_{s} = F_{c}$ . When stretched by a load, the net elongation $Δ L_{b} = Δ L_{c} + Δ L_{s}$ is found to be 0.70 mm $Δ L_{b} = 0.70$ mm. What is the load applied, in newtons?

The playground

Inputs

copper length

L_{c}

2.20 m

steel length

L_{s}

1.60 m

diameter

D_{c}

3 mm

elongation

Δ L_{b}

0.7 mm

Outputs

load

F_{c}

The solution

Solved with YoungsModulus

Given

$L_{c} = 2.2$ m
$L_{s} = 1.6$ m
$D_{c} = 3.0$ mm
$Δ L_{b} = 0.70$ mm
$D_{s} = D_{c}$ from “both of diameter”
$F_{s} = F_{c}$ from “connected end to end”
$Δ L_{b} = Δ L_{c} + Δ L_{s}$ from “the net elongation”

Assumed

$Y_{c} = 1.1 \times 10^{11}$ Pa
$Y_{s} = 2.0 \times 10^{11}$ Pa

Find $F_{c}$ .

Working

$D_{c} = 3.0$ mm $= 0.003$ m
$Δ L_{b} = 0.70$ mm $= 7 \times 10^{−4}$ m
radius is half the diameter(copper) $r_{c} = \frac{D_{c}}{2}$
$r_{c} = \frac{0.003}{2}$
$r_{c} = 0.0015$ m
a round rod s cross section(copper) $A_{c} = π \cdot {r_{c}}^{2}$
$A_{c} = 3.14159265 \cdot {0.0015}^{2}$
$A_{c} = 7.06858347 \times 10^{−6}$ m²
both of diameter $D_{s} = D_{c}$
$D_{s} = 0.003$
$D_{s} = 0.003$ m
radius is half the diameter(steel) $r_{s} = \frac{D_{s}}{2}$
$r_{s} = \frac{0.003}{2}$
$r_{s} = 0.0015$ m
a round rod s cross section(steel) $A_{s} = π \cdot {r_{s}}^{2}$
$A_{s} = 3.14159265 \cdot {0.0015}^{2}$
$A_{s} = 7.06858347 \times 10^{−6}$ m²
strain is stretch over length(steel) $ε_{s} = \frac{Δ L_{s}}{L_{s}}$
stress is load over area(copper) $σ_{c} = \frac{F_{c}}{A_{c}}$
hookes law(copper) $Y_{c} = \frac{σ_{c}}{ε_{c}}$
connected end to end $F_{s} = F_{c}$
stress is load over area(steel) $σ_{s} = \frac{F_{s}}{A_{s}}$
strain is stretch over length(copper) $ε_{c} = \frac{Δ L_{c}}{L_{c}}$
hookes law(steel) $Y_{s} = \frac{σ_{s}}{ε_{s}}$
the net elongation $Δ L_{b} = Δ L_{c} + Δ L_{s}$
$F_{c} = 176.7$ N
$F_{c} \approx 1.8 \times 10^{2}$ Nto 2 significant figures

Conclusion

One load pulls both wires; the stiffer steel stretches less, and the two stretches together make the elongation that is measured.

The Ruby

1require_relative "../physics"23question :copper_and_steel_end_to_end do4  kind :numerical5  describes "A copper wire of length %{copper_length} and a steel wire of length %{steel_length}, both of diameter " \6            "%{diameter}, are connected end to end. When stretched by a load, the net elongation is found to be " \7            "%{elongation}. %{load}, in newtons?"8  given copper_length: "2.2 m"9  given steel_length: "1.6 m"10  given diameter: "3.0 mm"11  given elongation: "0.70 mm"12  asks load: "What is the load applied"13end

1require_relative "../questions/copper_and_steel_end_to_end"2require_relative "../science/mechanical_properties_of_solids/youngs_modulus"34solve :copper_and_steel_end_to_end, using: YoungsModulus do5  objects :copper, :steel, :both67  take copper_length: copper.original_length8  take steel_length: steel.original_length9  take diameter: copper.diameter10  take elongation: both.extension1112  deduce_from "both of diameter", steel.diameter => copper.diameter13  deduce_from "connected end to end", steel.load => copper.load14  deduce_from "the net elongation", both.extension => copper.extension + steel.extension15  assume copper.youngs_modulus => "1.1e11 Pa", steel.youngs_modulus => "2.0e11 Pa"1617  load = copper.load18  answer load:, in: "N", figures: 21920  conclusion "One load pulls both wires; the stiffer steel stretches less, and the two stretches together make " \21             "the elongation that is measured."22end

1require_relative "../solutions/copper_and_steel_end_to_end"23play_with :copper_and_steel_end_to_end do4  title "Copper and Steel End to End"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 copper	$r_{c}$	m	0 … ∞
diameter of copper	$D_{c}$	m	—
thickness of steel	$r_{s}$	m	0 … ∞
diameter of steel	$D_{s}$	m	—
thickness of both	$r_{b}$	m	0 … ∞
diameter of both	$D_{b}$	m	—
cross section of ground	$A_{g}$	m²	—
cross section of copper	$A_{c}$	m²	—
cross section of steel	$A_{s}$	m²	—
cross section of both	$A_{b}$	m²	—
stress of ground	$σ_{g}$	Pa	—
load on ground	$F_{g}$	N	—
stress of copper	$σ_{c}$	Pa	—
load on copper	$F_{c}$	N	—
stress of steel	$σ_{s}$	Pa	—
load on steel	$F_{s}$	N	—
stress of both	$σ_{b}$	Pa	—
load on both	$F_{b}$	N	—
strain of ground	$ε_{g}$	—	—
extension of ground	$Δ L_{g}$	m	—
original length of ground	$L_{g}$	m	—
strain of copper	$ε_{c}$	—	—
extension of copper	$Δ L_{c}$	m	—
original length of copper	$L_{c}$	m	—
strain of steel	$ε_{s}$	—	—
extension of steel	$Δ L_{s}$	m	—
original length of steel	$L_{s}$	m	—
strain of both	$ε_{b}$	—	—
extension of both	$Δ L_{b}$	m	—
original length of both	$L_{b}$	m	—
youngs modulus of ground	$Y_{g}$	Pa	—
youngs modulus of copper	$Y_{c}$	Pa	—
youngs modulus of steel	$Y_{s}$	Pa	—
youngs modulus of both	$Y_{b}$	Pa	—
mass of copper	$m_{c}$	kg	0 … ∞
mass of steel	$m_{s}$	kg	0 … ∞
mass of both	$m_{b}$	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_{c} = \frac{D_{c}}{2}

radius is half the diameter(copper)

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

radius is half the diameter(steel)

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

radius is half the diameter(both)

A_{g} = π \cdot {r_{g}}^{2}

a round rod s cross section(ground)

A_{c} = π \cdot {r_{c}}^{2}

a round rod s cross section(copper)

A_{s} = π \cdot {r_{s}}^{2}

a round rod s cross section(steel)

A_{b} = π \cdot {r_{b}}^{2}

a round rod s cross section(both)

σ_{g} = \frac{F_{g}}{A_{g}}

stress is load over area(ground)

σ_{c} = \frac{F_{c}}{A_{c}}

stress is load over area(copper)

σ_{s} = \frac{F_{s}}{A_{s}}

stress is load over area(steel)

σ_{b} = \frac{F_{b}}{A_{b}}

stress is load over area(both)

ε_{g} = \frac{Δ L_{g}}{L_{g}}

strain is stretch over length(ground)

ε_{c} = \frac{Δ L_{c}}{L_{c}}

strain is stretch over length(copper)

ε_{s} = \frac{Δ L_{s}}{L_{s}}

strain is stretch over length(steel)

ε_{b} = \frac{Δ L_{b}}{L_{b}}

strain is stretch over length(both)

Y_{g} = \frac{σ_{g}}{ε_{g}}

hookes law(ground)

Y_{c} = \frac{σ_{c}}{ε_{c}}

hookes law(copper)

Y_{s} = \frac{σ_{s}}{ε_{s}}

hookes law(steel)

Y_{b} = \frac{σ_{b}}{ε_{b}}

hookes law(both)

D_{s} = D_{c}

both of diameter

F_{s} = F_{c}

connected end to end

Δ L_{b} = Δ L_{c} + Δ L_{s}

the net elongation