= Spring balance
A \b[spring balance] measures an applied <force> through the extension of a <spring>. In its linear operating range, <Hooke's law> relates the magnitude of the spring's restoring force to the extension from its unloaded reference:
$$
F=k\Delta x,
$$
where $k$ is the <spring constant>. The pointer or graduated spring position converts this extension into a force indication.<University Physics Volume 1 by William Moebs and others>[Section15.1, Characteristics of Simple Harmonic Motion, Hooke's-law force relation]{ref}
A vertical measurement illustrates how <mass> enters the apparatus. Suspend a load from the spring and allow it to come to rest, with no other support. Neglect air buoyancy and account for the hanger in the zero reference. <Static equilibrium> then gives
$$
k\Delta x=mg,
$$
where $g$ is local <gravitational acceleration>. The spring responds to the load's <gravitational weight>; a mass-marked scale additionally assumes a calibration value of $g$.<University Physics Volume 1 by William Moebs and others>[Section5.4, Mass and Weight, gravitational force versus mass]{ref}
For an original calibration example, choose $k=50\,\mathrm{N\,m^{-1}}$ and $g=9.8\,\mathrm{m\,s^{-2}}$. Known loads of $0.10\,\mathrm{kg}$ and $0.20\,\mathrm{kg}$ exert $0.98\,\mathrm N$ and $1.96\,\mathrm N$, producing extensions of $19.6\,\mathrm{mm}$ and $39.2\,\mathrm{mm}$. Recording unloaded position and several loaded positions tests both the reference and the linear response. Unloading can also reveal whether the reference returns or whether <hysteresis> is appreciable.
Now place the same idealized apparatus in an environment stipulated to have $g=1.6\,\mathrm{m\,s^{-2}}$, keeping $k$ unchanged. The $0.20\,\mathrm{kg}$ load gives only $0.32\,\mathrm N$ and $6.4\,\mathrm{mm}$ extension. A mass scale still calibrated through $m_{\rm indicated}=F/(9.8\,\mathrm{m\,s^{-2}})$ would indicate approximately $0.0327\,\mathrm{kg}$. The load's <mass> is unchanged; the assumed force-to-mass conversion is no longer appropriate.
The model also assumes a static load. With upward <acceleration> $a$, <Newton's second law> instead gives supporting force $F=m(g+a)$. A transient spring-balance reading therefore need not equal the static <gravitational weight>.
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