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Atmospheric pressure often provides the reference for a gauge pressure measurement: $p_g=p_{\mathrm{abs}}-p_{\mathrm{atm}}$. The absolute pressure is measured relative to an ideal vacuum, whereas zero gauge pressure means equality to the chosen reference. The actual atmospheric pressure can vary; it must be specified when converting between these descriptions. ^[ref].

The coefficient of kinetic friction is the dimensionless quantity ${\mu }_k=|f_k|/N$ for sliding contact with positive normal force $N$ in the ideal kinetic friction model. It characterizes the pair of contacting surfaces and their condition. A constant ${\mu }_k$ means the model neglects dependence on contact slip speed; unlike the coefficient of static friction, it specifies a sliding force magnitude rather than a no-slip upper bound.

The ratio compares the tangential kinetic friction magnitude with the perpendicular normal force:Here $\mathbf{w}$ is the tangential relative velocity defining contact slip. Both numerator and denominator have the same physical dimension, so the coefficient of kinetic friction has no units. It belongs to a contact pair under specified conditions, rather than to one material alone. The coefficient of static friction instead describes the threshold $|f_s{|}_{\max }/N$ before contact slip; a pull required to start motion cannot be substituted for a sliding measurement.^[ref]

As an original model example, stipulate ${\mu }_k=0.30$ and gravitational acceleration $g=9.8m/{\mathrm{s}}^2$. An already sliding, nonrotating block of total mass $m=2\mathrm{kg}$ is pulled horizontally across a stationary horizontal surface at constant velocity. With no other vertical force, Newton's second law gives $N=mg=19.6\mathrm{N}$. Zero horizontal acceleration then requires a pulling force $F=|f_k|={\mu }_{k}mg=5.88\mathrm{N}$. Doubling the total mass to $4\mathrm{kg}$ gives $N=39.2\mathrm{N}$ and $F=11.76\mathrm{N}$. The force doubles while the coefficient of kinetic friction stays $0.30$ by assumption. These numbers are model predictions, not observations or a calibration for a particular material pair.

A constant coefficient of kinetic friction is useful because one ratio can summarize many loads, as tested by load dependence of kinetic friction. It is an empirical approximation over a stated range of normal forces, speeds and surface conditions. It implies no universal independence from speed or apparent contact area: a change of face or sliding regime can change the result. Temperature, wear and lubrication are further conditions to report. Newton's second law still applies when the constant-coefficient approximation fails.^[ref]

Gauge pressure is pressure measured relative to the surrounding atmospheric pressure; absolute pressure is instead measured relative to an ideal vacuum. At an open surface $p_{\mathrm{surface}}=p_{\mathrm{atmosphere}}$, so the constant-mass density hydrostatic pressure law becomes $p_g={\rho }_{f}gh$. Here $p_g=0$ at the surface denotes equality with atmospheric pressure. It does not denote zero absolute pressure. ^[ref]

The magnitude of the gravitational force remains approximately $mg$ during a laboratory swing. Its tangential component changes with position, however. For simple pendulum length $\ell$ and angle $\theta$ measured from the downward vertical,This restoring component is zero at the lowest point and largest in magnitude at the turning points. It always accelerates the bob towards the lowest point. The radial gravitational component is $mg\cos \theta$; the string tension $S$ must additionally supply the centripetal acceleration, so $S-mg\cos \theta =m{v}^2/\ell$. Thus it is the radial resultant, not gravity alone, that bends the trajectory. The bob has maximum speed at the bottom even though its tangential acceleration is then zero.

For a small angular amplitude, the small-angle approximation gives simple harmonic motion with oscillation period $T=2\pi \sqrt{\ell /g}$. Measure $\ell$ from pivot to the bob's centre of mass, time many complete swings, divide the elapsed time by their number, and calculateRepeat with several lengths and plot $T^2$ against $\ell$; the ideal slope is $4{\pi }^2/g$. The first-order relative uncertainty obeys $\delta g/g\simeq \delta \ell /\ell -2\delta T/T$, with absolute uncertainty contributions combined according to their dependence. Keep the amplitude small: the finite-amplitude period is $T=T_0(1+{\theta }_0^2/16+\cdots )$, with ${\theta }_0$ in radians. Using the longer finite-amplitude period in the small-amplitude formula would underestimate $g$. Pivot friction, a finite bob and support motion are further experimental corrections.^[ref]

The moment of inertia about an axis is $I=\int r_{\perp }^{2}dm$. The parallel axis theorem states $I=I_{\mathrm{CM}}+M{a}^2$, where a is the perpendicular separation from a parallel axis through the centre of mass, and M is total mass.

Since $\rho =k{r}^2$, the density coefficient has units $\mathrm{kg}{\mathrm{m}}^{-5}$. The density-power cylindrical moment of inertia follows from cylindrical shells:For an offset pivot, the physical pendulum equation is $(I_{\mathrm{CM}}+M{a}^2)\ddot{\theta }=-Mga\sin \theta$. At small amplitude,For the minimum-period offset of a physical pendulum, minimizing $T^2$ is equivalent to minimizing $a+2R^2/(3a)$. Its derivative vanishes atThe second derivative is positive. The period diverges as $a^{-1/2}$ near zero and grows as $a^{1/2}$ for large offset, giving one minimum. This assumes the stated ideal pivot can pass inside the body; an externally constrained pivot would restrict the allowable a. ^[ref]

For a small-slope string element, the transverse net tension is $T[y_x(x+\delta x)-y_x(x)]\simeq T{y}_{xx}\delta x$. Its mass is $\rho \delta x$, so Newton's second law givesIts extra length is $[\sqrt{1+y_x^2}-1]\delta x\simeq y_x^2\delta x/2$. Work against the approximately constant tension gives the elastic energy of a small-slope string, while transverse motion gives its kinetic energy:For a single travelling wave $y=f(x\pm ct)$, $y_t^2=c^2{y}_x^2$, so $dK=dU$ pointwise to this order. A standing wave need not have pointwise or instantaneous kinetic–potential equality.

For a fixed-end string of length L, taking the stated note to be its fundamental gives $c=2Lf$. HenceIf98Hz were instead the nth harmonic, the required tension would be $16.5/n^2\mathrm{N}$; the usual fundamental interpretation is explicit.

For the specified standing wave, $\dot{y}=A\omega \sin kx\cos \omega t$. Every allowed nonzero fixed-end mode has ${\int }_0^L{\sin }^{2}kxdx=L/2$, yieldingAt this instant the ideal mode's elastic energy is zero; at maximum displacement all its energy is elastic. ^[ref]

A conservative force has work depending only on the endpoints, so it can be writtenThe arbitrary additive constant in potential energy has no effect on force. A smooth curl-free force is conservative on a simply connected domain; a hole or excluded singularity requires checking global circulation as well.

Uniform gravity has $U=mgz$, central gravity has $U=-GMm/r$ with zero at infinity, and a linear spring has $U=k{x}^2/2$ relative to equilibrium. If the potential is time independent and no other force does work, $K+U$ is conserved by the work-energy theorem. A time-dependent potential or nonconservative work can change mechanical energy; having an instantaneous gradient force alone is not sufficient for energy conservation. ^[ref]

Treat Earth as a stationary spherical source, with $\mu =G{M}_E$, and neglect drag and other bodies. The conserved quantities are total mechanical energy and the angular momentum vector, implying a fixed orbital plane and constant areal velocity. The Kepler problem additionally conserves the Laplace-Runge-Lenz vector. The orbiting object's linear momentum is not constant because gravity exerts a force.

For a circular orbit, $m{v}_c^2/r_0=\mu m/r_0^2$, so $K=\mu m/(2r_0)=-U/2$. HenceAt the closest approach of a noncircular orbit, the velocity is tangential and $v_0=k\sqrt{\mu /r_0}$. Its energy and specific angular momentum areThe pericentre-speed qualification of Kepler orbit classes requires $k\ge 1$ when $r_0$ is genuinely the closest approach. For an ellipse, conservation of h gives $v_m=v_0{r}_0/r_m$. Equating energies at the two turning points and writing $x=r_m/r_0$ givesThe distinct apocentre root is thereforeFor an unbound orbit, $v^2(r)=2E/m+2\mu /r$ decreases toward infinity. The asymptotic minimum speed of an unbound Kepler orbit isIt is the limiting infimum, reached only asymptotically: positive for a hyperbola and zero for a parabola. The turning-point interpretation of $r_0$ is essential; if $0<k<1$, the same instantaneous energy describes a bound orbit whose $r_0$ is the apocentre instead. ^[ref]

A U-tube differential manometer contains a connected column of liquid with one gas connection above each liquid meniscus. Write $p_L,p_R$ for the left and right pressures, and $z_L,z_R$ for the meniscus heights on an upward vertical coordinate. Assume constant mass density $\rho$ and downward gravitational acceleration of magnitude $g$; neglect the gas column's mass density and differences in capillary pressure. The signed vertical length difference $\Delta h=z_R-z_L$ then givesLarger left pressure means a lower left liquid meniscus: $p_L>p_R$ implies $z_L<z_R$. Equal pressures give equal meniscus heights.

The balance follows directly from hydrostatic pressure. At a common height $z_0$ below both surfaces, the two points in the connected homogeneous liquid have equal pressure:Subtracting cancels the common bottom depth and leaves $p_L-p_R=\rho g(z_R-z_L)$. The same-height rule follows from hydrostatic equilibrium, $\nabla p=-\rho g{\mathbf{e}}_z$: a horizontal displacement changes no pressure, and for constant $\rho$ the quantity $p+\rho gz$ is constant throughout the connected liquid. Equal height alone is insufficient for unrelated vessels or for fluid flow with significant acceleration.

For an original constructed example, choose water of stipulated mass density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, and vertical length difference $\Delta h=0.060\mathrm{m}$. ThenThe SI pressure unit is the pascal: the product has units $\mathrm{kg}{\mathrm{m}}^{-3}\times \mathrm{m}{\mathrm{s}}^{-2}\times \mathrm{m}=\mathrm{kg}{\mathrm{m}}^{-1}{\mathrm{s}}^{-2}=\mathrm{Pa}$. Swapping the two connections gives $-588.6\mathrm{Pa}$ and reverses which liquid meniscus is lower. In equal-bore arms, volume conservation makes each surface move $30\mathrm{mm}$ from its initial equal-height position; using only one surface's displacement would halve the inferred pressure. With unequal arm areas $A_L,A_R$, a left fall $a$ and right rise $b$ instead obey $A_{L}a=A_{R}b$, while the vertical length difference is still $a+b$.

If the right port is open to the local atmosphere, $p_L-p_R$ is the left gauge pressure. Only with that reference and known atmospheric pressure can one obtain absolute pressure fromFor an explicitly stipulated atmospheric pressure of $100000\mathrm{Pa}$, the example gives $100588.6\mathrm{Pa}$ absolute pressure. If the right port is connected to another vessel or a pressure tapping, the same reading is a differential pressure and supplies no absolute pressure without knowing that reference.^[ref]

NASA's Wind Tunnel Experiment Details, slide 27, describes a practical manometer measurement: partially fill a Tygon tube with water and connect its two ends to total pressure and static pressure ports in a wind tunnel. The total pressure connection pushes its liquid meniscus downward; the vertical length difference supplies the pressure difference. The slide predicts increasing separation as the tunnel speed increases. It supplies an apparatus and method, rather than a table of observed readings.^[ref]

A control for this practical is to expose both connections to the same atmospheric pressure before the run: settled liquid meniscus heights should agree in a symmetric column. Read each liquid meniscus at eye level, reconnect sealed tubing to the identified ports, and wait for settling. Recheck the equal-pressure zero afterwards; reverse connections to check the sign. These checks establish neither mass density nor ruler accuracy.^[ref]

The inference is a differential pressure from hydrostatic pressure, with stagnation pressure on one side and static pressure on the other in NASA's practical. Converting that difference to a flow speed additionally needs the hypotheses discussed under dynamic pressure and Pitot-static tube. A changing pressure can leave the column oscillating; leaks, blocked pressure tappings or trapped liquid in a gas connection can alter or delay the readout. The column should be read after settling, and temperature should be recorded when choosing the measuring liquid's mass density.

Two corrections clarify the limits. If both gas ports are at the same height and carry the same gas of constant mass density ${\rho }_g$, keeping its hydrostatic pressure gives $p_L-p_R=(\rho -{\rho }_g)g\Delta h$. Different port heights require their separate gas column heights. Also, a curved liquid meniscus can sustain a capillary pressure jump through surface tension. Write $J_L,J_R$ for liquid pressure minus gas pressure at the two surfaces. Neglecting gas weight, the balance becomes $p_L-p_R=\rho g\Delta h+J_R-J_L$. Identical meniscus shapes in matching clean bores cancel those jumps; unequal bores or different wetting conditions need not. Thus the useful simplicity of $\rho g\Delta h$ rests on identifiable physical conditions.