Force and translational dynamics

SYSTEMS AND CENTER OF MASS — TOPIC 2.1

A system is whatever you decide to draw the box around. Everything inside interacts with everything else inside through internal forces; everything outside pushes on the system through external forces. The whole unit turns on one consequence of that choice: internal forces cannot change the motion of the system’s center of mass. Two skaters who push each other apart move oppositely, but the point between them, weighted by their masses, stays put. A person cannot lift themselves by their own belt.

When a system is one object. If the parts do not matter for the question — a car’s engine, a person’s arms — treat the whole thing as one object located at its center of mass. This is the same object model as Unit 1, and it is why a free-body diagram is a dot: the dot is the center of mass, and every force is drawn from it. When the parts do matter — two blocks tied together and the question asks for the tension — the system comes apart into its objects, each with its own diagram.

Locating the center of mass. For a symmetrical shape it lies on the lines of symmetry — the center of a uniform rod, disc or box. For separate masses along an axis,

— a mass-weighted average of the positions, and the same formula on the $y$-axis gives $y_{\text{cm}}$. The center of mass sits closer to the heavier part, and it need not be inside any of the objects: a doughnut’s is in the hole. The course limits the calculation to five or fewer particles in two dimensions, or to highly symmetrical systems; anything more is qualitative.

Why it is here, at the start. Newton’s second law, stated properly, is about the center of mass: the acceleration of the system’s center of mass is proportional to the net external force. Everything that follows in this toolbox is that sentence applied to particular forces.

FORCES AND FREE-BODY DIAGRAMS — TOPIC 2.2, AND HOW THE EXAM SCORES THEM

A force is an interaction between two objects: one exerts it, one receives it. Every force in this course is either a contact force — the macroscopic result of atoms in one surface pushing on atoms in another: normal, friction, tension, a push, a spring — or gravity, the only force acting at a distance in AP Physics 1. An object cannot exert a net force on itself, which is why “the force of the ball’s motion” is not a force: nothing outside the ball is exerting it.

The free-body diagram is the tool for turning a picture into equations, and it has rules the exam scores directly:

• The object is a dot (its center of mass). Nothing else is drawn — not the surface, not the rope, not the ground.
• Each force is a straight arrow that starts on the dot and points in the direction of the force, labeled by what exerts it ($F_g$ or $\mathrm{mg}$, $F_n$, $F_T$, $F_f$, $F_{\text{push}}$).
• Two forces in the same direction are drawn side by side, never on top of each other.
• Components are never drawn on the diagram. The diagram shows the forces that exist; resolving them is the next step and belongs in the algebra. A diagram carrying $\mathrm{mg}\sin 𝜃$ and $\mathrm{mg}\cos 𝜃$ on the dot has drawn two forces that nothing exerts.
• Velocity and acceleration are not forces and do not go on it. If you want to note the direction of $\overrightarrow{a}$, put a small arrow beside the diagram, not on the dot.

The scored diagram is the left one. On the exam a force drawn as components, a velocity drawn as a force, or two arrows stacked so one is hidden each lose the point — the CED lists these rules in a boundary statement.

Then choose axes. Put one axis along the direction of the acceleration (or of the velocity, if $a=0$) and the other perpendicular to it. On an incline that means one axis along the slope, so that the normal force and the acceleration each live on one axis and only the weight needs resolving. Choosing horizontal–vertical on an incline resolves two forces instead of one and doubles the algebra for no gain.

NEWTON’S THIRD LAW — TOPIC 2.3: STRINGS, PULLEYS, AND THE PAIR YOU CANNOT CANCEL

Every force is one half of a pair: if A pushes on B, then B pushes on A with a force of equal size and opposite direction,

The two forces of a pair act on different objects, which is why neither can cancel the other inside a single object’s force sum, and why they never appear on the same free-body diagram. The Earth pulls the book down; the book pulls the Earth up. The table pushes the book up; the book pushes the table down. The weight and the normal force on the book are not a pair — they act on the same object, come from different partners (Earth, table), and happen to be equal only because the book is not accelerating.

The identification test. Name the pair by swapping the nouns: “the force of the hammer on the nail” pairs with “the force of the nail on the hammer.” If the swap changes the kind of force (gravity to normal), it is not a pair.

Inside a system, pairs are invisible. Internal forces come in pairs that add to zero, so they cannot move the system’s center of mass (topic 2.1 again). That is the whole justification for the “system first” move in the decision tree: pull the box around two connected blocks and the tension between them vanishes from the equation, leaving only external forces and the total mass.

Strings and tension. Tension is what the segments of a string exert on each other in response to an external pull. An ideal string has negligible mass and does not stretch, and its tension is the same at every point along it — including around an ideal pulley, which has negligible mass and turns without friction and therefore changes the string’s direction and nothing else. A string with mass is different: its tension is larger toward the end that supports more of it — a hanging chain is tightest at the top. The course asks about that only in words, never in numbers.

NEWTON’S FIRST LAW — TOPIC 2.4: EQUILIBRIUM, AND ONE AXIS AT A TIME

The net force is the vector sum of every force on the system. If it is zero, the velocity does not change — the object stays at rest, or keeps moving in a straight line at constant speed, and the two are the same state. That is translational equilibrium, and it is a statement about the sum, not about any single force: a block sliding at constant speed across a floor is in equilibrium with a pull, a friction force, a weight and a normal force all acting on it.

Balanced on one axis, unbalanced on another. Forces are added axis by axis. A projectile has zero net force horizontally (so $v_x$ is constant) and a net force $\mathrm{mg}$ vertically (so $v_y$ changes) — balanced along $x$, unbalanced along $y$, and the velocity changes only in the direction of the unbalanced part. That sentence is the first law and the second law in one breath, and it is why every problem is worked one axis at a time.

Inertial frames. Newton’s laws hold in a frame where an object with no net force stays at constant velocity — an inertial frame. A smoothly cruising train is one; a braking train is not, and in it a suitcase slides forward with no force pushing it. The course assumes every frame is inertial unless it says otherwise, and every frame that moves at a constant velocity relative to an inertial one is inertial too (Unit 1, topic 1.4: they all agree on accelerations).

The procedure for equilibrium is the general one with $a=0$: draw the diagram, choose axes, write $\mathrm{\Sigma }{F}_x=0$ and $\mathrm{\Sigma }{F}_y=0$, solve. Two unknowns need two axes; a hanging sign with two cables at different angles is the standard example, and it is Problem 4.

NEWTON’S SECOND LAW — TOPIC 2.5: THE ONE PROCEDURE

The acceleration of a system’s center of mass has the size of the net external force divided by the mass, and points the way the net force points. Three consequences the exam tests directly: the velocity of the center of mass changes only if a nonzero net external force acts; doubling the net force doubles the acceleration; doubling the mass halves it. And one consequence students resist: the acceleration points along the net force, not along the velocity. A car turning left at constant speed has a net force pointing left, toward the center of its turn, while its velocity points ahead.

The procedure, every problem:

1.

Decide the motion (decision tree 1–3): $a=0$, $a$ along a line, or $v^2∕r$ toward a center. Write what the right-hand side will be.

2.

Draw the free-body diagram by the rules.

3.

Choose axes with one along $\overrightarrow{a}$. Resolve only the forces that do not lie on an axis.

4.

Write $\mathrm{\Sigma }F=\mathrm{ma}$ on each axis, with every force carrying its sign. On the axis across the acceleration, the right-hand side is $0$.

5.

Solve the perpendicular equation first — it usually hands you the normal force, which friction needs.

6.

Solve for what was asked. Then read the answer back: sign, size, units, and the limiting case ($𝜃\to 0$, $\mu \to 0$, $m_1=m_2$).

Step 6 is where derivations earn their credit. An expression like $a=g(\sin 𝜃-{\mu }_k\cos 𝜃)$ should be checked at $𝜃=0$ (no motion down a flat floor) and at ${\mu }_k=0$ (the frictionless $g\sin 𝜃$) before it is trusted.

GRAVITATIONAL FORCE — TOPIC 2.6: THE FORCE, THE FIELD, AND WHAT A SCALE READS

The law. Any two objects with mass attract each other along the line joining their centers of mass, with a force proportional to each mass and inversely proportional to the square of the distance between the centers:

The force on each is the same size (a third-law pair); the Earth pulls the apple exactly as hard as the apple pulls the Earth. Only the accelerations differ, by the ratio of the masses. Double the distance and the force drops to a quarter; triple the mass of one body and the force triples — the functional dependence the “predict how it changes” questions live on.

The field. Rather than recompute the force for every test mass, define the gravitational field of a body $M$ at a point as the force per unit mass it would exert there:

Near Earth’s surface that number is about $10$ N/kg, and it is the same $10$ that appears as $10 {\text{m/s}}^2$ in Unit 1, because if gravity is the only force on an object, $a=F_g∕m=g$. Weight is the gravitational force an astronomical body exerts on a nearby object: $F_g=\mathrm{mg}$. It is a force, measured in newtons, and it changes with $g$; mass does not.

When “constant $g$” is allowed. If the object’s distance from the center of the Earth changes by a negligible fraction — anything you can throw, drop or drive — the field is the same at the start and the end and $\mathrm{mg}$ is exact enough. Satellites and moons are the exception: use $\mathrm{GMm}∕r^2$ with $r$ measured from the center, and expect $g$ to shrink with altitude.

Apparent weight is the normal force. A scale reads the force it exerts on you, which is the normal force, not $\mathrm{mg}$. On a level floor at rest the two are equal, and that coincidence is the source of most elevator errors. Accelerating upward, $F_n=m(g+a)$ and you feel heavier; accelerating downward, $F_n=m(g-a)$ and you feel lighter; in free fall $F_n=0$ and you are weightless — gravity has not vanished, it is merely the only force acting, which is why astronauts in orbit float. The equivalence principle says an observer in a closed accelerating box cannot tell apparent weight from real gravity: a rocket accelerating at $10 {\text{m/s}}^2$ in deep space feels exactly like standing on Earth.

Two masses that agree. Inertial mass is how much an object resists having its motion changed (the $m$ in $F=\mathrm{ma}$); gravitational mass is how strongly it is pulled by gravity (the $m$ in $F_g=\mathrm{mg}$). Nothing requires them to be the same number; experiment says they are, to extraordinary precision — which is why every object falls at the same $g$: the heavier one is pulled harder and resists harder, in exactly the same proportion.

KINETIC AND STATIC FRICTION — TOPIC 2.7: A FORCE THAT ADAPTS

Friction acts along the surfaces in contact, and it comes in two kinds that behave differently enough to be two different rules.

Kinetic friction acts while the surfaces slide relative to each other, opposes that relative motion, and has a fixed size:

where $F_n$ is the normal force — the perpendicular force the surface exerts, directed away from the surface — and ${\mu }_k$ depends on the two materials, not on the size of the contact area or on the speed. A wide block and a narrow block of the same mass and material feel the same friction.

Static friction acts while the surfaces do not slide, and it is not a fixed number. It takes whatever value and direction are needed to prevent slipping, up to a maximum:

Push a heavy box gently and static friction matches your push exactly; the box sits still. Push harder and it still matches, until your push exceeds ${\mu }_s{F}_n$; then the box breaks free, kinetic friction takes over at ${\mu }_k{F}_n$ — typically smaller than the maximum static value, since ${\mu }_s>{\mu }_k$ for most pairs of surfaces — and the box lurches forward. The graph of friction against applied force has a rising line, a peak, and a drop, and Problem 7 draws it.

Static friction climbs with the push, one to one, until the push reaches ${\mu }_s{F}_n$; then the block slides and friction falls to the smaller, constant ${\mu }_k{F}_n$. Below the break, friction is whatever equilibrium needs, not a formula.

Two habits. First, never write ${\mu }_s{F}_n$ for a static friction force unless the problem says the object is on the verge of slipping — below that point static friction is found from equilibrium, not from a coefficient. Second, find $F_n$ from the perpendicular equation before writing any friction — on an incline $F_n=\mathrm{mg}\cos 𝜃$, in an elevator $F_n\ne \mathrm{mg}$, under a downward push $F_n$ exceeds $\mathrm{mg}$. The coefficient multiplies the normal force, never the weight.

Which way does static friction point? Toward whatever prevents the slip: up the slope for a parked car, forward for a walking foot (the foot pushes back on the ground; the ground pushes the foot forward), toward the center for a car rounding a flat curve. If the direction is not obvious, draw it either way, solve, and read the sign.

SPRING FORCES — TOPIC 2.8: HOOKE’S LAW, AND $k$ AS A SLOPE

An ideal spring has negligible mass and exerts a force proportional to how far it has been stretched or compressed from its relaxed length:

where $\mathrm{\Delta }x$ is the change in length from the relaxed length (not the length itself) and $k$, the spring constant in N/m, says how stiff the spring is. The force always points toward the relaxed position: stretch it and it pulls back, compress it and it pushes back. Written with a sign along the spring’s axis, $F_s=-k\mathrm{\Delta }x$; the minus sign is that restoring direction, and it is why springs oscillate in Unit 7.

$k$ is a slope. Hang masses from a spring, measure the stretch, and the graph of force against stretch is a straight line through the origin whose slope is $k$. This is the exam’s favorite experimental-design question in the unit: what do you measure, what do you plot, and why does the slope beat any single pair of numbers? (Because one pair carries one measurement’s error; a line through many averages the errors out, and its intercept tells you whether the ruler was zeroed.) Say what the slope is: “the slope is $k$, because Hooke’s law makes $F$ proportional to $\mathrm{\Delta }x$ with $k$ as the constant.”

In equilibrium a hanging mass stretches the spring until $k\mathrm{\Delta }x=\mathrm{mg}$. Two springs, or a spring on an incline, are the same procedure with the spring force as one more arrow on the diagram; the diagram does not care that the force comes from a spring.

CIRCULAR MOTION — TOPIC 2.9: WHERE THE ACCELERATION POINTS

An object moving in a circle at constant speed is accelerating, because its velocity is changing direction (Unit 1, topic 1.2). The acceleration points toward the center and has size

— larger for faster motion, larger for tighter circles. It is called centripetal, “center-seeking,” and it is not a new force. There is no centripetal force on a free-body diagram. Centripetal acceleration is produced by whichever real forces happen to point toward the center: tension in a string, gravity on a satellite, the normal force on a banked track, static friction on a flat road, or a component of any of them. The procedure is the general one with a radial axis: choose “toward the center” as positive, add the real forces along it, and set the sum equal to $m{v}^2∕r$.

The velocity is always tangent to the circle; the centripetal acceleration always points at the center. If the speed is also changing, a tangential piece $a_t$ is added along the velocity, and the net acceleration is the vector sum of the two.

Speed changing too. If the object speeds up or slows as it circles, there is also a tangential acceleration $a_t$ along the velocity, and the total acceleration is the vector sum of $a_c$ and $a_t$. Uniform circular motion is the special case $a_t=0$.

Period and frequency. One trip around takes the period $T$; the number of trips per second is the frequency $f=1∕T$. At constant speed the circumference is covered in one period, so

The four standard sources of $m{v}^2∕r$, each a problem below:

• Over a hill, through a dip, at the bottom of a swing: gravity and the normal (or tension) force act along the vertical radius. At the top of a hill, $\mathrm{mg}-F_n=m{v}^2∕r$, so the car feels lighter; go fast enough that $F_n=0$ and it leaves the road at $v=\sqrt{\mathrm{gr}}$. At the top of a loop the same $\sqrt{\mathrm{gr}}$ is the minimum speed to stay on the track — there, gravity alone supplies the whole centripetal force. At the bottom of a dip or a swing, $F_n-\mathrm{mg}=m{v}^2∕r$ and the object feels heavier; a pendulum bob at the bottom of its arc is not in equilibrium even though it is neither speeding up nor slowing down at that instant.
• A banked curve, no friction: the normal force is the only force with a horizontal component, and that component is the whole centripetal force: $F_n\sin 𝜃=m{v}^2∕r$, $F_n\cos 𝜃=\mathrm{mg}$, so $v=\sqrt{\mathrm{gr}\tan 𝜃}$ — the one speed at which the bank alone holds the car. The course computes only this frictionless case; with friction, banked curves are discussed in words.
• A conical pendulum: a bob swinging in a horizontal circle on a string at angle $𝜃$ from the vertical. The horizontal component of the tension is the centripetal force, the vertical component holds the weight — the same two equations as the banked curve with tension in place of the normal force.

• A satellite: gravity is the only force, so $\mathrm{GMm}∕r^2=m{v}^2∕r$. Substitute $v=2\mathrm{\pi r}∕T$ and the mass of the satellite cancels:

  $$T^2=\frac{4{\pi }^2}{\mathrm{GM}}{r}^3$$

  — Kepler’s third law, derived rather than memorized. The period depends on the central mass and the orbit radius, and on nothing about the satellite. (Kepler’s first and second laws are not in the course.)

VERIFICATION HABITS — HOW TO CHECK WITHOUT REDOING

• Limiting cases. Set $𝜃=0$, $\mu =0$, $m_1=m_2$, $v=0$ in your derived expression and ask whether the physics you already know comes out. A formula that gives a nonzero acceleration on a flat frictionless floor with no push is wrong before any numbers go in.
• Direction. Does the acceleration point along the net force? Does static friction point toward preventing the slip? Does the normal force point away from the surface?
• The normal force check. Is $F_n$ larger than $\mathrm{mg}$ where the object should feel heavier (dip, upward acceleration, a downward push) and smaller where it should feel lighter (hill, downward acceleration, incline)?
• Units. N/kg and m/s${}^2$ are the same unit; N/m times m is N; kg$\cdot$m/s${}^2$ is N. A derived $v=\sqrt{\mathrm{gr}\tan 𝜃}$ has units $\sqrt{\left({\text{m/s}}^2\right)\left(\text{m}\right)}=$ m/s.
• A second route. System-then-part gives the tension; part alone gives it again with a different equation. If the two agree, the answer is defended.