Spring balance

A 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:where $k$ is the spring constant. The pointer or graduated spring position converts this extension into a force indication.^[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 giveswhere $g$ is local gravitational acceleration. The spring responds to the load's gravitational weight; a mass-marked scale additionally assumes a calibration value of $g$.^[ref]

For an original calibration example, choose $k=50\mathrm{N}{\mathrm{m}}^{-1}$ and $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$. 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 $g=1.6\mathrm{m}{\mathrm{s}}^{-2}$, 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 $m_{\mathrm{indicated}}=F/(9.8\mathrm{m}{\mathrm{s}}^{-2})$ 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.