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Translation and rotation. For particles constituting a rigid body, the total linear momentum isThus the total external force governs the motion of the centre of mass. About that centre of mass, the angular momentum and applied torque arewhere primes denote relative positions and velocities. The derivative of the vector is taken relative to nonrotating axes. The inertia tensor relates angular momentum to angular velocity, ${\mathbf{L}}_{\mathrm{CM}}=\mathsf{I}\mathbf{\omega }$. For rotation about a fixed unit direction $\mathbf{n}$, the moment of inertia is $I_{\mathbf{n}}=\int r_{\perp }^{2}dm=\mathbf{n}\cdot \mathsf{I}\mathbf{n}$; it measures the resistance to changing that rotation and gives rotational kinetic energy $I_{\mathbf{n}}{\omega }^2/2$. The simpler vector relation $\mathbf{L}=I\mathbf{\omega }$ applies when the rotation axis is a principal axis, not for every orientation of an arbitrary rigid body. These translational and rotational balances must be applied together; a nonzero torque does not replace the net-force equation.

Gyroscope precession. A rapidly spinning gyroscope has a large angular momentum approximately along its spin axis. The torque component parallel to this vector changes its magnitude, while the perpendicular component turns its direction. If the vector undergoes precession with angular velocity $\mathbf{\Omega }$, then over $dt$ its change is the infinitesimal rotation $d\mathbf{L}=(\mathbf{\Omega }dt)\times \mathbf{L}$. HenceThis describes direction-changing precession at constant $|\mathbf{L}|$; any spin-up contribution must be added separately. For perpendicular torque and angular momentum, the precessional component perpendicular to $\mathbf{L}$ is ${\mathbf{\Omega }}_{\perp }=\mathbf{L}\times \mathbf{G}/L^2$. The familiar tendency of a supported gyroscope to turn sideways is this vector balance, not the absence of gravitational force.

Wheel-lift threshold without the flywheel. Let $h=0.50\mathrm{m}$ be the centre of mass height, $a=1.0\mathrm{m}$ half the wheel separation, and $R_b=10\mathrm{m}$ the bend radius. The required inward force is $F=M{v}^2/R_b$, provided by static friction. Let $N_o,N_i$ be the summed outer and inner wheel normal forces. Vertical balance gives $N_o+N_i=Mg$. Moment balance about the centre of mass givesThe inner wheels lift when $N_i=0$. Thus the vehicle rollover threshold isThe stated mass cancels. This assumes the rigid geometry and sufficient static friction specified in the paper; the threshold is set by loss of contact, not by sliding.

Engine flywheel. The moment of inertia of a uniform disc givesIts spin angular speed, spin angular momentum, and turn angular speed at $40\mathrm{km}{\mathrm{h}}^{-1}$ areThe transverse flywheel axis is perpendicular to the vertical turn axis, soTo specify its direction, use right-handed axes $+x$ forward, $+y$ to the car's left, and $+z$ upward, and consider a left turn. Then $\mathbf{\Omega }=+\Omega \hat{\mathbf{z}}$. Choose the flywheel angular momentum to the car's right, ${\mathbf{L}}_f=-L_f\hat{\mathbf{y}}$. The car must applyto obtain the required precession. This gyroscopic roll torque in a turning vehicle opposes the outward roll tendency in a left turn. By the right-hand rule, the flywheel must turn counterclockwise viewed from the car's right side toward its centre, opposite to the rotation of wheels rolling forward. Reversing flywheel spin would aggravate the outward roll. Reversing the turn reverses both roll and gyroscopic torques, so the same flywheel spin remains helpful.

For comparison, the outward inertial roll moment at the stated speed is $Mh{v}^2/R_b=6173\mathrm{N}\mathrm{m}$, and the limiting gravitational restoring moment is $Mga=9810\mathrm{N}\mathrm{m}$. The gyroscopic reaction is about $30\%$ of the former and $19\%$ of the latter, so it is appreciable for this large flywheel. If this ideal model is extended to include the favorable gyroscopic reaction in the wheel-lift condition, it gives $Mh{v}^2/R_b-L_{f}v/R_b=Mga$, or about $56.8\mathrm{km}{\mathrm{h}}^{-1}$ rather than $50.4\mathrm{km}{\mathrm{h}}^{-1}$. This quantifies a useful but finite increase in the rollover threshold.

Pressure measures the compressive normal force per unit surface area. For a uniformly loaded surface of surface area $A$, the magnitude of the perpendicular force is $F_{\perp }=PA$; locally, $P=d{F}_{\perp }/dA$. Its SI unit is the pascal, $1\mathrm{Pa}=1\mathrm{N}{\mathrm{m}}^{-2}$. A force parallel to the surface contributes shear stress, rather than pressure. In a gas, the steady macroscopic force comes from many microscopic transfers of momentum; ideal gas pressure from wall collisions makes this connection explicit. ^[ref]

For a computed example, a uniform pressure of $100\mathrm{kPa}$ on a piston of surface area $10\mathrm{c}{\mathrm{m}}^2=10^{-3}{\mathrm{m}}^2$ produces a perpendicular force of $100\mathrm{N}$. If the opposite face is exposed to $80\mathrm{kPa}$, the net force due to the two pressures is $(100-80)\times 10^3\times 10^{-3}=20\mathrm{N}$, directed toward the lower-pressure side. The pressure on either face and their difference answer different questions; gauge pressure records a difference from a specified reference, often atmospheric pressure.

A scalar pressure describes a gas in local thermodynamic equilibrium with isotropic random particle velocities: every wall orientation has the same normal force per unit surface area. Without this isotropy, normal stresses can depend on direction and a pressure tensor is needed. Thus random motion alone is insufficient; a preferred direction matters.

In thermodynamics, pressure and volume are thermodynamic conjugate variables. For a simple compressible system, reversible expansion delivers work $\delta W=PdV$, and at fixed entropy $S$ and particle number $N$,where $U$ is internal energy. A large pressure therefore means both a large wall force per unit surface area and a large amount of expansion work per small increase in volume.

For a rigid body rolling on a stationary support, the instantaneous contact point $P$ has zero velocity: ${\mathbf{v}}_P={\mathbf{v}}_G+\mathbf{\omega }\times \overrightarrow{GP}=0$. For a wheel of radius $R$ moving horizontally and a counterclockwise angle $\varphi$, this gives $\dot{x}+R\dot{\varphi }=0$, hence $x+R\varphi$ is constant. Maintaining this mechanical constraint generally requires sufficient static friction and a nonnegative normal force.

First test rest. With no other tangential forces, the required uphill static friction is $f_s=mg\sin \theta$. Its maximum is ${\mu }_{s}N$, where ${\mu }_s$ is the coefficient of static friction. ThusThe tangent function is increasing on this angle range, and its inverse tangent determines the last permitted resting angle ${\theta }_c$. At equality, the ideal model still permits limiting rest; a larger angle excludes it. Below the threshold, setting $f_s={\mu }_{s}N$ would incorrectly impose the maximum static friction even when less is required. Static friction on an inclined plane and inclined-plane measurement of static friction develop this onset criterion further.

For a block already sliding downhill, uphill kinetic friction has magnitude $f_k={\mu }_{k}N$, where ${\mu }_k$ is the coefficient of kinetic friction. Taking downhill as positive, Newton's second law becomesEvery term in the force balance scales with mass; division by $m>0$ removes it. Doubling mass doubles both gravitational weight and the contact forces, while leaving this acceleration unchanged if the coefficients remain unchanged. A mass-independent result is a prediction of these assumptions, rather than a universal property of every ramp.

For an original numerical model, stipulate $m=3\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$, ${\mu }_s=0.45$ and ${\mu }_k=0.30$. The critical angle is ${\theta }_c\simeq 24.23^{\circ }$. At $20^{\circ }$, the normal force is $27.627\mathrm{N}$; rest requires $10.055\mathrm{N}$ of uphill static friction, below its $12.432\mathrm{N}$ bound. A resting block can stay there. If it is already sliding downhill, however, the kinetic friction is only $8.288\mathrm{N}$, giving downhill acceleration $a\simeq 0.5891m/{\mathrm{s}}^2$. Specifying the initial state matters.

At $30^{\circ }$, the downhill component of gravitational weight is $14.700\mathrm{N}$, while $N\simeq 25.461\mathrm{N}$ and the static friction bound is only $11.458\mathrm{N}$. Rest is impossible. During downhill sliding,Released from rest on a sufficiently long ramp, this ideal block covers its first $1.00\mathrm{m}$ in $t=\sqrt{2L/a}\simeq 0.9218\mathrm{s}$, reaching speed $v=\sqrt{2aL}\simeq 2.1697m/s$. As an independent energy calculation, the lost gravitational potential energy is $mgL\sin \theta =14.700\mathrm{J}$ and kinetic friction removes $f_{k}L\simeq 7.638\mathrm{J}$ of mechanical energy. The remaining $7.062\mathrm{J}$ equals $m{v}^2/2$ by the work-energy theorem.

At $\theta =\arctan {\mu }_k$, downhill sliding has zero acceleration and therefore constant velocity. Below that angle, a downhill-moving block slows; when it stops, test the static friction bound again. For uphill sliding, kinetic friction points downhill, so the signed downhill acceleration instead is $g(\sin \theta +{\mu }_k\cos \theta )$. The sign follows contact slip, and the downhill formula cannot be used unchanged after a reversal.

An ideal contact mechanical constraint supplying a normal reaction and no tangential friction. For a stationary constraining surface, the permitted velocity is a tangent vector, so the normal force has zero power and does no work. For a moving surface, normality alone does not imply zero power in the laboratory frame.

A uniform solid sphere released from the top of a fixed spherical surface has $v^2=10g(R+r)(1-\cos \theta )/7$ under the ideal rolling constraint. The normal force is $Mg(17\cos \theta -10)/7$, giving the title's ideal detachment angle. The required static friction is $2Mg\sin \theta /7$. Its ratio to the normal force diverges at detachment, so any finite coefficient of static friction causes prior slipping; a physical subsequent trajectory requires a sliding-friction model.

Static friction is the tangential contact force that maintains zero contact slip. “Static” describes the contact state: both surfaces can move together in the chosen inertial frame. In the ideal dry-contact model, the actual static friction adjusts to the equations of motion, subject to a bound, rather than always taking its maximum value.^[ref]Here $N$ is the magnitude of the normal force and ${\mu }_s$ the coefficient of static friction. Equality means limiting static friction; it is a threshold, not an instruction to replace every unknown contact force by ${\mu }_{s}N$.

For a stipulated example, place a nonrotating $2\mathrm{kg}$ block initially at rest on a stationary horizontal surface. Take gravitational acceleration $g=9.8m/{\mathrm{s}}^2$ and coefficient of static friction ${\mu }_s=0.40$, with no other vertical force. The vertical force balance gives $N=mg=19.6\mathrm{N}$, so the largest allowed static friction magnitude is $7.84\mathrm{N}$. Choose right as positive and apply a horizontal force $F$. Remaining at rest requires Newton's second law to giveAt $F=2\mathrm{N}$, the static friction is $-2\mathrm{N}$; at $F=6\mathrm{N}$, it is $-6\mathrm{N}$. Both states have zero acceleration. At $F=10\mathrm{N}$, balancing would require $f_s=-10\mathrm{N}$, outside the bound, so this initially resting block cannot remain at rest. Its subsequent acceleration requires a kinetic friction law; setting $f_s=-7.84\mathrm{N}$ after contact slip begins would still use the wrong contact state. At exactly $F=7.84\mathrm{N}$, the ideal inequality still permits rest at the threshold. These are model calculations, not measured material properties.

The toy model used in kinetic friction makes the contact-state distinction explicit: $m=1\mathrm{kg}$, $N=10\mathrm{N}$, ${\mu }_s=0.5$ and coefficient of kinetic friction ${\mu }_k=0.3$. On stationary ground, an initially resting box pushed right with $4\mathrm{N}$ has $f_s=-4\mathrm{N}$. At $6\mathrm{N}$ it cannot remain at rest; once it slides right, $f_k=-3\mathrm{N}$ and $a=3m/{\mathrm{s}}^2$. At $5\mathrm{N}$, rest is allowed with $f_s=-5\mathrm{N}$, whereas an already right-sliding box has $f_k=-3\mathrm{N}$ and $a=2m/{\mathrm{s}}^2$. The applied force alone does not specify the friction branch: contact slip and motion history matter.

For a flat contact with two tangential directions, the bound applies to a vector norm, not separately to each component:For the stipulated $7.84\mathrm{N}$ bound, balancing horizontal components $4\mathrm{N}$ and $4\mathrm{N}$ requires a force magnitude $\sqrt{32}\mathrm{N}\simeq 5.657\mathrm{N}$, which is allowed. Components $6\mathrm{N}$ and $6\mathrm{N}$ require $\sqrt{72}\mathrm{N}\simeq 8.485\mathrm{N}$, which is forbidden even though each component separately is below $7.84\mathrm{N}$. This circular bound assumes the same coefficient of static friction in every tangential direction. Near impending contact slip, the limiting static friction opposes the relative contact motion that would otherwise begin. Once actual contact slip exists, kinetic friction instead opposes its tangential relative velocity, which need not have the direction of the body's center of mass motion.

The normal force must also come from the equations of motion. On horizontal ground, an upward pull of magnitude $P$ at angle $\alpha$ above horizontal gives, for a block remaining at rest with no other forces,The sine component unloads the contact and the cosine component demands tangential support. These equations require $N>0$; if the supporting contact opens, this dry-contact model cannot supply a negative normal force. For the stipulated $2\mathrm{kg}$ block and $\alpha =30^{\circ }$, the positive threshold is $P={\mu }_{s}mg/(\cos \alpha +{\mu }_s\sin \alpha )\simeq 7.3544\mathrm{N}$, with $N\simeq 15.9228\mathrm{N}$ and $|f_s|\simeq 6.3691\mathrm{N}$. A downward pull increases $N$ instead. An inclined plane gives another reason why $N$ is not always $mg$.

To solve a new problem, impose zero contact slip, find the required static friction and normal force from the equations of motion, then check the bound. A demand inside it permits that contact state; a demand outside it rules the state out. For an ideal driven tire undergoing rolling without slipping on stationary ground, a forward static friction can accelerate the vehicle. Its role is to prevent contact slip, rather than to oppose all motion of the whole body.

Nor does the word “static” by itself make friction do zero work. The contact power delivered to a body is $P_s={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, a dot product with its material contact-point velocity in the chosen inertial frame. At an ideal contact undergoing rolling without slipping on stationary ground, ${\mathbf{v}}_c=0$, so this power is zero. For a rigid body, adding the translational and rotational contributions gives ${\mathbf{f}}_s\cdot {\mathbf{V}}_{\mathrm{CM}}+{\mathbf{\tau }}_{\mathrm{CM}}\cdot \mathbf{\omega }={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, where ${\mathbf{\tau }}_{\mathrm{CM}}$ is the contact torque about the center of mass and $\mathbf{\omega }$ its angular velocity. Counting only the first term omits the rotational contribution.

At a moving contact with zero contact slip, the two material contact points share a velocity ${\mathbf{v}}_c$ that can be nonzero. For instance, a nonrotating block moving with an accelerating belt receives positive contact power $f_{s}U$ when the belt's rightward velocity $U$ and rightward static friction $f_s$ are positive. The belt receives $-f_{s}U$ by Newton's third law. These contact contributions sum to zero in the ideal no-slip model, while the block's kinetic energy can increase. Zero contact slip and zero work on each body are different conditions.

An inclined plane tilted slowly from horizontal turns a coefficient of static friction into a critical angle. Assume a nonrotating block on a stationary, rigid, dry surface, with gravitational weight and the contact forces as its only forces. Take $0\le \theta <\pi /2$ to be the angle between the uphill surface direction and horizontal. Let ${\mathbf{e}}_t$ be a unit vector pointing uphill and ${\mathbf{e}}_n$ a perpendicular unit vector pointing outward from the surface. Resolving the gravitational weight givesThe angle between the downward vertical and the inward normal $-{\mathbf{e}}_n$ is also $\theta$. Thus the inward component uses cosine and the downhill component uses sine. These are components of one force, not two additional forces.

Before downhill contact slip, zero acceleration and Newton's second law requireThe required static friction points uphill. Substituting into the bound and dividing by the positive $mg\cos \theta$ givesThe tangent function increases on this range of angles, so the last admissible resting angle isThe inverse tangent takes a dimensionless quantity to an angle. At equality the ideal model still permits a limiting resting state; increasing $\theta$ further excludes it. At a smaller angle, $f_s/N=\tan \theta$ is the demand ratio, not the coefficient of static friction.^[ref]

For the stipulated $m=2\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$ and ${\mu }_s=0.40$, the critical angle is approximately $21.8014^{\circ }$. At $\theta =15^{\circ }$, $N\simeq 18.932\mathrm{N}$ and $f_s\simeq 5.073\mathrm{N}$, comfortably below the bound ${\mu }_{s}N\simeq 7.573\mathrm{N}$. At $\theta =25^{\circ }$, maintaining rest would require $f_s\simeq 8.283\mathrm{N}$, but the bound is only $7.105\mathrm{N}$: rest is impossible. At ${\theta }_c$, $N\simeq 18.198\mathrm{N}$ and $f_s\simeq 7.279\mathrm{N}$. Notice that the limiting tangential force here is smaller than the horizontal $7.84\mathrm{N}$ limit because tilting reduces the normal force.

Cancellation of mass makes this relation useful: the ideal critical angle tests a contact coefficient without needing a calibrated force gauge. Doubling the block's mass doubles both demanded static friction and its bound, leaving ${\theta }_c$ unchanged if ${\mu }_s$ is unchanged. This is a testable model prediction. It need not survive a change in surface condition, a direction-dependent coefficient, appreciable deformation, or vibration.

The derivation also assumes continued contact and that sliding occurs before tipping or rolling without slipping. The resultant contact force must act within the supporting footprint so that the block can satisfy torque balance; a tall block may tip first. No extra tension, imposed tangential pull, or lift is included. In a slowly raised apparatus, quasistatic process means that motion of the board contributes negligible inertial forces while locating the threshold. Once downhill contact slip begins, a constant-velocity sliding angle would concern kinetic friction, not the onset of static friction.

A third-law interaction pair consists of the two forces belonging to the same mutual interaction, with their agent and recipient exchanged. The names action-reaction pair do not imply a delay: in the instantaneous model, the two forces occur together. Test a proposed pair by naming both bodies and the interaction, rather than merely noticing equal magnitudes and opposite directions.

For a book resting on a table, the table's upward normal force and Earth's downward gravitational weight both act on the book. They can balance in the book's free-body diagram, but they are not a third-law interaction pair. The partner of the table's normal force on the book is the book's downward contact force on the table. The partner of Earth's gravitational force on the book is the book's gravitational force on Earth. Their equality does not require equal accelerations: Earth's much greater mass makes its response correspondingly smaller for a given net force.

Similarly, friction on a sliding block from a surface is paired with friction on that surface from the block. For the collection containing only the block, that friction is an external force; including the contacting surface makes both members internal forces. Changing the collection changes the bookkeeping of momentum, not the physical interaction.

For a rigid vehicle turning on a level road, with centre of mass height $h$, half track width $a$ and turn radius $R_b$, sufficient static friction supplies centripetal acceleration $v^2/R_b$. The inner and outer wheel normal forces obey $N_i+N_o=Mg$ and $a(N_o-N_i)=Mh{v}^2/R_b$. The inner contact force first vanishes at the displayed speed. The approximation neglects suspension movement and extra torques; inadequate static friction can cause sliding before rollover.