Copper and Steel End to End

solved YoungsModulus

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?

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}

How it's solved

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 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 :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

solution.rb

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

playground.rb

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 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 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