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[3] 17.22 A brass rod long and in diameter. What force must be applied to each end of the rod to prevent it from contracting when it is cooled from to ?

by admin|Published March 11, 2026

Thermodynamics: [17.4] Thermal Expansion

[3] 17.22 A brass rod 185 cm long and 1.60 cm in diameter. What force must be applied to each end of the rod to prevent it from contracting when it is cooled from 120.0°C to 10.0°C?

Identify Significant Figure: The least sig fig count is 3
Identify Keywords: contracting, cooled, rod, force

[1st – Extract Data] 50% of the credit on exams.

Given (directly give)	Need (indirectly give)	Seeking
$L_0 = 185\text{ cm}$ $d = 1.60\text{ cm}$ $T_0 = 120.0^\circ C$ $T = 10.0^\circ C$ $\Delta T = 10.0^\circ C - 120.0^\circ C = -110.0^\circ C$ $A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{1.60\text{ cm}}{2}\right)^2$ $\alpha = 2.0\times10^{-5}K^{-1} = 2.0\times10^{-5}(C^\circ)^{-1}$ $Y_{Brass} = 9.0\times10^{10}\text{ Pa}$	$\Delta L = \alpha L_0 \Delta T$ $\frac{F}{A} = -T\alpha\Delta T$ $\Delta T = -110.0^\circ C = -110.0K$	$F = ?$ $\frac{F}{A} = -Y\alpha\Delta T$ $\Rightarrow F = -AY\alpha\Delta T$

NOTE The interval in Kelvin is the same as Celsius

Initial Length / Final Length

$\alpha \text{ units are } K^{-1} \text{ or } (C^\circ)^{-1}$

From [3], “If an object has length $L_0$ at temperature $T_0$ , then its length $L$ at temperature $T = T_0 + \Delta T$ is

$L = L_0 + \Delta L = L_0 + \alpha L_0 \Delta T = L_0(1 + \alpha\Delta T)$

[17.7]

Thermal Stress

$\frac{F}{A} = -Y\alpha\Delta T$

$F$ , is the force needed to keep the length of the rod constant.

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