--- title: Newton's Laws module: Dynamics moduleNumber: 2 lessonNumber: 1 order: 201 summary: > What makes a body change its motion, and in which frames does the answer take its simplest form? Newton's three laws settle both: inertial frames are the ones where a force-free body coasts, force is whatever changes momentum, and every interaction pushes back on its source. We write the second law as $\sum\vec F=\d\vec p/\d t$, reduce it to $m\vec a$ at constant mass, and separate what a scale actually reads — the support force — from the weight $m\vec g$ it is so often mistaken for. topics: [Dynamics] draft: false sources: - book: Tipler & Mosca ref: "Ch. 4 — Newton's Laws; §§4-1–4-4, 4-7" --- ## Inertial frames and the first law Newton's first law identifies the frames in which the laws have their simplest form. In an inertial frame, a body with zero net force has constant velocity: $$ \sum\vec F=\vec 0 \quad\Longrightarrow\quad \vec v=\text{constant}. $$ Rest is the special case of zero constant velocity. A frame moving at constant velocity relative to an inertial frame is itself inertial; an accelerating or rotating frame requires added inertial-force terms when Newton's equations are written in it. A laboratory frame is inertial to the extent that any residual acceleration of a nearly isolated body stays smaller than the acceleration being measured. When a body's velocity changes with no identified interaction, the frame is accelerating, not a hidden force: hold the motion fixed and change to an accelerating origin, and a straight constant-velocity trace becomes a curved one. $$ % caption: Position versus time for three bodies that share the same start point and the % same initial velocity. Zero acceleration (force-free, inertial) traces a straight % constant-velocity line; a positive acceleration curves the trace upward as the body % speeds up, and a negative acceleration curves it downward as the body slows. Any % departure from the straight line is the visible signature of a nonzero acceleration. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \definecolor{posacc}{HTML}{2E9E4F} \definecolor{negacc}{HTML}{D6453D} \draw[->, black] (0,0) -- (6.6,0) node[right] {time}; \draw[->, black] (0,0) -- (0,4.0) node[above] {position}; \draw[posacc, very thick, domain=0.4:5.0, samples=90] plot (\x,{0.3772+0.414*\x+0.045*\x*\x}) node[right] {positive a}; \draw[acc, very thick] (0.4,0.55) -- (5.0,2.62) node[right] {inertial}; \draw[negacc, very thick, domain=0.4:5.0, samples=90] plot (\x,{0.3628+0.486*\x-0.045*\x*\x}) node[right] {negative a}; \fill[black] (0.4,0.55) circle (1.8pt); \end{tikzpicture} $$ ## Force, mass, and the second law Force is an interaction that changes momentum. For a constant-mass particle, $$ \sum\vec F=m\vec a. $$ The equation is vectorial. Component equations follow only after a coordinate basis is chosen: $$ \sum F_x=ma_x, \qquad \sum F_y=ma_y, \qquad \sum F_z=ma_z. $$ Mass measures inertia: for a fixed net force, larger mass gives smaller acceleration. The SI force unit follows from the second law, $$ 1\ \mathrm N=1\ \mathrm{kg\,m\,s^{-2}}. $$ Net force alone sets acceleration. Velocity and position fix the instantaneous state but not the acceleration: a body can move fast with zero net force, or accelerate while momentarily at rest. For equal horizontal net forces, cart accelerations are inversely proportional to mass. The diagram compares the resulting acceleration vectors after the common interaction has been specified. Cartesian components provide independent scalar equations. A horizontal net force leaves vertical acceleration unchanged unless another vertical interaction is present, and a vertical force balance leaves horizontal dynamics unresolved. The component diagram represents one force in a chosen Cartesian basis; its component arrows support the scalar equations without adding interactions. **Weight and gravitational interaction.** Near Earth's surface, the gravitational force on mass $m$ is $$ \vec W=m\vec g. $$ Weight is a force with units of newtons; mass is a scalar with units of kilograms. The local field $\vec g$ points toward Earth's centre. A scale measures a contact force. Its reading can differ from $mg$ whenever the supported object accelerates. An elevator passenger with upward positive obeys $$ N-mg=ma. $$ Upward acceleration gives $N>mg$, downward acceleration gives $N=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[thick] (1.0,1.2) circle (0.32); \draw[->, acc, very thick] (1.0,1.52) -- (1.0,2.65) node[above] {N}; \draw[->, black, very thick] (1.0,0.88) -- (1.0,-0.25) node[below] {mg}; \node[black, below] at (1.0,-0.6) {supported}; \draw[thick] (4.4,1.2) circle (0.32); \draw[->, black, very thick] (4.4,0.88) -- (4.4,-0.25) node[below] {mg}; \node[black, below] at (4.4,-0.6) {free fall}; \end{tikzpicture} $$ ## Interaction pairs and system boundaries Every interaction produces a pair of forces on different objects: $$ \vec F_{A\to B}=-\vec F_{B\to A}. $$ The equality is simultaneous and exact for the interaction. Each member acts on a different body, so the pair does not enter one object's force sum. A book's weight and a table's normal force can cancel in the book's net force; they arise from different interactions. The normal force on the book and the force exerted by the book on the table form the pair. Third-law pairs are identified by an interaction and by reversed object labels. The diagram separates the force on the book from the force on the table. The arrows have equal magnitude and opposite direction, but they belong to different body diagrams and each belongs to a different individual second-law equation. $$ % caption: The book rests on the table, meeting it along one contact surface. At that shared % point the table pushes the book up and the book pushes the table down. The two forces are % equal and opposite and act on different bodies, so they never cancel. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black, thick] (-0.9,1.6) -- (3.7,1.6); \draw[thick] (0.75,1.6) rectangle (2.05,2.8); \node at (1.4,2.5) {book}; \draw[->, very thick] (1.4,1.6) -- (1.4,2.12); \draw[->, black, very thick] (1.4,1.6) -- (1.4,1.2); \fill[black!70] (1.4,1.6) circle (1.5pt); \node[left] at (0.6,2.0) {table on book}; \node[black, right] at (1.72,1.33) {book on table}; \end{tikzpicture} $$ **Interaction analysis.** Apply Newton's laws after identifying objects connected by contact or field interactions. Gravity, contact, strings, springs, and fluid forces each have a physical source. A correct dynamical model distinguishes these actual interactions from kinematic labels such as “centripetal force.” This label denotes the required net inward force for a curved path. A turning vehicle illustrates the frame dependence of the outward inertial term. An inertial-ground analysis identifies a real inward contact force that changes an occupant's velocity direction. The outward term belongs to the rotating vehicle frame and has no interaction source on another body. ## Momentum form and frame transformations When momentum is measured directly, use Newton's second law in the form $$ \sum\vec F=\frac{\d\vec p}{\d t}. $$ A constant-mass particle has $\vec p=m\vec v$, so differentiation gives $$ \frac{\d\vec p}{\d t}=m\frac{\d\vec v}{\d t}=m\vec a. $$ The familiar $m\vec a$ equation is the constant-mass form of the momentum law. The distinction matters when the modeled system gains or loses matter. A rocket, a leaking cart, and a falling raindrop require a system definition and momentum-flow analysis; writing $m\vec a$ for the changing material collection without that analysis omits momentum carried across the system boundary. A fixed-mass cart moving along one horizontal axis has a measured force-time record that determines the momentum change over an interval: $$ \Delta p_x=\int_{t_i}^{t_f}\sum F_x\,\d t. $$ The area under a signed net-force graph gives the change in momentum. Positive and negative intervals can cancel even when the cart experiences large forces at separate times. A net force that reverses sign first increases and then decreases the same momentum component; a force sign reversal can occur while the velocity retains its direction. $$ % caption: Signed net force versus time for one cart. The positive area (a push) and the % later negative area (a check) contribute opposite momentum changes; the net change is the signed area. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (6.0,0) node[right] {time}; \draw[->, black] (0,-1.4) -- (0,1.8) node[above] {net force}; \draw[acc, thick, fill=acc!12] (0.7,0) rectangle (2.7,1.15); \draw[black, thick, fill=black!8] (3.5,0) rectangle (5.1,-0.75); \node[acc] at (1.7,0.57) {positive}; \node[black] at (4.3,-0.4) {negative}; \end{tikzpicture} $$ > **Worked example.** A $2.0\ \mathrm{kg}$ cart on a level, near-frictionless track > feels $+6.0\ \mathrm N$ for $0.50\ \mathrm s$, then $-2.0\ \mathrm N$ for > $0.75\ \mathrm s$ (the signed force record above). The impulse is the signed area, > $$ > \Delta p_x=(6.0\ \mathrm N)(0.50\ \mathrm s)+(-2.0\ \mathrm N)(0.75\ \mathrm s) > =1.5\ \mathrm{N\,s}, > $$ > so the velocity change is > $$ > \Delta v_x=\frac{\Delta p_x}{m} > =\frac{1.5\ \mathrm{N\,s}}{2.0\ \mathrm{kg}}=0.75\ \mathrm{m\,s^{-1}}. > $$ > Averaging the two force magnitudes without weighting by duration gives the wrong > momentum change; the signed contributions must be added separately. Force perpendicular to instantaneous momentum changes its direction at that instant while leaving its magnitude unchanged to first order. Circular motion is an example: the inward net force continually rotates the momentum vector. A tangential net force changes the momentum magnitude. Most motion has both components, so force analysis uses vectors together with speed information. $$ % caption: Momentum change from a transverse net force. Two equal-length momentum % vectors differ in direction; their difference points inward, along the average net force. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (-0.2,0) -- (4.8,0) node[right] {$x$}; \draw[->, black] (0,-0.2) -- (0,3.6) node[above] {$y$}; \draw[->, acc, very thick] (0,0) -- (3.6,0.75) node[right] {$p_i$}; \draw[->, acc, very thick] (0,0) -- (2.8,2.4) node[above] {$p_f$}; \draw[->, black, thick] (3.6,0.75) -- (2.8,2.4) node[midway, right] {change}; \end{tikzpicture} $$ The derivative in Newton's law is local in time. A rapid force variation can produce the same finite momentum change as a weaker force acting longer, but the intermediate trajectory differs. A force sensor sampled at discrete times estimates the integral by summing areas over short intervals. Sampling that misses a narrow force pulse can underestimate the impulse even if all recorded force values are accurate. **Inertial frames and frame transformations.** Inertial frames are related by constant relative velocity. If frame $S'$ moves at constant velocity $\vec V$ relative to inertial frame $S$, the particle coordinates obey $$ \vec r'=\vec r-\vec Vt-\vec r_0, $$ and differentiation gives $$ \vec v'=\vec v-\vec V, \qquad \vec a'=\vec a. $$ All inertial observers agree on acceleration and net force for a given particle. They can report different positions and velocities because those quantities depend on the chosen origin and relative motion. The second law has the same form in each inertial frame because $m\vec a'=m\vec a$. An accelerating frame has $\vec a'\ne\vec a$. If its origin accelerates at $\vec A$, then $\vec a'=\vec a-\vec A$. Writing a second-law equation in that frame introduces the inertial term $-m\vec A$ so that $$ \sum\vec F-m\vec A=m\vec a'. $$ The extra term belongs to the frame description, whereas gravity, contact, tension, and electromagnetic forces arise from interactions. Keeping those categories separate prevents a coordinate acceleration from being placed on a free-body diagram as a force from another object. ## Mass, weight, and support forces Mass is defined operationally through a body's response to a specified net force. If two isolated carts experience equal horizontal net forces, their accelerations satisfy $$ \frac{a_1}{a_2}=\frac{m_2}{m_1}. $$ The comparison permits either cart to be moving initially. Identify the force on each cart over the same interval and control or measure the other force components. A cart moving steadily before the applied force begins has the same acceleration response as an initially stationary cart of the same mass. At fixed mass, a graph of net-force magnitude against acceleration magnitude is a straight line through the origin: $$ F_{\rm net}=ma. $$ The slope is the inertial mass. A systematic offset at zero applied force suggests an unmodeled friction force, a sensor offset, or an inclined track. A curved force- acceleration graph signals that the assumed constant mass or force model needs review. With a calibrated force sensor, a $1.50\ \mathrm{kg}$ cart on a level nearly frictionless track subject to a net $3.00\ \mathrm N$ force has $$ a=\frac{3.00}{1.50}=2.00\ \mathrm{m\,s^{-2}}. $$ Repeating the test with $1.00\ \mathrm{kg}$ of added mass while retaining the same net force gives $a=1.20\ \mathrm{m\,s^{-2}}$. The ratio $$ \frac{2.00}{1.20}=\frac{2.50}{1.50} $$ matches the inverse-mass prediction. The experiment compares response ratios, so an uncertainty in the force calibration cancels when the same force is used in both runs. Mass is scalar and frame independent in Newtonian mechanics. Changing coordinates can change the components of velocity, acceleration, and force, while the cart's inertial mass remains fixed. The same mass appears in each component equation: $$ F_x=ma_x, \qquad F_y=ma_y, \qquad F_z=ma_z. $$ An anisotropic acceleration measurement usually signals different unmodeled forces or a constrained direction of motion. Inertial mass remains direction independent. Inertial mass and gravitational mass enter different force statements. Near Earth, the gravitational force is proportional to gravitational mass, while the acceleration response is controlled by inertial mass. The observed equality of free-fall acceleration for different objects means their gravitational-to-inertial mass ratio is the same to high precision. Within elementary Newtonian problems, the common symbol $m$ is used for both roles after that empirical equivalence has been established. The mass parameter belongs to the body selected for the second-law equation. For a pair of carts joined by a light coupling, the mass in each individual equation is that cart's mass; the mass in an equation for the two-cart system is the total mass. The force inventory changes with the chosen body or system, but the inertial response of each material object remains attached to its own mass. **Third-law pairs, system boundaries, and momentum transfer.** The third law labels forces by both source and recipient: $$ \vec F_{A\to B}=-\vec F_{B\to A}. $$ The order of the subscripts matters. The force of a hand on a cart acts on the cart. The force of the cart on the hand acts on the hand. Equal magnitude and opposite direction act on bodies with different masses and may be accompanied by other forces on those bodies, so their accelerations can differ. At every instant of a contact interval, the interaction forces form a pair. Their impulses over the same interval also form a pair: $$ \int\vec F_{A\to B}\,\d t =-\int\vec F_{B\to A}\,\d t. $$ For two objects considered together as one system, those internal forces cancel in the system momentum equation. Adding the individual second-law equations gives $$ \frac{\d}{\d t}(\vec p_A+\vec p_B) =\vec F_{\rm ext}. $$ Only forces exerted by objects outside the selected system remain on the right side. The cancellation follows from the third law and the system boundary. Each object still experiences its contact force. > **Worked example.** Two skaters, $m_1=45\ \mathrm{kg}$ and $m_2=60\ \mathrm{kg}$, > start at rest on near-frictionless ice and push apart. The horizontal external > impulse is negligible, so the total horizontal momentum stays zero: > $$ > m_1v_{1x}+m_2v_{2x}=0. > $$ > If the heavier skater recoils at $|v_{2x}|=1.5\ \mathrm{m\,s^{-1}}$, the lighter one > leaves in the opposite direction at > $$ > |v_{1x}|=\frac{m_2}{m_1}\,|v_{2x}|=\frac{60}{45}(1.5\ \mathrm{m\,s^{-1}}) > =2.0\ \mathrm{m\,s^{-1}}. > $$ > Equal and opposite momentum changes act on unequal masses, so the lighter skater > carries the larger speed. $$ % caption: Two skaters after a mutual push. With negligible external impulse their % momenta stay equal and opposite, so the lighter skater carries the larger speed. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black] (-3.0,0) -- (3.0,0); \draw[fill=acc!12, draw=acc, thick] (-0.7,0.3) circle (0.3); \draw[fill=black!10, draw=black, thick] (0.7,0.45) circle (0.45); \draw[->, acc, very thick] (-1.05,0.3) -- (-2.45,0.3) node[left] {$p_1$}; \draw[->, black, very thick] (1.2,0.45) -- (2.6,0.45) node[right] {$p_2$}; \node[black, below] at (-0.7,-0.05) {light}; \node[black, below] at (0.7,-0.05) {heavy}; \end{tikzpicture} $$ Third-law pairs differ from forces that balance in a single equation. A book resting on a table has an upward table-on-book force and a downward Earth-on- book force. Both act on the book and can sum to zero, although they arise from different interactions. The partner of the table-on-book force is the book-on-table force. The partner of the Earth-on-book force is the book-on-Earth gravitational force. System choice separates external and internal forces. In a system containing only the book, table contact and gravity are external. In a system containing book and table, their contact forces are internal while Earth's gravity remains external. Expanding the system to include Earth makes the book-Earth gravity pair internal as well. The momentum equation changes with the chosen system boundary. Forces transmitted through a string, rod, or spring still occur as interaction pairs at the two ends. A massless ideal string can have the same tension magnitude along a straight segment in an ideal model, but those equal tensions are forces on different objects or different string segments. The detailed constraint equations belong in the free-body-diagram lesson; the third-law requirement applies before any such simplification is made. **Weight, support forces, and apparent weight.** Near Earth's surface, the gravitational field can be treated as uniform over ordinary laboratory distances. With upward as the positive vertical direction, $$ \vec g=-g\hat y, \qquad \vec W=-mg\hat y. $$ Weight is the gravitational force $mg$, whereas apparent weight is a contact-force measurement. A bathroom scale, a load cell, or the floor of an elevator registers the normal force exerted on the supported body. The two quantities agree only when the vertical acceleration is zero. An elevator passenger of mass $m$ standing on a horizontal scale has $$ N-mg=ma_y, $$ so $$ N=m(g+a_y). $$ An upward acceleration raises the scale reading. A downward acceleration of magnitude less than $g$ lowers it. Constant upward or downward velocity has $a_y=0$ and gives $N=mg$. The sign of velocity therefore has no direct place in the scale equation. $$ % caption: Scale reading N = m(g+a) versus upward acceleration. It equals mg at zero % acceleration and falls to zero at a = -g, the free-fall condition where support is lost. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (-2.3,0) -- (2.7,0) node[right] {acceleration}; \draw[->, black] (0,0) -- (0,3.0) node[above] {scale reading}; \draw[acc, very thick] (-1.8,0) -- (2.2,2.8); \draw[black, dashed] (-1.9,1.26) -- (0,1.26) node[pos=0, left] {mg}; \draw[black, dashed] (0,0) -- (0,1.26); \draw[fill=acc, draw=acc] (-1.8,0) circle (1.6pt); \node[black, below] at (-1.8,-0.05) {free fall}; \end{tikzpicture} $$ > **Worked example.** A $70\ \mathrm{kg}$ passenger stands on a scale in an elevator > ($g=9.8\ \mathrm{m\,s^{-2}}$, upward positive). Accelerating upward at > $1.5\ \mathrm{m\,s^{-2}}$, the scale reads > $$ > N=m(g+a_y)=(70\ \mathrm{kg})(9.8+1.5)\ \mathrm{m\,s^{-2}}=7.9\times10^2\ \mathrm N. > $$ > Accelerating downward at the same rate, > $$ > N=(70)(9.8-1.5)=5.8\times10^2\ \mathrm N. > $$ > The gravitational force is $mg=6.9\times10^2\ \mathrm N$ in both cases; only the > support force changes. Reporting the scale reading as "the force of gravity" > conflates apparent weight with weight and fails whenever the support accelerates. The contact condition imposes a limit. A floor can push upward on a passenger but cannot pull the passenger downward. Thus $$ N\geq0. $$ If the elevator, vehicle, or platform has downward acceleration $a_y=-g$, then $N=0$ and the passenger is in free fall relative to the supporting structure. A formal calculation giving $N<0$ indicates loss of contact; after that instant the normal force is set to zero and the passenger's acceleration follows from the remaining forces. $$ % caption: Contact-loss threshold. As a platform's downward acceleration approaches g the % normal force shrinks to zero; beyond it the surface cannot pull, so the body separates. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black, thick] (0.2,0) -- (2.6,0); \draw[thick] (1.1,0) rectangle (1.9,0.7); \draw[->, acc, very thick] (1.5,0.7) -- (1.5,1.7) node[above] {N}; \draw[->, black, very thick] (1.5,0.35) -- (1.5,-0.9) node[below] {mg}; \node[black, below] at (1.5,-1.2) {supported}; \draw[black, thick] (4.0,0) -- (6.4,0); \draw[thick] (4.9,0.45) rectangle (5.7,1.15); \draw[->, black, very thick] (5.3,0.8) -- (5.3,-0.45) node[below] {mg}; \draw[->, very thick] (5.3,1.35) -- (5.3,2.15) node[above] {a}; \node[black, below] at (5.3,-0.75) {contact lost}; \end{tikzpicture} $$ The local constant-$g$ model is an approximation. At altitude or across planetary scales, use the gravitational field appropriate to position. The normal-force method uses the actual gravitational force vector and contact force in the vertical second- law component. Treat $g$ with its local magnitude and the chosen coordinate sign. Weight and support also differ in a rotating system. A rider in a turning vehicle can have a large seat or wall contact force even when the gravitational force is unchanged. The contact direction is then set by the acceleration required for the curved path, while the scale-reading idea still refers to the component of contact force along the sensor's sensitive direction. ## Model tests and superposition Newton's first law identifies the reference frames in which a force-free body has constant velocity. The observation is stronger than the statement that a body at rest remains at rest. A puck gliding at constant speed, a spacecraft coasting between thruster firings, and a cart moving uniformly on a level low-friction track all satisfy the same zero-net-force condition: $$ \sum\vec F=\vec 0 \quad\Longrightarrow\quad \frac{\d\vec v}{\d t}=\vec 0. $$ Measured nonzero acceleration of a constant-mass body in an inertial frame requires a nonzero net force. The individual forces may be large and cancel partly; only their vector sum determines acceleration. A passenger standing still in a parked elevator experiences gravity and a normal force, yet has zero net force. A satellite in a circular orbit can have nearly constant speed while its velocity direction and net gravitational force change continuously. Physical experiments approach isolation by reducing or modeling interactions until any residual acceleration agrees with the estimated residual net force. Air drag, track tilt, magnetic coupling, and sensor bias can each mimic a failure of constant velocity. Assess the frame from the complete interaction model. Inertial frames form a large class. If $S$ is inertial, any frame $S'$ whose origin moves with constant velocity and whose axes do not rotate relative to $S$ is also inertial. The coordinate transformation shifts velocity by a constant vector while preserving acceleration: $$ \vec v'=\vec v-\vec V, \qquad \vec a'=\vec a. $$ Road-side and train-seat observers can disagree on the velocity of a tossed ball yet obtain the same acceleration in inertial coordinates. A frame with changing velocity or rotating axes requires extra terms. A frame whose origin has acceleration $\vec A$ measures $$ \vec a'=\vec a-\vec A. $$ An observer who writes the second law entirely in that accelerating frame uses $$ \sum\vec F+\vec F_{\rm inertial}=m\vec a', \qquad \vec F_{\rm inertial}=-m\vec A. $$ The added term produces the same coordinate prediction as an inertial-frame analysis, but it has no interaction partner on another body. A passenger in a bus that accelerates forward appears to accelerate backward relative to the bus unless the inward contact force from the seat is included in the ground-frame analysis. In the bus frame, the term $-m\vec A$ accounts for the same relative motion. Rotating frames introduce further position- and velocity-dependent inertial terms. A turning vehicle frame, a carousel, and Earth's rotating surface are common examples. The local inertial approximation is often adequate for short laboratory experiments, but it fails when the frame's rotation or translational acceleration is comparable with the acceleration being measured. The frame choice belongs in the model statement before component equations are interpreted. **Net force, superposition, and interaction models.** Newton's laws specify how a net force changes motion. Force laws supply the interaction terms; geometry alone does not. Each term in $$ \sum\vec F =\vec F_{\rm gravity} +\vec F_{\rm contact} +\vec F_{\rm spring} +\vec F_{\rm drag} +\cdots $$ must be tied to an interaction and to a force model appropriate to that interaction. Add interaction forces component by component after all applicable interactions have been identified. This vector addition is the superposition of forces. A particle acted on by two perpendicular forces has $$ \vec F_1=6.0\hat\imath\ \mathrm N, \qquad \vec F_2=-8.0\hat\jmath\ \mathrm N, $$ the net force is $$ \vec F_{\rm net}=6.0\hat\imath-8.0\hat\jmath\ \mathrm N, \qquad F_{\rm net}=10\ \mathrm N. $$ A $2.0\ \mathrm{kg}$ particle has acceleration $$ \vec a=3.0\hat\imath-4.0\hat\jmath\ \mathrm{m\,s^{-2}}. $$ The acceleration points along the net-force resultant. The velocity can point in a different direction at the same instant because it records the particle's prior motion. Force determines the local change in velocity. $$ % caption: Vector superposition before a second-law step. Two perpendicular forces add to % the diagonal net force; dividing by mass gives an acceleration in the same direction. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (4.4,0) node[right] {$x$}; \draw[->, black] (0,0) -- (0,-3.6) node[below] {$y$}; \draw[->, black, very thick] (0,0) -- (3.0,0) node[midway, above] {$F_1$}; \draw[->, black, very thick] (3.0,0) -- (3.0,-3.0) node[midway, right] {$F_2$}; \draw[->, acc, very thick] (0,0) -- (3.0,-3.0) node[midway, below left] {net}; \draw[black, dashed] (0,-3.0) -- (3.0,-3.0); \end{tikzpicture} $$ The physical source of a force constrains its direction and range. A normal force acts perpendicular to the contacting surfaces in an ideal smooth-contact model. Tension in an ideal taut string acts along the string. A spring force depends on deformation and acts along the spring axis. Gravity near Earth is approximately vertical and downward. Drag depends on the relative motion through a fluid. These models are introduced and tested in later dynamics lessons; Newton's second law combines their vector effects. Some frequently used words describe a required net-force direction. "Centripetal" refers to the inward component of the net force needed for curved motion. A string tension, normal force, friction force, gravity, or electric force can supply that component in different physical situations. Adding a separate centripetal-force arrow to an interaction inventory double counts the force unless a new physical source has been identified. $$ % caption: The same inward net-force requirement is met by different interactions: string % tension, track normal force, or gravity. The inward arrows share a role; the labels name distinct sources. \begin{tikzpicture}[>=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[black] (0,0) circle (0.9); \fill[black] (0,0) circle (1.2pt); \draw[thick] (0.64,0.64) circle (0.15); \draw[->, acc, very thick] (0.6,0.6) -- (0.14,0.14); \node[black, below] at (0,-1.15) {tension}; \draw[black] (4.919,0.233) arc[start angle=15, end angle=75, radius=0.9]; \draw[thick] (4.686,0.636) circle (0.15); \draw[->, acc, very thick] (4.646,0.596) -- (4.208,0.158); \node[black, below] at (4.05,-1.15) {normal}; \draw[fill=white, draw=black, thick] (7.0,-0.5) circle (0.28); \draw[thick] (7.85,0.6) circle (0.15); \draw[->, acc, very thick] (7.8,0.55) -- (7.2,-0.2); \node[black, below] at (7.35,-1.15) {gravity}; \end{tikzpicture} $$ Static equilibrium is the special condition $$ \sum\vec F=\vec 0, \qquad \vec a=\vec 0. $$ It permits rest or constant-velocity motion. Zero net force also permits a body to coast, so force balance alone leaves its velocity unspecified. A body at an instant of zero velocity can have nonzero net force and begin moving immediately. Position, velocity, acceleration, and force are separate state variables linked by the laws. When an interaction force depends on position, velocity, or time, the acceleration can change during the motion. The equation $$ m\vec a(t)=\sum_i\vec F_i\bigl(\vec r(t),\vec v(t),t\bigr) $$ is then a differential equation for the trajectory. Constant-acceleration kinematics applies only when the net force and mass remain constant over the interval. The force model determines whether that simplification is justified. **Quantitative tests of a Newton-law model.** A numerical test of the second law requires force, mass, and acceleration to refer to the same selected body and time interval. Experimental comparisons therefore use synchronized measurements. A force sensor gives the interaction or net-force components at a stated time. A motion sensor or position record gives velocity and acceleration over a stated interval. A mass measurement identifies the inertial parameter for the selected body or system. For sampled one-dimensional data at equal time spacing $\Delta t$, a centered acceleration estimate is $$ a_x(t_i)\approx\frac{v_x(t_{i+1})-v_x(t_{i-1})}{2\Delta t}. $$ The model residual is $$ R_i=F_{{\rm net},x}(t_i)-ma_x(t_i). $$ Residuals that fluctuate around zero within measurement uncertainty support the chosen force model. A residual with a consistent sign or position dependence suggests a missing interaction, an incorrect force direction, a calibration offset, or a mass assignment that does not match the modeled system. For example, an unmodeled constant track friction produces a nearly constant negative residual when positive direction is chosen along the pull. The time interval used for acceleration should match the time resolution of the force measurement. A narrow contact pulse can change momentum substantially while producing an acceleration peak that is missed by a slow position sensor. In that case, compare the time-integrated quantities instead: $$ \int_{t_i}^{t_f}F_{\rm net}\,\d t =m\bigl[v(t_f)-v(t_i)\bigr]. $$ The left side uses the signed area under the measured force curve. The right side uses the velocity difference across the same endpoints. The impulse form retains a rapidly changing force without assigning it one representative acceleration. Dimensional analysis provides an independent check on every component equation. The only SI combination of mass and acceleration with force dimensions is $$ [m a]=\mathrm{kg}\,\mathrm{m\,s^{-2}}=\mathrm N. $$ An expression such as $m/v$ or $mv$ cannot equal force because its dimensions are not newtons. Dimensions cannot detect a wrong sign or a missing force term, but they expose many algebraic substitutions before numerical values are inserted. The choice of coordinate direction affects component signs, not the physical prediction. Reversing the $x$ axis changes $F_x$, $a_x$, and $v_x$ to their negatives, leaving $F_x=ma_x$ true. A calculation that gives different acceleration magnitude after an axis reversal has changed more than notation. It has usually applied a force sign from a diagram without transforming the coordinate convention. The component cases below apply at one stated instant. | Measured result | Dynamical interpretation | |---|---| | $\sum F_x=0$, $v_x\ne0$ | Constant horizontal velocity is allowed. | | $\sum F_x\ne0$, $v_x=0$ at one instant | The body has nonzero horizontal acceleration at that instant. | | $\sum F_y=0$, $\sum F_x\ne0$ | Vertical acceleration is zero while horizontal acceleration is nonzero. | | $F_{\rm net}$ fixed, $m$ doubled | Acceleration magnitude is halved. | These rows concern components and instants. They do not assume that the same force model remains valid after contact changes, a string goes slack, a body leaves a surface, or a force-producing source moves. Uncertainty in a force prediction can be estimated from the inputs. For $a=F/m$ with independent small uncertainties, $$ \frac{\delta a}{|a|} \approx\sqrt{\left(\frac{\delta F}{|F|}\right)^2 +\left(\frac{\delta m}{m}\right)^2}. $$ If force has $2\%$ uncertainty and mass has $1\%$ uncertainty, the acceleration uncertainty is about $2.2\%$. Reporting many digits from a calculator cannot improve the precision set by the force and mass measurements. Report the residual or the measured-versus-predicted difference together with that uncertainty. ## Scope and two-dimensional applications The equation $\sum\vec F=m\vec a$ treats the selected object as a particle or describes the translational motion of a body when rotational effects can be separated. An extended rigid body has center-of-mass acceleration set by net force, while the distribution of forces can also produce torque and angular acceleration. A force through a body's center and an equal force applied off center can produce the same net translation but different rotation. Rotational dynamics requires its own equations and is developed in the rotation module. The constant-mass particle form has clear boundary conditions. It applies directly to a body whose material content remains fixed during the interval. For an open system, such as a rocket ejecting propellant or a cart collecting sand, momentum crosses the system boundary. The correct starting point is the momentum balance for the selected system, including the momentum flow of entering or leaving material. Treating the instantaneous mass as a number in $m\vec a$ without the flow terms yields an incomplete equation. Newtonian mechanics also uses a low-speed, weak-gravity approximation. At speeds much smaller than the speed of light and at ordinary laboratory scales, mass is effectively constant and time is shared by inertial observers to the precision of the model. At relativistic speeds, momentum and energy require the relativistic relations. At atomic scales, quantum mechanics replaces a deterministic single-particle trajectory with a different description. These limits do not weaken Newton's laws within their domain; they specify the conditions under which the model is being used. Force laws themselves may be approximations. Near Earth's surface, writing $\vec W=m\vec g$ with constant $g$ is accurate over small height changes. Across planetary distances, the gravitational field changes with radius and direction. A linear spring law applies over a deformation range where the material response is proportional to extension. A drag law may change form between slow and fast flow. The second law remains the equation of motion while the interaction term is updated to the appropriate physical model. The distinction between a mathematical constraint and a force remains important. A body constrained to a circular path has radial acceleration, but the constraint does not identify whether tension, normal contact, gravity, or another interaction supplies the required inward force. A condition such as $x^2+y^2=R^2$ restricts possible positions; it does not appear as an additional physical force. Constraint forces are identified by their source and then included in the net force. Newtonian predictions are tested through changes in motion. A force model predicts acceleration, momentum change, or a trajectory. Agreement requires the model's stated frame, body selection, interaction terms, initial conditions, and parameter values. Changing any of those inputs changes the prediction. A short equation is reliable only when those physical choices remain visible alongside it. > **Worked example.** A $0.50\ \mathrm{kg}$ instrument package moves in a horizontal > plane under two measured forces, > $$ > \vec F_A=(4.0\hat\imath+1.0\hat\jmath)\ \mathrm N,\qquad > \vec F_B=(-1.0\hat\imath+3.0\hat\jmath)\ \mathrm N. > $$ > Add the forces first, then divide by mass: > $$ > \vec F_{\rm net}=(3.0\hat\imath+4.0\hat\jmath)\ \mathrm N,\qquad > \vec a=\frac{\vec F_{\rm net}}{m}=(6.0\hat\imath+8.0\hat\jmath)\ \mathrm{m\,s^{-2}}, > $$ > so $|\vec a|=10\ \mathrm{m\,s^{-2}}$ along the $3$–$4$–$5$ direction of the net > force. With initial velocity $\vec v_0=(5.0\hat\imath-2.0\hat\jmath)\ \mathrm{m\,s^{-1}}$ > and the force held constant for $0.30\ \mathrm s$, > $$ > \begin{aligned} > v_x&=5.0+(6.0)(0.30)=6.8\ \mathrm{m\,s^{-1}},\\ > v_y&=-2.0+(8.0)(0.30)=0.40\ \mathrm{m\,s^{-1}}. > \end{aligned} > $$ > The vertical velocity passes through zero at $t=2.0/8.0=0.25\ \mathrm s$, where the > vertical acceleration is still positive: a zero velocity component is not a zero > acceleration component. The displacement uses the full constant-acceleration form, > $$ > \Delta\vec r=\vec v_0\,\Delta t+\tfrac12\vec a(\Delta t)^2 > =(1.77\hat\imath-0.24\hat\jmath)\ \mathrm m, > $$ > negative in $y$ even though $v_y$ ends positive, because the package first moved > downward. The momentum change equals the impulse, an independent check: > $$ > m(\vec v-\vec v_0)=(0.90\hat\imath+1.20\hat\jmath)\ \mathrm{N\,s} > =\vec F_{\rm net}\,\Delta t=(3.0\hat\imath+4.0\hat\jmath)(0.30\ \mathrm s). > $$ If the force changes before the interval ends, split the motion at the change time and carry the first interval's final position and velocity into the second as initial conditions. The interaction model, the second-law step, and the kinematics step stay separate: the force model gives $\vec F_A$ and $\vec F_B$; Newton's law converts their sum to $\vec a$; kinematics converts $\vec a$ and the initial data to a trajectory. **Separating force, momentum, velocity, and acceleration.** The four quantities most often conflated in elementary dynamics have different roles. Force is an interaction contribution. Momentum is a mass-weighted velocity. Velocity describes the instantaneous rate of position change. Acceleration describes the instantaneous rate of velocity change. Newton's second law connects net force to the time derivative of momentum; it does not identify force with velocity or momentum. A body can have large momentum and zero net force. A freight train moving uniformly on a straight level track has momentum $m\vec v$ but no acceleration if its driving force balances resistance. A small object at rest can have zero momentum and a large net force; its acceleration begins immediately after the force is applied. A particle with velocity perpendicular to net force has changing momentum direction while its speed remains momentarily unchanged. The distinction can be expressed through successive time derivatives: $$ \vec r(t) \xrightarrow{\ \d/\d t\ } \vec v(t) \xrightarrow{\ \d/\d t\ } \vec a(t), \qquad \sum\vec F=m\vec a $$ for constant mass. Integration runs in the other direction, each step carrying an initial condition: $$ \vec v(t)=\vec v(t_0)+\int_{t_0}^{t}\vec a(t')\,\d t', \qquad \vec r(t)=\vec r(t_0)+\int_{t_0}^{t}\vec v(t')\,\d t'. $$ A specified acceleration and initial velocity determine the later velocity; that velocity and an initial position determine the later position. Initial conditions are therefore required for a trajectory even when the net force is known exactly. A net force parallel to velocity changes speed. A net force antiparallel to velocity reduces speed. A net force perpendicular to velocity turns the velocity vector. A general net force has parallel and perpendicular components, changing both speed and direction. Resolve those components against the instantaneous velocity direction, which can change during the motion. The decomposition gives separate observable consequences for one force vector. It should be made before a magnitude is used in a speed or curvature calculation. | Force component | Kinematic effect | Check | | --- | --- | --- | | $F_{\parallel}$ | changes speed | compare with $\d(v^2)/\d t$ | | $F_{\perp}$ | changes direction | compare with local curvature | | $F_{\parallel}=0$ | speed is instantaneously stationary | velocity may still turn | | $F_{\perp}=0$ | direction is instantaneously fixed | speed may still change | Mass controls the response to a fixed net force but does not change the force supplied by a specified interaction pair. During a collision, a light ball and a heavy wall exert equal and opposite contact forces. Their acceleration magnitudes can differ by many orders of magnitude because their masses differ. The momentum change of the complete isolated ball-wall system still follows from the external impulse, regardless of the individual accelerations. Separate these variables to avoid recurring algebra errors: using speed in place of acceleration in $F=ma$; treating zero velocity as a zero-force condition; adding force magnitudes before resolving directions; and assigning an interaction force to a body that does not receive it. Each error changes the physical statement before the numerical calculation begins. Record the frame, selected body or system, time interval, and force model in a Newton-law statement. For example, "the cart has acceleration $2.0\ \mathrm{m\,s^{-2}}$" is incomplete until the frame and direction are stated. "The horizontal net force on the $1.5\ \mathrm{kg}$ cart is $3.0\ \mathrm N$ in the laboratory frame during the interval" states the information needed to predict the acceleration component. Stating these elements keeps later changes in contact, constraints, or frame explicit. The force model also requires a stated time scale. A constant-net-force model predicts a linear velocity change only over the interval during which that force remains constant. A contact can begin or end, a spring can change extension, a vehicle can enter a curved section of track, or a source can move. At each such event, update the interaction terms and continue from the position and velocity reached at the event. Piecewise Newton-law models preserve the same laws while allowing the physical configuration to change. A piecewise model uses continuity conditions to connect its intervals. Position is continuous for ordinary finite-speed motion. Velocity is also continuous unless an idealized impulsive interaction is being used; a narrow but finite contact force changes velocity rapidly rather than discontinuously. The momentum form of the second law determines that change from the impulse. State these transition assumptions whenever a force graph, contact event, or constraint changes during a calculation. Experimental records need the same bookkeeping as a force diagram. A force-sensor trace must be associated with its calibration, the body at the point of attachment, the direction defined as positive, and the time interval used for the calculation. A motion sensor requires a stated coordinate origin and an account of whether its data refer to the object, the laboratory, or a moving support. A scale reading gives a contact-force measurement, so its interpretation changes if the support accelerates. Those details determine whether a force component, a velocity component, and a time derivative refer to one coherent model. The calculation record can be checked before any numerical substitution. Each row identifies a different physical object in the model; none can be supplied by a force magnitude alone. | Model entry | Representation | Independent check | | --- | --- | --- | | Selected body | stated boundary and mass | every force arrow acts on that body | | Motion state | $\vec r\rightarrow\vec v\rightarrow\vec a$ | initial conditions and time interval | | Force balance | $\sum\vec F=m\vec a$ for fixed mass | component units and signs | | Constraint | position or velocity condition | force source identified separately | A calculation based on several measurements should retain units and signs in the intermediate component equations. The total horizontal force may be obtained from a spring balance and a friction estimate, while the vertical component may come from a scale reading and a weight model. Their uncertainties and observation times need not be identical. Combining them requires an explicit approximation: for example, that the forces remained effectively constant over the stated interval. A disagreement between predicted and measured acceleration may then be traced to a missing interaction, a changed contact condition, a timing mismatch, or measurement uncertainty rather than hidden by premature rounding. Coordinate components should also be retained through the final force sum. A negative component carries directional information and cannot be replaced by its magnitude before vector addition. Once the net components are known, the acceleration components follow by division by the same selected mass. Magnitude and direction may then be reported together, with the coordinate convention stated alongside the result. [^tipler4]: Tipler and Mosca, _Physics for Scientists and Engineers_, 6th ed., §§4-1–4-4, 4-7.