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A centripetal force is the inward resultant force needed for circular motion. Newton's second law and centripetal acceleration give magnitude $m{v}^2/r$ for a particle of mass $m$, speed $v$, and path radius $r$. It is a role played by an actual interaction, such as tension or an electric force, rather than an additional interaction to add to a free-body diagram. A purely radial centripetal force is perpendicular to the velocity, so it does no work and leaves the speed unchanged.

A contact force is an interaction transmitted between bodies through a region where their material surfaces meet. A support pushing on a resting object and friction acting along a sliding interface are examples. This macroscopic classification does not mean there is a new fundamental interaction of exactly zero range; the underlying material response involves electromagnetic and quantum interactions.

At a surface, resolve the force into a normal contribution and a tangential contribution:For a stipulated nonadhesive contact, $N\ge 0$ describes compression. In a dry static friction model, $\Vert {\mathbf{F}}_t\Vert \le {\mu }_{s}N$ is a bound, not a claim that friction always equals its maximum. The contact force adjusts to the force and moment balance while that bound and the other model constraints can be met.^[ref]

For an original inclined-plane example, a $2\mathrm{kg}$ block rests on a plane at $20^{\circ }$. With $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the required normal and tangential forces areA chosen ${\mu }_s=0.40$ permits up to $7.38\mathrm{N}$, so this static balance is feasible in the stated model. The actual friction is $6.71\mathrm{N}$, not $7.38\mathrm{N}$. The resultant contact force balances gravitational weight, although its normal part alone does not.

Nonpenetration constraints describe a further feature of rigid contact. If the gap is $g_n\ge 0$, a simple unilateral model imposes $N\ge 0$ and $g_{n}N=0$: separated surfaces do not push on one another, while touching surfaces may carry a compressive force. Adhesion, finite deformation and fluid films require different or extended assumptions.

A force sensor between two bodies can measure the transmitted interaction while motion tracking tests the resultant-force balance. The equal and opposite contact force on the other body belongs on that other body's free-body diagram. Those two forces do not cancel within the force sum for just one chosen body.

A force describes an interaction's contribution to the rate of change of momentum. In an inertial frame, Newton's second law relates the vector sum of external forces on a particle to $d\mathbf{p}/dt$. For constant Newtonian mass, this becomes $\sum \mathbf{F}=m\mathbf{a}$.

A cart pulled along a nearly frictionless track illustrates the distinction between one applied force and the resultant. A spring scale measures the pull, while video tracking measures acceleration; friction and track slope must also enter the balance. Opposing pulls can cancel, leaving zero acceleration even though each pull is nonzero.

A force's point or line of application also matters for an extended body, because it determines the torque. The same resultant force can produce different rotational motion when applied at different locations. A free-body diagram makes both the chosen body and its external interactions explicit.

A free-body diagram isolates a body and shows the forces exerted on it. A third-law interaction pair acts on different bodies and belongs in separate body diagrams.

Choose a unit vector ${\mathbf{e}}_d$ pointing downhill and a perpendicular unit vector ${\mathbf{e}}_n$ pointing outward from the inclined plane. The gravitational weight points vertically downward. Its scalar projections onto these directions giveThe angle between the downward vertical and the inward normal is $\theta$: the inward component uses cosine, and the downhill component uses sine. These are two components of one force. They replace its vertical arrow in a component calculation; adding them to that same arrow would count gravitational weight twice.

Continued contact with a stationary straight inclined plane gives zero perpendicular acceleration. The perpendicular component of Newton's second law therefore givesThis is why the normal force generally differs from $mg$: it balances the inward component, rather than the whole gravitational weight. An extra lifting force, a curved surface, or motion of the support would require a new perpendicular balance.^[ref]

The same geometry explains the ramp's usefulness. A frictionless block pulled uphill at constant velocity needs a parallel force $P=mg\sin \theta$. Travelling a distance $L$ raises it through height $h=L\sin \theta$, so the pulling work is $PL=mgh$, its gain in gravitational potential energy. The smaller force costs a larger displacement; the work-energy theorem keeps the accounting consistent.

An idler support reaction is the force exerted by the shaft and bearings that balances the mesh loads and other external loads on an idler gear. A vanishing net transmitted torque does not imply a vanishing resultant force: moments can cancel while forces add. A free-body diagram must retain each tooth-contact force and its point of application. The tangential load in each ideal mesh follows from $F_t=|\tau |/r_w$.^[ref]

Consider a steady collinear train with input, idler and output reference radii20,30 and50mm. An input torque $2\mathrm{N}\mathrm{m}$ gives a100N tangential mesh component. On the idler, choose contact positions at $x=-30$ and $x=30\mathrm{mm}$ relative to its center, with both tangential forces upward. Their moments are $-3$ and $+3\mathrm{N}\mathrm{m}$, but their vector sum is200N upward. For equal20degree pressure angles, the two radial mesh components have magnitude $100\tan 20{}^{\circ }\approx 36.4\mathrm{N}$ and opposite directions, so they cancel. Neglecting other loads, the net support reaction is200N downward.

This is the combined support resultant, not automatically the load on each bearing. For a hypothetical shaft supported at axial positions $z=0$ and $z=100\mathrm{mm}$, with the200N upward gear resultant acting at $z=40\mathrm{mm}$, static equilibrium givesThus $R_B=-80\mathrm{N}$ and $R_A=-120\mathrm{N}$. Idler bearing load sharing depends on placement even when the total reaction is fixed.

Noncollinear shaft layouts rotate the mesh-force directions, and helical gears add gear axial force. Acceleration also changes the balance: the sum of torques must supply the idler's angular acceleration, and unequal mesh loads need not cancel radially. The zero-net-torque example therefore establishes a steady ideal force balance, rather than a universal bearing-load formula.

For a nonzero force applied at position ${\mathbf{r}}_0$, its line of action is the straight line through that point parallel to the force:The force direction and line position together determine its torque about a chosen origin. They contain more mechanical information than the magnitude alone.

The moment about the origin is $\mathbf{\tau }={\mathbf{r}}_0\times \mathbf{F}$. Moving the application point by a displacement parallel to $\mathbf{F}$ leaves the moment unchanged because $\Delta \mathbf{r}\times \mathbf{F}=0$. The principle of transmissibility uses this equivalence for the external motion or equilibrium of an ideal rigid body. The lever-arm formulation expresses the same geometry: the torque magnitude is $Fd$, where $d$ is perpendicular distance from the origin to the line.^[ref]

For a constructed two-dimensional example, a force $(0,10)\mathrm{N}$ acts at $(0.30,0.20)\mathrm{m}$. Its line is $x=0.30\mathrm{m}$ and its moment is ${\tau }_z=0.30(10)=3.0\mathrm{N}\mathrm{m}$. Moving the application point to $(0.30,0.80)\mathrm{m}$ changes neither force nor moment. Moving it instead to $x=0.50\mathrm{m}$ gives $5.0\mathrm{N}\mathrm{m}$, so the unchanged force magnitude does not imply the same rigid-body effect.

An equal and opposite pair on different parallel lines forms a couple: the resultant force is zero but a moment remains. Consequently, not every loading system can be replaced by one force with one line of action. A free-body diagram should retain the relevant forces and couples.

Transmissibility is an external rigid-body statement. Applying a load at a different location on a deformable object can change internal stress, deformation and local contact even if the resultant and moment are preserved. A static load-cell comparison and a deformation measurement can therefore answer different questions about apparently equivalent loading arrangements.

In an inertial frame, a body with zero net external force remains at rest or moves with constant velocity. For a point mass of constant positive mass, write the net external force as the vector sum of all external forces on that body:Here ${\mathbf{r}}_0$ and ${\mathbf{v}}_0$ are the position and velocity at time $t_0$. The zero net external force must hold throughout the interval, not merely at one instant. Both the magnitude and direction of the velocity stay fixed: constant speed along a curved trajectory still involves acceleration. Rest is the case ${\mathbf{v}}_0=0$. For an extended closed system, the analogous statement concerns its center of mass; zero net external force need not prevent its parts from moving relative to one another.

The short mathematical argument is to use Newton's second law with $m>0$, then take the integral of $d\mathbf{v}/dt=0$ and of $d\mathbf{r}/dt={\mathbf{v}}_0$. This gives the two constants fixed by the initial position and velocity. The physical content also includes the existence of inertial frames in which these relations apply; an arbitrary accelerating reference frame need not show constant velocity for a body with zero actual net external force.

The useful question is therefore which forces change the motion, rather than which force keeps it going. A free-body diagram may show several nonzero forces whose vector sum is zero. Equal opposing forces on the same body can balance; a third-law interaction pair acts on different bodies and cannot be cancelled within that one body's diagram.

In 1687, Isaac Newton stated Law I in Philosophiae Naturalis Principia Mathematica: a body's rest or uniform straight motion persists unless impressed forces compel a change. The following paragraph identifies air resistance and gravity as influences on the trajectory of a thrown body. The vector equations above are a modern expression of the law, not a quotation of his notation.^[ref]

In the pair-interaction model of Newtonian mechanics, two bodies exert equal and opposite forces on each other:Both vectors are expressed in the same inertial frame at the same time. The subscripts name the agent and the recipient: $A$ acts on $B$, and $B$ acts on $A$. These are forces on different bodies, so they cannot cancel in the net force on either body. Newton's second law still uses all the forces on the particular body whose acceleration is sought. Identifying this third-law interaction pair does not require the bodies to be isolated: additional external forces can act at the same time.

The distinction makes Newton's third law useful both for drawing a free-body diagram and for deriving momentum conservation. A body-by-body calculation retains one member of each third-law interaction pair; a calculation for a collection containing both bodies cancels that pair in the total momentum balance. Equal interaction forces consequently coexist with nonzero, generally unequal accelerations.

Isaac Newton stated the equality and opposition of mutual actions in Law III of his 1687 Philosophiae Naturalis Principia Mathematica. The explanation distinguishes equal opposite changes of momentum from changes of velocity, which are inversely proportional to the bodies' masses when other influences are absent. Corollary III then explains why mutual interactions preserve the signed total momentum. The equations here are modern notation for that distinction. ^[ref]

The instantaneous body-pair equation has a scope: use it for the mechanical interactions assumed by this Newtonian mechanics model. In electromagnetism, bodies can exchange momentum with an electromagnetic field; their instantaneous mutual forces need not sum to zero. Momentum conservation then accounts for electromagnetic field momentum as well as matter, and for momentum crossing the chosen boundary. This complication occurs in classical electromagnetism, without invoking quantum mechanics.^[ref]

The free-body diagram shows a meridional section of the right circular cone. On the first particle, gravity acts vertically downwards, tension points along the straight line to the apex, and the normal reaction is perpendicular to the surface. The smooth mechanical constraint supplies no tangential friction. The second particle has only its downward gravitational weight and the upward tension.

Adding the forces on each particle and applying Newton's second law givesAt $z_1=-r_1$, a normal vector is ${\mathbf{e}}_r+{\mathbf{e}}_z$. Its sign depends on the supporting face of the cone; either sign has zero dot product with every permitted velocity. The diagram chooses the reaction towards the upper side of the surface.

The inclined parcel has tension $T$ up the slope, weight $mg$ vertically downward, normal reaction $N=mg\cos \theta$ and friction $\mu mg\cos \theta$ down the slope. The hanging mass has weight downward and tension upward. The ideal string makes the two acceleration magnitudes equal.

The velocity slope is $a=2.4/1.5=1.60\mathrm{m}{\mathrm{s}}^{-2}$, up the track for the parcel and downward for the hanging mass. The free-body diagrams show gravity itself as one vector; its slope-parallel and normal components are used in the equations rather than counted as additional forces.

The principle of transmissibility states that moving a force along its line of action preserves its external mechanical effect on an ideal rigid body. The force vector is unchanged, and its torque about every reference point is unchanged. The statement concerns external rigid-body motion or equilibrium; it does not assert unchanged internal deformation.

For an original proof, a nonzero force $\mathbf{F}$ is applied at $\mathbf{r}$. Move it to $\mathbf{r}+s\hat{\mathbf{F}}$, where $\hat{\mathbf{F}}=\mathbf{F}/\Vert \mathbf{F}\Vert$ and $s$ is a signed distance. The new moment isThe last cross product vanishes because its vectors are parallel. This is the same moment geometry used in the primary torque definition.^[ref]

For a numerical example, let $\mathbf{F}=(0,10,0)\mathrm{N}$ act at $(0.30,0.20,0)\mathrm{m}$. Its moment is $(0,0,3)\mathrm{N}\mathrm{m}$. Moving the application point to $(0.30,0.80,0)$ preserves that moment because the displacement follows the force direction. Moving it instead to $(0.50,0.20,0)$ gives $(0,0,5)\mathrm{N}\mathrm{m}$. The second move needs an additional compensating couple to preserve the original wrench.

A one-dimensional elastic counterexample shows the rigid-body limitation. Consider a uniform bar along $0\le x\le L$, fixed at the left end, with a rightward axial force $F$ applied to a cross-section at $x=a$. In a small-strain axial model, internal force is $F$ on $0<x<a$ and zero on $a<x<L$. Its right-end displacement iswhere $E$ is Young's modulus and $A$ the section area. Applying the same force at $a=L$ instead gives $FL/(EA)$, although both applied forces share one line of action and give the same external resultant and moment.

With stipulated $F=100\mathrm{N}$, $L=1\mathrm{m}$, $E=200\mathrm{GPa}$ and $A=1\mathrm{m}{\mathrm{m}}^2$, the end displacement is $0.25\mathrm{mm}$ for $a=L/2$ and $0.50\mathrm{mm}$ for $a=L$. The linear stress–strain relation underlying that calculation is described in the primary elasticity section.^[ref]

A proposed comparison therefore measures support reaction and body deformation separately when changing load position along a common line. Equal reactions and moments can coexist with different local stress and displacement. The external free-body diagram may be equivalent while the internal loading model differs.

Static equilibrium is a state in which a mechanical system remains at rest in a specified reference frame. For a rigid body in an inertial frame, zero resultant external force and zero resultant external torque are the balance conditions:The body must also initially be at rest: zero acceleration alone does not exclude uniform translation or steady rotation.^[ref]

A free-body diagram identifies the body being analyzed and includes its external loads and support reactions. Internal action–reaction pairs cancel when both interacting parts belong to the chosen system. Choosing a different point $O$ for the torques changes their sum by a term proportional to the resultant force. Thus, when the resultant force is zero, a vanishing resultant torque about one point also vanishes about any other point.^[ref]

For an original example, take an ideal massless horizontal beam of length $4\mathrm{m}$, supported at its ends, with a downward $120\mathrm{N}$ load $1\mathrm{m}$ from the left support. Let the upward support reactions be $R_L$ and $R_R$. Vertical force balance and torque balance about the left end giveso $R_R=30\mathrm{N}$ and $R_L=90\mathrm{N}$. The nearer support carries more load because the farther reaction has the larger moment arm. Two equal and opposite forces alone would not guarantee equilibrium: a force couple can have zero resultant force and a nonzero torque.

A simple apparatus test suspends a beam from two spring balances and moves a known load along it. Account separately for the beam's own gravitational weight and its center of mass before comparing the measured reactions with the ideal calculation. The sum of the readings should track total downward load, while their difference changes with load position. Finally, static equilibrium does not imply dynamical stability: an exactly upright inverted pendulum balances at rest but departs from that state after a small disturbance.

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.