| Learning Objective | Essential Knowledge |
|---|---|
2.1.A |
|
2.1.B |
|
Force and Translational Dynamics
AP Physics C: Mechanics · Topic 2
9:06
Force and Translational Dynamics
A cart runs around a vertical loop. At the very top it is completely upside down, with nothing holding it up. Why does it not fall off? The answer is not some…
English narration · English + 中文 subtitles burned in
2.1
Systems and Center of Mass
Syllabus
Source: College Board AP Course and Exam Description
On 2 August 1971, astronaut David Scott stood on the Moon and dropped a hammer and a falcon feather at the same moment. With no air to slow the feather, both hit the dusty ground together – exactly as Galileo had predicted nearly 400 years earlier. That one test captures this unit: the same few force laws, applied with care through Newton's laws, explain falling feathers, cars on curves, skydivers and orbiting moons.
A system 系统 is the object or objects you analyze, treated as a point at its center of mass 质心. For a set of masses, $x_{\text{cm}}=\dfrac{\sum m_i x_i}{\sum m_i}$; for a continuous body, $x_{\text{cm}}=\dfrac{1}{M}\int x\,dm$. Only external forces move the center of mass.
Choosing the system is your decision, and it changes the analysis. If the parts' internal details do not matter, treat the whole system as one object at its center of mass. If they do – two blocks joined by a string, a person in an elevator – you can look inside it, and the parts may move differently from each other and from the system as a whole. A system can also exchange energy or mass with its surroundings (a rocket losing fuel, a car losing heat). For any body with a symmetrical mass distribution, the center of mass lies on every line of symmetry – the center of a uniform disk or sphere, the midpoint of a uniform rod.
Worked example (a non-uniform rod). A thin rod of length $L=2.0\ \text{m}$ lies along the $x$-axis from $x=0$. Its linear mass density increases along it, $\lambda=kx$ with $k=3.0\ \text{kg/m}^2$. Find its mass and its center of mass.
- A slice of width $dx$ at position $x$ has mass $dm=\lambda\,dx=kx\,dx$.
- Mass: $M=\displaystyle\int_0^L kx\,dx=\frac{kL^2}{2}=\frac{(3.0)(2.0)^2}{2}=6.0\ \text{kg}$.
- Center of mass: $x_{\text{cm}}=\dfrac{1}{M}\displaystyle\int_0^L x\,(kx\,dx)=\dfrac{1}{M}\cdot\dfrac{kL^3}{3}=\dfrac{(3.0)(2.0)^3/3}{6.0}=1.33\ \text{m}$, which is $\tfrac23L$.
Check the answer against the physics: a uniform rod would balance at its midpoint, $1.0\ \text{m}$. More mass sits toward the right end, so the center of mass must be to the right of the midpoint.

Centre of mass of a system
Tap the parts to see where the centre of mass of a two-body system sits.
| English |
|---|
| system/ˈsɪstəm/ |
| center of mass/ˈsentə ɒv mæs/ |
2.2
Forces and Free-Body Diagrams
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.2.A |
|
2.2.B |
Boundary statement: AP Physics C: Mechanics and AP Physics C: Electricity and Magnetism only expect students to depict the forces exerted on objects, not the force components on free-body diagrams. On the AP Physics exams, individual forces represented on a free-body diagram must be drawn as individual straight arrows, originating on the dot and pointing in the direction of the force. Individual forces that are in the same direction must be drawn side by side, not overlapping. |
Source: College Board AP Course and Exam Description
A force 力 is a push or pull (a vector, in newtons). A free-body diagram 受力图 draws one object with an arrow for every force on it – weight, normal, tension, friction, applied, drag. Draw it first; it sets up every dynamics equation.
- Draw the object as a dot representing its center of mass, with every force arrow starting from that dot.
- Show only forces exerted on the object by its environment – never forces the object exerts on other things, and never "$ma$" or "centripetal force".
- Contact forces (normal force, friction, tension) are the large-scale result of the electric forces between atoms at the surfaces in contact.
- Choose one axis parallel to the acceleration (along an incline, or toward the centre of a circle). Then only one equation contains $a$.

Balance the forces on a free-body diagram
A free-body diagram shows every force on one object as an arrow. The object accelerates only if the forces don't cancel — the net force sets $a=F/m$.
| English |
|---|
| force/fɔːs/ |
| free-body diagram/friː ˈbɒdi ˈdaɪəɡræm/ |
2.3
Newton's Third Law
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.3.A |
|
Source: College Board AP Course and Exam Description
Newton's third law 牛顿第三定律: A pushes B, B pushes back equally and oppositely. The pair acts on different objects, so it never cancels within one free-body diagram:
The two forces are always the same type (both gravitational, both contact) and act at the same moment. When a truck hits a fly, the force on the fly is exactly as large as the force on the truck; the fly suffers more because its mass is tiny. Inside a system, third-law pairs are internal forces: they cancel in pairs, so they cannot change the motion of the system's center of mass.
Tension in a string or cable is the net result of each tiny segment of the string pulling on its neighbours. For an ideal (massless) string over a frictionless pulley, the tension is the same all the way along.

Third-law pairs
Newton's third law pairs act on different objects. Sort each statement.
| English |
|---|
| Newton's third law/ˈnjuːtnz θɜːd lɔː/ |
2.4
Newton's First Law
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.4.A |
|
Source: College Board AP Course and Exam Description
Newton's first law (inertia 惯性): with zero net force 合力, velocity stays constant – the object is in translational equilibrium 平动平衡. The net force is the vector sum of all forces on the system, so equilibrium means $\sum F_x=0$ and $\sum F_y=0$ separately.
Forces can be balanced in one direction but not another: a hockey puck sliding on frictionless ice has balanced vertical forces (weight and normal force), so its vertical velocity stays zero, while any unbalanced horizontal force changes only its horizontal velocity. An inertial reference frame is exactly a frame in which Newton's first law is observed to hold – one that is not accelerating.
Worked example (equilibrium). A 4.0 kg lamp hangs from two cords, each at $30^\circ$ above the horizontal. Vertical: $2T\sin30^\circ=mg=40$ N, so $T=40$ N. Each cord carries a tension equal to the lamp's whole weight – shallow cords need large tensions.
Balanced or unbalanced?
By Newton's first law, decide whether the forces on each object are balanced.
| English |
|---|
| inertia/ɪˈnɜːʃə/ |
| net force/net fɔːs/ |
| translational equilibrium/trænˈsleɪʃənl ˌiːkwɪˈlɪbrɪəm/ |
2.5
Newton's Second Law
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.5.A |
|
Source: College Board AP Course and Exam Description
The general form uses momentum:
Worked example. A $2.0\ \text{kg}$ block slides down a frictionless incline 斜面 at $30^{\circ}$. Only the along-ramp component of gravity drives it, so $a=g\sin30^{\circ}=9.8(0.5)=4.9\ \text{m/s}^2$, independent of the mass.
Worked example (two bodies). A $6.0\ \text{kg}$ cart on a frictionless table is pulled by a string over a light pulley to a hanging $2.0\ \text{kg}$ mass. Treat the pair as one system: only the hanging weight drives it, so $a=\dfrac{2.0(9.8)}{8.0}=2.45\ \text{m/s}^2$. Then isolate the cart alone to find the string tension 张力: $T=6.0(2.45)\approx15\ \text{N}$. System first for $a$, single body for internal forces – that two-step is the standard pattern.

Worked example (elevator and apparent weight). A 60 kg student stands on a scale in an elevator that accelerates upward at $2.0\ \text{m/s}^2$. Taking up as positive, $N-mg=ma$, so $N=60(10+2.0)=720$ N. The scale reads the normal force, so the student's apparent weight is 720 N, though the gravitational force is still 600 N. Accelerating downward at $2.0\ \text{m/s}^2$ gives $N=480$ N; in free fall, $N=0$.
Free-body diagram: F = ma
The resultant of the forces, divided by the mass, gives the acceleration. Change a force and watch the acceleration respond.
| English |
|---|
| differential equation/ˌdɪfəˈrenʃl ɪˈkweɪʒn/ |
| incline/ɪnˈklaɪn/ |
| tension/ˈtenʃn/ |
2.6
Gravitational Force
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.6.A |
|
2.6.B |
|
2.6.C |
|
2.6.D |
|
2.6.E |
Boundary statement: AP Physics C: Mechanics does not expect students to mathematically prove or derive Newton's shell theorem. |
Source: College Board AP Course and Exam Description
Near a surface, weight is $F_g=mg$. In general, Newton's law of gravitation 万有引力定律:
Apparent weight is the magnitude of the normal force on a system (what a bathroom scale measures). When the system accelerates, apparent weight differs from $mg$ (see the elevator example). An object appears weightless when gravity is the only force on it, as for astronauts in an orbiting space station – they are in free fall around Earth, not beyond its gravity. The equivalence principle says that an observer inside a closed accelerating room cannot tell the difference between that acceleration and a gravitational field.
Mass appears in two roles. Inertial mass measures how strongly an object resists changes in its motion ($m$ in $F=ma$); gravitational mass measures how strongly it attracts and is attracted by other masses ($m$ in $Gm_1m_2/r^2$). Experiments show the two are equal to better than one part in $10^{14}$ – which is why all objects fall with the same acceleration, as the hammer and feather showed.
Spheres and the shell theorem. The force from a large body is the sum of the forces from all its small pieces $dm$. For a uniform spherical shell, Newton showed that:
- outside the shell, it attracts as if all its mass were at its centre;
- inside the shell, the forces from all its parts cancel – the net force is zero.
So outside a uniform solid sphere, $g=GM/r^2$. Inside it, at radius $r$, only the mass closer to the centre pulls; that mass is $M\dfrac{r^3}{R^3}$, so $g=\dfrac{GMr}{R^3}$ – proportional to $r$, and zero at the centre.

Worked example (field at altitude). At what height above Earth's surface ($R=6.4\times10^6$ m) is $g$ one quarter of its surface value? $g\propto1/r^2$, so $r=2R$ and the height is $R=6.4\times10^6$ m. Halfway to the centre, by contrast, $g$ is one half of its surface value.

Gravity falls off with distance
The gravitational force between two masses gets weaker as they move apart (an inverse-square law).
| English |
|---|
| Newton's law of gravitation/ˈnjuːtnz lɔː ɒv ˌɡrævɪˈteɪʃn/ |
2.7
Kinetic and Static Friction
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.7.A |
|
2.7.B |
|
Source: College Board AP Course and Exam Description
Friction 摩擦力 opposes sliding along a surface: kinetic 动摩擦 $f_k=\mu_k N$ while sliding, and static 静摩擦 $f_s\le\mu_s N$ up to a maximum before sliding. $N$ is the normal force 法向力. Note the inequality: static friction takes whatever value and direction are needed to stop the surfaces sliding, and $\mu_s N$ is its ceiling, not its value. For a given pair of surfaces $\mu_s$ is usually larger than $\mu_k$, which is why a heavy box is harder to start moving than to keep moving.
Worked example (static friction). A 20 kg crate rests on a floor with $\mu_s=0.50$ and $\mu_k=0.40$. A worker pushes horizontally with 60 N. The maximum static friction is $0.50(200)=100$ N, more than 60 N, so the crate stays put and static friction is exactly 60 N, not 100 N. The crate starts moving only above 100 N; once it slides, friction drops to $0.40(200)=80$ N.
Worked example. The same block on the same $30^{\circ}$ incline, now with $\mu_k=0.20$. Perpendicular to the ramp: $N=mg\cos30^{\circ}$. Along the ramp: $ma=mg\sin30^{\circ}-\mu_k mg\cos30^{\circ}$, so $a=g(\sin30^{\circ}-0.20\cos30^{\circ})=9.8(0.500-0.173)=3.2\ \text{m/s}^2$ – the mass cancels again.
Worked example (two blocks at rest). Block 1 (mass $M=2.0$ kg) sits on a table, joined by a string over a light, frictionless pulley to block 2 (mass $m=0.60$ kg) hanging over the edge. What is the smallest $\mu_s$ that keeps them at rest?
- At rest, the tension equals the hanging weight: $T=mg$.
- Block 1 stays put if static friction can reach $T$: $\mu_s Mg\ge mg$.
- So $\mu_s\ge\dfrac{m}{M}=\dfrac{0.60}{2.0}=0.30$. The answer does not depend on $g$.
Finding $\mu_k$ from a graph. In the same set-up, students keep the total mass $M_{\text{tot}}$ fixed but move small masses from block 1 to the hanger, so the hanging mass $m$ changes. Newton's second law for the whole system:
A graph of $a$ against $m$ is a straight line. Its vertical intercept is $-\mu_k g$, so $\mu_k=-\dfrac{\text{intercept}}{g}$. If an intercept of $-2.0\ \text{m/s}^2$ is read off, $\mu_k\approx0.20$. If the pulley actually has friction, the measured $a$ values are too small, so this method gives a $\mu_k$ that is too large.
Slide a block down a slope with friction
Friction opposes motion up to a maximum $\mu N$. Tilt the slope until gravity's pull along it beats static friction and the block starts to slide.
| English |
|---|
| Friction/ˈfrɪkʃn/ |
| kinetic/kɪˈnetɪk/ |
| static/ˈstætɪk/ |
| normal force/ˈnɔːml fɔːs/ |
2.8
Spring Forces
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.8.A |
|
2.8.B |
Boundary statement: AP Physics C: Mechanics only expects students to find the effective spring constant of systems of springs that are arranged either in series or in parallel and does not expect students to find the effective spring constant of a system in which springs are arranged in both series and parallel. |
Source: College Board AP Course and Exam Description
An ideal spring has negligible mass and exerts a force proportional to the change in its length from its relaxed length. (A real spring may have significant mass, or stop obeying the proportional rule when stretched too far.) It obeys Hooke's law 胡克定律 $\vec F_s=-k\Delta\vec x$, always directed toward the equilibrium position – a restoring force set by the spring constant 弹簧劲度系数 $k$. Its stored energy is $U=\tfrac12 kx^2$.

Combined springs act as one equivalent spring of constant $k_{\text{eq}}$. In parallel (side by side, sharing the load) the constants add, $k_{\text{eq}}=k_1+k_2$ – stiffer than either. In series (end to end) they combine as $\dfrac{1}{k_{\text{eq}}}=\dfrac{1}{k_1}+\dfrac{1}{k_2}$ – softer than the softest one. Two springs pulling a mass from opposite sides also act in parallel, giving a restoring constant $k_1+k_2$. Whichever you have, the SHM period uses $k_{\text{eq}}$: $T=2\pi\sqrt{m/k_{\text{eq}}}$.
Stretch a spring (Hooke's law)
A spring's force is proportional to its extension, $F=kx$ (Hooke's law). Pull harder and the extension grows in step — until the spring's limit.
| English |
|---|
| Hooke's law/hʊks lɔː/ |
| spring constant/sprɪŋ ˈkɒnstənt/ |
2.9
Resistive Forces
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.9.A |
|
Source: College Board AP Course and Exam Description
A resistive force 阻力 (drag) opposes motion through a fluid and grows with speed, often modeled as $F=-bv$ or $F=-cv^2$. Newton's second law then gives a differential equation, e.g. $m\dfrac{dv}{dt}=mg-bv$ for a falling object. As speed rises, drag builds until it balances the driving force; the object then stops accelerating and moves at a constant terminal velocity 终极速度, found by setting the net force (and $dv/dt$) to zero.
Worked example. For a falling object with $m\dfrac{dv}{dt}=mg-bv$, the terminal velocity is where $\dfrac{dv}{dt}=0$: $mg=bv_T$, so $v_T=\dfrac{mg}{b}$. With $m=0.10\ \text{kg}$ and $b=0.50\ \text{kg/s}$, $v_T=\dfrac{0.10(9.8)}{0.50}=2.0\ \text{m/s}$.
To get the full motion $v(t)$, separate the variables 分离变量 and integrate from rest:


Resistive forces and terminal velocity
As a falling body speeds up, drag grows until it balances weight and the body stops accelerating.
| English |
|---|
| resistive force/rɪˈzɪstɪv fɔːs/ |
| terminal velocity/ˈtɜːmɪnl vəˈlɒsɪti/ |
| separate the variables/ˈseprət ðə ˈveərɪəblz/ |
2.10
Circular Motion
Syllabus
| Learning Objective | Essential Knowledge |
|---|---|
2.10.A |
|
2.10.B |
Boundary statement: AP Physics C: Mechanics does not expect students to know Kepler's first or second laws of planetary motion. |
Source: College Board AP Course and Exam Description
Uniform circular motion has a centripetal acceleration 向心加速度 $a_c=\dfrac{v^2}{r}$ pointing to the center, requiring a net inward force $F_c=\dfrac{mv^2}{r}$ supplied by a real force (tension, gravity, friction, normal). There is no outward force. Centripetal acceleration can come from a single force, from several forces, or from components of forces. For vertical circles and banked curves, decompose the real forces to find which provides the centripetal requirement.
Worked example. A $0.50\ \text{kg}$ ball is whirled on a $1.0\ \text{m}$ string in a horizontal circle at $3.0\ \text{m/s}$. The string tension supplies the whole centripetal force: $T=\dfrac{mv^2}{r}=\dfrac{0.50(3.0)^2}{1.0}=4.5\ \text{N}$.
Worked example (vertical circle). At the top of a vertical loop of radius $r$, gravity and the normal force both point toward the center: $N+mg=\dfrac{mv^2}{r}$. The slowest possible speed at the top is where the track pushes with nothing at all ($N=0$): $v_{\min}=\sqrt{gr}$. For $r=2.5\ \text{m}$: $v_{\min}=\sqrt{9.8(2.5)}=4.9\ \text{m/s}$. Any slower and the cart leaves the track before the top.
If the speed is also changing, there is a tangential acceleration $a_t=\dfrac{dv}{dt}$ along the path as well. The object's total acceleration is the vector sum $\vec a=\vec a_c+\vec a_t$, with magnitude $\sqrt{a_c^2+a_t^2}$.
Uniform circular motion repeats, so it is described by its period $T$ (time for one revolution) and frequency $f=1/T$; the speed is $v=\dfrac{2\pi r}{T}$.
For a satellite in a circular orbit, gravity alone supplies the centripetal force:
This is Kepler's third law: the square of the period is proportional to the cube of the orbit radius, and the satellite's own mass cancels.
Worked example (geostationary orbit). A satellite that stays above one point on the equator needs $T=24$ h $=8.64\times10^4$ s. With $GM=4.0\times10^{14}\ \text{m}^3/\text{s}^2$: $r^3=\dfrac{GMT^2}{4\pi^2}=\dfrac{4.0\times10^{14}(8.64\times10^4)^2}{39.5}=7.6\times10^{22}$, so $r=4.2\times10^7$ m – about 6.6 Earth radii from Earth's centre.

What provides the centripetal force?
Uniform circular motion needs a net force toward the centre. Match each case to its source.
| English |
|---|
| centripetal acceleration/senˈtrɪpɪtl əkˌseləˈreɪʃn/ |
2.10
Exam tips
- Locate the center of mass with $x_{cm}=\tfrac{1}{M}\int x\,dm$ (or $\tfrac{\sum m_i x_i}{\sum m_i}$ for point masses) and use $\lambda,\sigma,\rho$ for the mass element.
- The net external force moves the center of mass as if all mass sat there: $\vec F_{net}=M\vec a_{cm}$.
- Define your system clearly — internal forces cancel, so only external forces change its momentum.
- Exploit symmetry to shortcut a center-of-mass integral.
- Distinguish center of mass from center of gravity (identical in a uniform field).
Common mistakes
- Writing $N = mg$ on every surface. On a ramp $N = mg\cos\theta$ (if nothing else pushes), and in an elevator accelerating upward at $a$, $N = m(g + a)$.
- Using $f_s = \mu_s N$ as the static friction. $\mu_s N$ is only the maximum; static friction is whatever the equilibrium needs, up to that limit.
- Drawing a "centripetal force" or a "centrifugal force" on a free-body diagram. Draw only real forces; their net inward component equals $mv^2/r$.
- Forgetting that $N$ and $mg$ both point to the centre at the top of a vertical loop: $N + mg = mv^2/r$.
- Quoting $v(t) = v_T(1 - e^{-bt/m})$ without the separation of variables when a derivation is asked for.
Interactive lessons on this topic
Work through it step by step, with instant-check exercises.