Unit 2 - Force and Translational Dynamics

1. Force and Translational Dynamics

Title

AP Physics C: Mechanics — Unit 2

Newton's three laws, the free-body diagram, friction, connected bodies, circular motion and drag

2. What this deck buys you

Objectives

Unit 1 described motion. It could never predict motion, because the acceleration always had to be handed to you in the problem statement. This deck supplies the missing half.

• Draw a free-body diagram that is complete, correctly oriented, and free of invented forces
• Apply Newton's second law as two component equations rather than one vague statement
• Handle normal force, tension and friction — including the cases where each is not what you expect
• Solve connected-body problems by choosing between system and individual analysis
• Treat circular motion as an acceleration problem rather than a new kind of force
• Set up and solve velocity-dependent drag with a differential equation

The single most valuable habit in this unit is mechanical: draw the diagram before writing any equation. Almost every lost mark in dynamics traces to a force that was omitted, invented, or pointed the wrong way — and all three are visible on a diagram and invisible in algebra.

Everything in the kinematics deck described motion. None of it could predict motion, and the reason is worth naming: the acceleration always had to be given to you in the problem statement. Nothing in kinematics explains where an acceleration comes from.

This deck supplies that. Newton's second law says the acceleration is determined by the forces, so once you can identify the forces on an object you can compute its acceleration, and once you have the acceleration all of Unit 1 applies.

The mechanical skill that carries this unit is the free-body diagram, and it is genuinely worth taking seriously rather than treating as a formality. Almost every mark lost in dynamics comes from one of three things: a force left off the diagram, a force invented that does not exist, or a force pointed the wrong way. All three are visible in a drawing and invisible in algebra.

So the discipline for the whole deck is: draw first, write equations second. It feels slow at first and it is not — it is the fastest route to a correct answer, and it stays fast when the problems get hard.

3. Newton's Three Laws

Section

Section 1

4. What a force is, and the short list you will actually meet

Concept

A force is a push or a pull exerted by one object on another. That definition contains a requirement people forget: every force has an agent, a specific other object doing the pushing.

Only two entries in that table have a formula you can write down immediately. Normal force and tension are almost never given — they are unknowns you solve for, and assuming a value for them is one of the most common errors in this unit.

If you cannot name the agent of a force you have drawn, it is not a force. This is the test that removes centrifugal force, 'the force of motion', and 'the force of the throw' from your diagrams.

The definition to hold onto is that a force is a push or a pull exerted by one object on another. The clause people drop is the last one, and it is the useful part: every force has an agent, a specific other object doing the pushing.

That clause is a filter. If you have drawn an arrow and cannot name the object exerting it, it is not a force and it should come off the diagram. This single test eliminates centrifugal force, the force of motion, the force of the throw, and every other invented arrow that shows up on student diagrams.

Look carefully at the magnitude column of the table, because the pattern there is important. Weight, friction and the spring force all have formulas you can write down immediately. Normal force and tension do not — they are unknowns you solve for from the equations of motion.

That distinction is where a great deal of trouble originates. Assuming the normal force equals mg, or that the tension equals the hanging weight, feels natural and is usually wrong. Those two quantities take whatever value the situation demands, and finding out what that value is is often the whole problem.

5. Newton's first law defines what a valid frame is

Concept

An object with zero net force keeps a constant velocity — which for an object at rest means staying at rest, and for a moving one means continuing in a straight line at constant speed.

\[ \sum \vec{F} = 0 \quad \Longleftrightarrow \quad \vec{v} = \text{constant} \]

Stated that way it looks like a special case of the second law, and algebraically it is. Its real content is subtler: it declares which reference frames the second law is allowed to be used in.

A frame where the first law holds is called inertial. In an accelerating car or a spinning carousel it does not hold — objects accelerate with nothing pushing them — and F = ma gives wrong answers there.

For everything in this course, the ground is inertial and that is the frame to work in. When a problem is set inside an accelerating elevator, analyse it from outside, in the ground frame.

It is fair to wonder why the first law exists at all, given that it looks like the second law with a equal to zero. Algebraically that is exactly what it is.

Its real content is about reference frames. The first law says: there exist frames in which an object with no net force keeps a constant velocity. Those frames are called inertial, and they are the only frames in which the second law is true.

The figure shows why this matters. From the platform, a cart on a frictionless track moves in a straight line, and nothing needs explaining. From inside an accelerating train, that same cart appears to curve — and there is no object anywhere you could blame for the curving. No agent, no force, yet an acceleration. The second law has failed.

For everything in this course, treat the ground as inertial and work in that frame. When a problem is set inside an accelerating elevator or a turning car, analyse it from outside. That single habit removes the need for fictitious forces entirely, which is why you will never need centrifugal force in AP Physics C.

6. An object needs a force to keep it moving

Trap

7. Newton's second law is two equations, not one

Concept

\[ \sum \vec{F} = m\vec{a} \]

This is a vector equation, which means it is shorthand for one independent equation per direction. In two dimensions you always get two, and you must write both.

\[ \sum F_x = ma_x \qquad \sum F_y = ma_y \]

Three things follow immediately, and each one is worth stating separately.

• The net force matters, not any individual force. A 100 N push against 100 N of friction produces no acceleration at all.
• Acceleration is parallel to the net force — always, exactly. Not parallel to the velocity, and not parallel to the largest single force.
• The equation is instantaneous. It relates the forces right now to the acceleration right now, so a force that changes with time produces an acceleration that changes with it.

The mass in the denominator is the object's inertia — its resistance to being accelerated. Double the mass and the same net force produces half the acceleration, which is the whole content of the law.

The arrow over the F and over the a is doing a lot of work, and skipping over it is how people end up trying to solve two-dimensional problems with one equation.

A vector equation is shorthand for one independent scalar equation per direction. In two dimensions that means two equations, and you must write both — even when one of them turns out to say that things balance. In fact the balancing one is often where the useful information is, because it is what determines the normal force.

Three consequences follow, and each is worth holding separately. The net force is what matters, so equal and opposite pushes produce nothing. The acceleration is parallel to the net force exactly — not to the velocity, and not to whichever single force happens to be largest. And the relationship is instantaneous, so a force that varies in time produces an acceleration that varies with it.

The mass in the denominator has a name worth using: inertia. It is the object's resistance to being accelerated, and it is why the same push moves a shopping trolley easily and a car barely at all.

8. Reading the second law properly

Notation

The phrase on the object is doing more work than its size suggests. A free-body diagram shows forces on one chosen body, drawn as if that body were alone in the universe with a few arrows attached.

Everything else — what the body pushes on, what holds the table up, what the Earth is doing — belongs to a different diagram, and mixing them is what produces diagrams with too many arrows.

Four details in this equation, and the second is the one that reorganises how you think about the whole unit.

The sum means every force acting, added as vectors. Missing one is the most common error in dynamics, and the free-body diagram exists precisely to make that error visible.

On the object is the crucial phrase. Forces that the object exerts on other things never appear in this sum. This is exactly why third-law pairs are irrelevant to a single-body analysis: the partner force acts somewhere else and belongs on a different diagram.

The m is the mass of the object whose diagram you drew — not the total mass of the scene. In a connected-body problem this changes depending on which body you are analysing, so saying out loud which object you are considering is not pedantry, it is bookkeeping.

And the a is that same object's acceleration, which is zero both when the object is at rest and when it moves at constant velocity. Those two cases are indistinguishable to the second law, which is worth remembering when a problem describes something as moving steadily.

9. Newton's third law: the pair acts on different objects

Concept

\[ \vec{F}_{A\,\text{on}\,B} = -\,\vec{F}_{B\,\text{on}\,A} \]

The two forces are equal in magnitude, opposite in direction, of the same type, and acting on different objects. That last clause is the one that matters, and it is why third-law pairs can never cancel each other out.

Cancellation requires two forces on the same object. A third-law pair is by definition split between two objects, so it appears on two different free-body diagrams — one arrow each.

The practical rule for this unit: third-law pairs are irrelevant to a single-body diagram. When you draw the crate, you draw the person's push on the crate and stop. The crate's push back on the person belongs on the person's diagram.

The usual phrasing — equal and opposite reaction — is accurate and hides the important part. The important part is that the two forces act on different objects.

That is what makes third-law pairs unable to cancel. Cancellation requires two forces on the same object, and a third-law pair is by definition split between two, so it shows up as one arrow on each of two separate diagrams.

If this were not true, nothing could ever accelerate. Push a crate and the crate pushes back on you with an equal force. If those cancelled, no push would ever move anything, and the fact that crates do move is the everyday evidence that they act on different bodies.

For the purposes of this unit, the working rule is: third-law pairs are irrelevant to a single-body diagram. When you draw the crate, you draw the push on the crate and stop. The crate's push back on you goes on your diagram, and you will only draw that if the problem asks about you.

10. Which of these are genuine third-law pairs?

Discrimination

Every pair below involves two real forces. Only some of them are third-law partners.

The test takes two seconds: swap the two nouns. If 'Earth pulls book' becomes 'book pulls Earth', you have a pair. If 'table pushes book' becomes 'Earth pulls book', the nouns did not swap and it is not a pair.

There is a two-second test for this, and it is worth using every time rather than reasoning it out from scratch.

Swap the two nouns. If Earth-pulls-book becomes book-pulls-Earth, you have a genuine pair. If table-pushes-book becomes Earth-pulls-book, the nouns did not swap and you do not.

The items in the no bucket are worth understanding rather than just sorting. Those pairs are equal in this situation, often because the object is in equilibrium, but that equality is a fact about the circumstances rather than a law. Tilt the table, add an acceleration, and the balance breaks immediately.

A genuine third-law pair, by contrast, stays exactly equal no matter what. Put the book in a lift, drop it, throw it — the Earth still pulls the book exactly as hard as the book pulls the Earth. That robustness is the signature of the third law, and it is a good way to check yourself when the swap test feels ambiguous.

11. The normal force is the third-law partner of the weight

Trap

12. The Free-Body Diagram

Section

Section 2

13. What belongs on the diagram, and what does not

Concept

A free-body diagram shows one object, reduced to a dot or a simple box, with an arrow for every force acting on it — and nothing else at all.

The last entry in the right-hand column causes real damage. Once you resolve a force into components, cross the original out. Leaving both on the diagram double-counts it, and the error is invisible in the algebra that follows.

The test for whether an arrow belongs: name the object exerting it. 'The Earth', 'the table', 'the rope' are all fine. 'Its motion', 'inertia' and 'the throw' are not forces and must be erased.

A free-body diagram is a picture of one object, reduced to a box or a dot, with an arrow for every force acting on it and absolutely nothing else. The nothing-else part is what most people get wrong.

Velocity and acceleration arrows do not belong on the body. They are not forces, and drawing them among the force arrows is how they end up in the force sum. If you want to mark them, mark them off to one side.

Forces the object exerts on other things do not belong. That is the whole content of the previous slide, applied.

And the last row of the table is the one that does quiet damage. When you resolve a force into components, cross the original out. Leaving both the vector and its components on the diagram double-counts that force, and once it is in the algebra the error is invisible.

The test that catches all of these at once: for each arrow, name the object exerting it. The Earth, the table, the rope, the hand — fine. Its motion, its inertia, the throw — not forces, and off they come.

14. How to draw a free-body diagram

Pattern

1. Choose one object and say out loud which one. Circle it on the picture if it helps.
2. Draw it isolated — a box or a dot, nothing else from the scene.
3. Weight first, always, pointing straight down. It is the force people forget when the object is on a slope.
4. Walk the perimeter: for every surface in contact, add a normal force perpendicular to it and a friction force along it. For every rope, add a tension pulling away along the rope.
5. Add named applied forces. If you cannot name the agent, do not draw it.
6. Choose axes — and tilt them if the acceleration is along a slope.
7. Resolve anything not along an axis, and cross out the originals.
8. Write one equation per axis, and only then start solving.

Step three is worth its own line because of how often it is skipped. On an incline the weight is the only force not aligned with the tilted axes, so it is the one that has to be resolved — and it is also the one most likely to be left off entirely.

Step eight matters as much as the rest. Writing equations while still drawing arrows is how forces get missed. Finish the picture, then start the algebra.

This is a procedure rather than an art, and following it in order is the point. Improvising the order is where forces get missed.

Step three deserves its own mention. Weight always goes on the diagram, always points straight down regardless of what the object is resting on, and is the force most likely to be forgotten when the object sits on a slope — precisely because on a slope it is the only force not aligned with anything.

The walk-the-perimeter step in the middle is what makes the procedure reliable. Rather than trying to recall which forces are present, go round the object physically: every surface in contact contributes a normal force and possibly a friction force, every rope contributes a tension. If nothing touches a side, nothing acts on it.

Step eight matters as much as any of the others. Do not start writing equations while you are still drawing arrows. Finish the picture, look at it, and only then begin the algebra. The temptation to start early is strongest under time pressure, which is exactly when the missed-force error is most likely.

15. A block pushed along a rough floor

Worked example

A 12 kg crate is pushed across a level floor by a 90 N force directed 30 degrees below the horizontal. The coefficient of kinetic friction is 0.25. Find the acceleration.

\[ F_x = 90\cos 30^\circ = 77.9\,\text{N} \qquad F_y = 90\sin 30^\circ = 45.0\,\text{N down} \]

\[ N - mg - F_y = 0 \;\Rightarrow\; N = (12)(9.8) + 45.0 = 162.6\,\text{N} \]

\[ f = \mu_k N = (0.25)(162.6) = 40.7\,\text{N} \]

\[ a = \frac{F_x - f}{m} = \frac{77.9 - 40.7}{12} = 3.1\,\text{m/s}^2 \]

The interesting physics in this problem is in the vertical equation, which is easy to skip past on the way to the horizontal one.

The push is angled downward, so it has a vertical component pressing the crate into the floor. That raises the normal force well above mg — 162.6 N instead of 117.6 N — and since friction is proportional to the normal force, it raises the friction too.

So the angled push is fighting itself. Its horizontal component drives the crate forward while its vertical component increases the resistance. That competition is the whole content of the problem.

The comparison at the end quantifies it. The same 90 N applied horizontally would produce an acceleration of 5.1 rather than 3.1 metres per second squared. And angling the push upward instead would do better still, by reducing the normal force. This is not an abstraction — it is why people instinctively pull suitcases and wheelbarrows rather than pushing them down into the ground.

The general lesson: whenever a force has a component perpendicular to the surface, do the perpendicular equation first. The normal force it determines feeds straight into the friction.

16. Which normal force?

Check

A 5.0 kg box sits on a level floor. Someone pulls upward on it with a rope at 20 N, but the box does not leave the floor.

This question exists to break the reflex that N equals mg, and it is worth checking whether the reflex fired before you read the options.

The box is in vertical equilibrium, so all the vertical forces must balance. Two of them point up — the normal force and the tension — and one points down. So N plus T equals mg, and N is 29 N.

Physically the rope is carrying part of the box's weight, so the floor has less to support. The box is still touching the floor, so the normal force has not vanished, but it has been reduced by exactly the amount the rope is taking.

Worth extending the idea: as the tension grows toward 49 N, the normal force shrinks toward zero, and at exactly 49 N the box is on the verge of lifting off. Beyond that the floor cannot pull down, so N stays at zero and the box accelerates upward. The normal force can be reduced to zero but never made negative — surfaces push, they do not pull.

17. Tilt your axes to match the acceleration

Concept

You may point your coordinate axes wherever you like. The useful choice is almost always: put one axis along the direction the object actually accelerates.

On an incline the normal force is already perpendicular to the surface and friction is already along it. Tilting the axes puts both of them on an axis for free, leaving only the weight to be resolved.

\[ mg_{\parallel} = mg\sin\theta \qquad mg_{\perp} = mg\cos\theta \]

Which gets the sine and which the cosine is worth settling permanently rather than re-deriving under pressure. Check it at the extremes: a flat surface has theta = 0, where nothing should slide, and sin 0 = 0. So the down-slope component carries the sine.

You are free to point your axes anywhere. The useful choice is to put one axis along the direction the object actually accelerates, and on an incline that means tilting them to match the slope.

Look at what the tilt buys. The normal force is already perpendicular to the surface and friction is already along it, so both land on an axis for free. That leaves the weight as the only force needing components — one force to resolve instead of two.

Compare with keeping the axes level. Now the normal force must be resolved, friction must be resolved, and the acceleration itself must be resolved into horizontal and vertical parts, since the block accelerates along the slope rather than along either axis. Three resolutions instead of one, and the algebra is correspondingly worse.

As for which component gets the sine: settle it once by checking a limit rather than re-deriving it under pressure. On a flat surface theta is zero and nothing should slide, and sine of zero is zero. So the down-slope component carries the sine. That check takes two seconds and is reliable.

18. A block on a frictionless incline

Worked example

A block of mass m is released on a frictionless incline of angle theta. Find its acceleration and the normal force.

\[ N - mg\cos\theta = 0 \;\Rightarrow\; N = mg\cos\theta \]

\[ mg\sin\theta = ma \;\Rightarrow\; a = g\sin\theta \]

Notice that the mass cancelled. Every object slides down a frictionless incline with the same acceleration regardless of mass — the same cancellation that makes free fall mass-independent, since a frictionless incline is just a diluted version of free fall.

This is the template problem for inclines, and every incline problem in the course is a variation on it.

Notice that the perpendicular equation is solved first, and that it is an equilibrium equation — the block does not accelerate into or off the surface, so the normal force balances the perpendicular component of the weight. That gives N equals mg cos theta, which is smaller than mg. On any slope the surface supports less than the full weight.

The along-slope equation then has only one force in it, and the acceleration falls straight out.

The limits are worth checking properly. At zero degrees you get zero acceleration and N equal to mg, which is a block sitting on a level floor. At ninety degrees you get an acceleration of g and N equal to zero, which is free fall alongside a vertical wall. Both are situations you already understand, and the formula reproduces them.

And the mass cancelled. Every object slides down a frictionless incline at the same rate, whatever its mass — which is the same cancellation that makes free fall mass-independent. That is not a coincidence: a frictionless incline is free fall with the gravitational pull diluted by a factor of sine theta.

19. Find the errors in this free-body diagram

Error analysis

A student draws this diagram for a block sliding up a rough incline, having been given a shove and then released.

The corrected equation has only two forces along the slope, both pointing down it, so the block decelerates at a rate greater than g sin theta.

\[ -mg\sin\theta - \mu_k mg\cos\theta = ma \;\Rightarrow\; a = -g(\sin\theta + \mu_k\cos\theta) \]

The general lesson: a force that acted in the past is not on the diagram. Free-body diagrams are snapshots of this instant, and the only trace of the past is the object's current velocity.

Two of the five arrows on this diagram correspond to nothing, and both errors are common enough to be worth naming.

The force of the push does not exist. The hand finished pushing before the block was released, and a force that is no longer being applied does not appear on any diagram. What the block retained from the shove is velocity, not force. This is the Aristotelian misconception from earlier in the deck, dressed up in an arrow.

Centrifugal force does not exist either, and here it does not even have the excuse of a rotating frame — the block is moving in a straight line up a slope. No agent, no third-law partner, no reason for it to be there.

The friction direction, by contrast, is right, and it is worth saying why. The block is sliding up the slope, so kinetic friction acts down the slope. Friction opposes the relative sliding of the surfaces, not the applied force and not gravity. When the block later slides back down, friction will reverse and point up the slope.

The corrected physics is worth noting too: with both remaining forces pointing down the slope, the block decelerates faster than g sin theta would suggest, which is why it stops sooner going up than a frictionless block would.

The transferable rule: a free-body diagram is a snapshot of this instant. A force that acted in the past leaves no trace on it except through the object's current velocity.

20. Normal Force and Tension

Section

Section 3

21. The normal force is not always mg

Concept

The normal force is whatever it needs to be to stop the object sinking into the surface. It has no formula. It is an unknown you solve for, and it changes whenever anything else vertical changes.

Only the first row gives N = mg, and it is the row people generalise from. Writing N = mg by reflex is the most reliable way to get a friction problem wrong, because friction is proportional to N.

The correct procedure never varies: write the perpendicular-direction equation from the free-body diagram and solve it for N. If nothing accelerates perpendicular to the surface, that equation sets the forces equal — but which forces are in it depends entirely on the situation.

This table is worth reading as six counterexamples to a single bad habit.

The normal force has no formula. It is whatever it needs to be to prevent the object sinking into the surface, and that value changes whenever anything else in the perpendicular direction changes — an extra push, a rope, a tilt, an acceleration.

Only the first row gives N equal to mg, and it is unfortunately the row everyone meets first and generalises from. Writing N equals mg by reflex is the single most reliable way to get a friction problem wrong, because friction is proportional to N, so the error propagates into every subsequent line.

The last row is worth a thought. In free fall the normal force is zero — nothing is pressing on anything. That is not an exotic special case; it is what happens to a coffee cup in a falling lift, and it is what an astronaut experiences continuously.

The correct procedure never varies: write the equation for the direction perpendicular to the surface and solve it for N. Which forces appear in that equation depends entirely on the situation, and the diagram is what tells you.

22. Standing on a scale in an elevator

Prediction

Work these by acceleration, never by velocity. Acceleration up means the scale reads high; acceleration down means it reads low. Whether the elevator is going up or down does not enter the calculation at all.

Elevator problems are a favourite because they can be got wrong by reasoning from the wrong quantity, and this one is set up to catch exactly that.

The elevator is moving downward, which pulls the eye toward the answer that the scale reads low. But the elevator is slowing down, and slowing while moving downward means the acceleration points upward.

Once the acceleration points up, the second law demands a net upward force, so the normal force must exceed the weight, and the scale reads high. The direction of motion never entered the calculation at all.

The general rule is worth stating in exactly this form: acceleration up means the scale reads high, acceleration down means it reads low, and the velocity is irrelevant. An elevator starting upward and an elevator stopping on the way down feel identical, because they are — the acceleration is the same in both.

The reason the scale reads the normal force at all is worth a moment too. Scales measure the force pressing on their surface, which by the third law equals the force the surface exerts on you. What you feel as heaviness is that normal force, not your weight, which is why you can feel heavy or light while your weight is unchanged.

23. Apparent weight in an accelerating elevator

Worked example

A 65 kg person stands on a scale in an elevator that accelerates upward at 2.4 m/s squared. What does the scale read? What if the cable snapped?

\[ N - mg = ma \;\Rightarrow\; N = m(g + a) \]

\[ N = 65(9.8 + 2.4) = 793\,\text{N} \]

\[ N = m(g - g) = 0 \]

This is exactly the situation of an astronaut in orbit. Weightlessness is not the absence of gravity — it is the absence of a normal force, because everything nearby is falling at the same rate you are.

The algebra here is two lines, and the payoff is a distinction that matters for the rest of the course.

The result N equals m times g plus a covers every case at once. Accelerating upward gives a positive a and a scale reading above the true weight. Accelerating downward gives a negative a and a reading below it. Constant velocity gives a equal to zero and the true weight.

The snapped cable is the case worth dwelling on. With no cable, gravity is the only force, so a equals minus g, and N comes out exactly zero. The scale reads nothing and the person floats.

But note what has not changed. The weight is still 637 newtons, because weight is mg and neither m nor g has altered. What vanished is the normal force. Weightlessness is not the absence of gravity; it is the absence of a normal force.

And that is precisely the situation of an astronaut in orbit, which is why the astronaut slide later in this deck is really this problem continued indefinitely. An orbit is a fall that keeps missing the ground, and everything inside falls together.

24. Tension in an ideal rope

Concept

Physics C makes two idealisations about ropes, and both are stated so rarely that people forget they are assumptions.

• The rope is massless, so the tension is the same everywhere along it.
• The rope is inextensible, so objects tied to it share a single magnitude of acceleration.

The rope in that figure has 50 N of tension, not 100 N. Tension is not the sum of the pulls — it is what each end feels. Tie one end to a wall instead and nothing changes: the wall pulls with 50 N and the tension is still 50 N.

The second idealisation is the more useful one in problems. Inextensible means one acceleration: if two blocks are joined by a taut rope, the magnitudes of their accelerations are equal, which is the extra equation that makes connected-body problems solvable.

Two idealisations are being made here, and they are stated so rarely in problems that it is easy to forget they are assumptions rather than facts.

Massless means the tension is the same everywhere along the rope. A real rope has weight, so the tension at the top of a hanging rope exceeds the tension at the bottom by the weight of the rope in between. Problems that say light rope or massless string are telling you to ignore that.

Inextensible means the objects tied to the rope share one acceleration magnitude. This is the more useful of the two, because it supplies the extra equation that makes connected-body problems solvable.

The figure addresses a specific confusion. Two people pulling with 50 newtons each on opposite ends produce a tension of 50 newtons, not 100. Tension is what each end feels, not the sum of the pulls. Replace one person with a wall and nothing changes — the wall pulls back with 50 newtons and the tension is still 50.

If that seems odd, notice that the rope is in equilibrium: it has 50 newtons pulling each way, which sums to zero, which is exactly why it does not accelerate. A tension of 100 would require 100 newtons at each end.

25. A pulley redirects tension without changing it

Concept

An ideal pulley is massless and frictionless. Its only job is to change the direction of the rope, and the tension is unchanged as it passes over.

This is what makes the standard table-and-pulley problem tractable: one unknown tension serves both blocks, and the rope's inextensibility gives both blocks the same acceleration magnitude.

Be careful with directions, though. The hanging block accelerates downward while the table block accelerates horizontally. The magnitudes match; the vectors do not. Set up each block's axis so that positive means the direction it actually moves, and the signs take care of themselves.

A real pulley has mass, which means the tensions on the two sides differ — the difference is what spins it up. That case waits until the rotation deck, and until then 'ideal pulley' means you may use one T.

The word ideal is carrying two assumptions: the pulley is massless and frictionless. Under those assumptions its only function is to change the direction of the rope, and the tension passes through unchanged.

That is what makes the standard table-and-pulley problem tractable. One unknown tension serves both blocks, and the rope's inextensibility gives them equal acceleration magnitudes. Two equations, two unknowns, done.

Be careful with directions, though, because the equal accelerations are equal only in magnitude. The hanging block accelerates downward while the table block accelerates horizontally — different directions entirely. The reliable trick is to set up each block's positive axis along the direction it actually moves, so both accelerations come out positive and the signs stop fighting you.

A real pulley has mass, and then the tensions on the two sides are genuinely different — the difference is the net torque that spins it up. That case is waiting in the rotation deck. Until then, ideal pulley means you may use a single T, and problems will say so.

26. The Atwood machine

Worked example

Two masses hang from a rope over an ideal pulley: m1 = 3.0 kg and m2 = 5.0 kg. Find the acceleration and the tension.

\[ m_2 g - T = m_2 a \qquad T - m_1 g = m_1 a \]

\[ (m_2 - m_1)g = (m_1 + m_2)a \;\Rightarrow\; a = \frac{(m_2 - m_1)g}{m_1 + m_2} \]

\[ a = \frac{(5.0 - 3.0)(9.8)}{8.0} = 2.45\,\text{m/s}^2 \]

This is one of the most-tested setups in the entire course, so it is worth knowing the structure rather than the answer.

The sign convention is the thing to copy. Rather than picking one global up-is-positive axis, define positive for each block as the direction it actually moves — down for the heavy one, up for the light one. Then both accelerations are positive numbers and the tension appears with opposite signs in the two equations.

That is what makes adding the equations eliminate T immediately. With a global convention the same result appears, but only after more sign-wrangling, and sign-wrangling under exam pressure is where errors live.

Look at the structure of the answer. The numerator is the difference of the weights — the unbalanced part, which is what drives the system. The denominator is the total mass, which is what resists. That is the second law applied to the system as a whole, and it is the same pattern you will see in every connected-body problem in this deck.

The final check is the one to internalise. The tension comes out at 36.8 newtons, between the two weights of 29.4 and 49. It must be: the rope has to pull harder than the light block's weight in order to lift it, and less hard than the heavy block's weight in order to let it fall. A tension outside that bracket means a sign error, and this check will find it faster than re-reading the algebra.

27. Match each setup to its acceleration

Matching

Four standard configurations, all frictionless. Match each to its acceleration.

Seeing the common structure is more valuable than memorising four formulas. Acceleration equals the driving force divided by the total mass in motion — every one of these is that sentence with different words for 'driving force'.

Rather than memorising four formulas, notice that they are all the same formula with different fillings.

In every case, the acceleration is the net driving force divided by the total mass being accelerated. What changes between the cases is only what counts as driving force.

For the Atwood machine, the driving force is the difference of the two weights, because they oppose each other. For the table-and-pulley, it is the hanging weight alone — the table block's weight points perpendicular to its motion and contributes nothing along it. For the incline, it is the down-slope component of the weight. For the pushed pair, it is simply the applied force.

Seeing the shared structure is worth far more than the four results. It means that when a problem presents a configuration you have not seen, you can still write down the acceleration by asking two questions: what is driving this, and what mass is being moved.

28. Friction

Section

Section 4

29. Static and kinetic friction are different rules

Concept

\[ f_s \le \mu_s N \qquad \qquad f_k = \mu_k N \]

Look at the operators. Static friction obeys an inequality; kinetic friction obeys an equation. That difference is the whole of this section.

Static friction is not usually at its maximum. It takes whatever value is needed to prevent sliding, up to a ceiling. A crate that will not budge under a 30 N push has exactly 30 N of static friction, not mu-s N.

The maximum only appears at the moment of slipping, which is why problems asking for the minimum force to start it moving or the steepest angle before it slides are the ones where you set the inequality to an equality.

Notice that mu-k is smaller than mu-s in the graph. That is why a skidding car takes longer to stop than one braking at the edge of grip, and why anti-lock brakes exist at all.

Look at the operators before anything else. Static friction has a less-than-or-equal sign; kinetic friction has an equals sign. That difference is the entire content of this section, and most friction errors come from applying the wrong one.

Static friction is not a fixed value. It takes whatever value is needed to prevent sliding, up to a ceiling of mu-s times N. A crate that will not budge under a 30 newton push has exactly 30 newtons of static friction on it. Push with 50 and it has 50, provided that is still below the ceiling.

The ceiling only becomes relevant at the moment of slipping. That is why questions asking for the minimum force to start something moving, or the steepest angle before it slides, are precisely the questions where you turn the inequality into an equality.

The graph is worth reading carefully. The diagonal section is the static regime, where friction simply matches whatever you apply. The peak is the ceiling. And then it drops — because mu-k is smaller than mu-s — to a constant value that no longer depends on how hard you push.

That drop has a familiar consequence. Once something starts sliding it becomes easier to keep sliding, which is why a heavy box lurches when it finally gives way, and why a skidding car has less grip than one braking at the edge of traction. Anti-lock brakes exist entirely to keep you on the left side of that peak.

30. Friction is always mu times N

Trap

31. Will the block slide, and how fast?

Worked example

A 4.0 kg block sits on a 25-degree incline with mu-s = 0.55 and mu-k = 0.40. Does it slide? If so, find its acceleration.

\[ f_{\text{needed}} = mg\sin\theta = (4.0)(9.8)\sin 25^\circ = 16.6\,\text{N} \]

\[ N = mg\cos\theta = 35.5\,\text{N} \qquad f_{s,\text{max}} = (0.55)(35.5) = 19.5\,\text{N} \]

\[ a = g(\sin\theta - \mu_k\cos\theta) = 9.8(\sin 35^\circ - 0.40\cos 35^\circ) = 2.41\,\text{m/s}^2 \]

This problem shows the check-your-assumption procedure working in both directions, which is why it is worth following closely.

Start by assuming the block stays put and asking how much friction that requires. It needs 16.6 newtons. Then ask how much is available: mu-s times N, where N is mg cos theta and not mg. That gives 19.5 newtons. Available exceeds required, so the block stays.

And note what the actual friction is in that case — 16.6 newtons, not 19.5. Static friction supplies what is needed and no more. Reporting 19.5 would be a real error.

Now raise the angle to 35 degrees and the comparison reverses: 22.5 needed, only 17.7 available. Not enough, so the block slides, and only now does friction become kinetic with the definite value mu-k times N.

The final check on the answer is worth copying. If mu-k were zero, the expression reduces to g sin theta, the frictionless incline. If mu-k equalled tan theta, the acceleration would be zero and the block would slide at constant speed. Both limits behave correctly, which is decent evidence the algebra is sound.

32. The angle of repose

Concept

Tilt a surface gradually and there is one particular angle at which the object finally lets go. It depends on nothing but the coefficient of static friction.

\[ mg\sin\theta_c = \mu_s\,mg\cos\theta_c \]

\[ \tan\theta_c = \mu_s \quad \Longrightarrow \quad \theta_c = \arctan\mu_s \]

The mass has vanished, which is genuinely surprising and experimentally true: a heavy crate and a light one slip at the same angle. More weight means more down-slope force, but it means exactly proportionally more normal force and therefore more friction.

This gives you a way to measure a coefficient of friction with no equipment beyond a protractor: tilt until it slides, take the tangent. It is a standard AP lab, and the fact that mass drops out is usually the point being tested.

This is the tilt-until-it-slips experiment, and its result is genuinely surprising the first time you meet it.

Set the down-slope weight component equal to the maximum static friction, and both mass and g cancel. The critical angle depends on nothing but the coefficient of static friction, through a tangent.

So a heavy crate and a light one slip at exactly the same angle. The reason is that adding mass adds down-slope force and normal force in exactly the same proportion, so the two effects cancel completely. This is the same structural reason that mass cancelled on the frictionless incline and will cancel again for the car on a curve.

Practically, this gives you a way to measure a coefficient of friction with nothing but a protractor: tilt until it slides and take the tangent of the angle. It is a standard AP lab, and the mass-independence is usually the point being examined rather than the number itself.

You can also see the result in the world. A pile of dry sand always forms a cone with the same slope angle regardless of how much sand you pour, and that angle is the angle of repose for sand on sand.

33. Static or kinetic?

Sorting

In each case, decide which kind of friction is acting on the object described.

Item c is the one that surprises people. A rolling tyre is not sliding on the road — the contact patch is instantaneously at rest — so the friction that drives a car forward is static. That is exactly why losing traction is so dangerous: mu-k is lower than mu-s, so a skidding car has less grip than a rolling one.

The only test that matters is whether the two surfaces are sliding across each other at the point of contact. Not whether the object is moving — whether the surfaces are sliding.

That distinction is what makes item c surprising. A car driving normally has tyres that are rolling, not skidding, which means the patch of rubber touching the road is instantaneously at rest against it. The friction that drives a car forward is therefore static friction, even at motorway speed.

This is not a technicality. Because mu-s is larger than mu-k, a rolling tyre has more grip available than a skidding one. Losing traction genuinely reduces your ability to stop or steer, and that is the whole reason anti-lock braking systems exist — they prevent the transition from static to kinetic.

Item e is worth checking too. A box riding on an accelerating truck bed without slipping is held there by static friction, and that friction must point forward to accelerate the box along with the truck. When the truck brakes, the same static friction points backward. It always points whichever way is needed to keep the box from sliding.

34. Static versus kinetic, side by side

Comparison

The row that generates exam questions is the magnitude row. Static friction is a range; kinetic friction is a value. Everything else follows from that.

And the direction row deserves care on inclines. Friction opposes the relative sliding of the surfaces, not the applied force and not the object's velocity through space. A block sliding up an incline has friction pointing down it; the same block sliding back down has friction pointing up it.

The magnitude row is where the exam questions come from, so if you only remember one row, remember that one.

Static friction is a range and kinetic friction is a value. Every other difference on this table follows from that, including why static problems require a check step and kinetic ones do not.

The direction row deserves attention on inclines, where it is easy to get turned around. Friction opposes the relative sliding of the surfaces — not the applied force, and not the object's velocity through space. A block sliding up an incline has friction pointing down the slope. The same block on the way back down has friction pointing up it. Same surfaces, same coefficients, opposite directions.

The final row is worth noticing as well. Static friction can be zero, and often is; kinetic friction cannot be zero while sliding is happening. That asymmetry is another consequence of one being a range and the other a value.

35. Systems and Connected Bodies

Section

Section 5

36. System analysis versus individual analysis

Concept

When several objects move together, you have two legitimate ways to proceed, and choosing well saves a great deal of algebra.

The rule that makes the system approach legal: internal forces cancel in pairs by Newton's third law, so they contribute nothing to the net force on the system as a whole.

So the strategy writes itself. If a question asks only for the acceleration, use the system. If it asks for a tension or a contact force between two blocks — an internal force — you must then isolate one body and analyse it individually.

Most connected-body problems want both, in that order: system first for a, then one individual diagram to get the internal force. Doing it in the other order works but usually means solving simultaneous equations you did not need.

When several objects move together you have two legitimate routes, and picking the right one first can halve the work.

The system approach treats everything as one combined mass. It is legal because internal forces cancel in pairs by the third law — every push one block gives another is matched by an equal push back — so they contribute nothing to the net force on the system as a whole.

That cancellation is also its limitation. Because the internal forces vanish from the system equation, the system approach can never tell you what any of them are. If a question asks for a tension or the contact force between two blocks, you must isolate a body.

So the strategy is usually two-stage: system first for the acceleration, then one individual diagram for whatever internal force is wanted. Doing it in the reverse order works but tends to leave you solving simultaneous equations you did not need.

One warning about the system approach: only external forces along the direction of motion count. In a table-and-pulley setup, the table block's weight and normal force are external but perpendicular to its motion, so they do not enter the system equation, while the friction does.

37. The connected-body recipe

Pattern

1. Find the acceleration with the system approach. Add the masses; identify only the external forces along the direction of motion; divide.
2. Check the constraint. Objects joined by an inextensible rope share one acceleration magnitude — but not necessarily one direction.
3. Isolate one body to get an internal force. Choose the one with the fewest forces on it; that choice can halve the algebra.
4. Write that body's second law with the acceleration you already found, and solve for the unknown.
5. Check the internal force lies between the extremes it should. A tension should sit between the two weights it connects; a contact force should be smaller than the applied force.

Step three is where the time is won or lost. In a three-block train pushed from behind, the force between blocks two and three is most easily found by isolating block three alone — it has exactly one horizontal force on it, so the equation is a single line.

Step five is a genuine error-catcher, not decoration. An internal force outside its bracketing range means a sign went wrong somewhere, and you will find it faster from this check than by re-reading the algebra.

Five steps, and the value is concentrated in the third and the fifth.

Step three is where the time is won. When you need an internal force, isolate the body with the fewest forces acting on it. In a three-block train pushed from behind, the force between the second and third blocks is most easily found by looking at the third block alone, which has exactly one horizontal force on it. One equation, one line, done.

The instinct to isolate the block nearest the applied force is natural and usually the expensive choice, because that block has both the applied force and a contact force on it.

Step five is a genuine error-catcher rather than decoration. A tension should always lie between the two weights it connects. A contact force should always be smaller than the applied force driving the system. If your answer falls outside that bracket, a sign is wrong somewhere, and this check will locate the problem faster than re-reading the algebra.

Step two, the constraint, is the one that gets forgotten and then causes a mysterious shortage of equations. Objects joined by an inextensible rope share one acceleration magnitude. Write that down explicitly rather than assuming it.

38. Block on a table, connected to a hanging mass

Worked example

A 6.0 kg block on a table (mu_k = 0.20) is connected over an ideal pulley to a 4.0 kg hanging block. Find the acceleration and the tension.

\[ N = m_1 g = 58.8\,\text{N} \qquad f = \mu_k N = 11.8\,\text{N} \]

\[ a = \frac{m_2 g - f}{m_1 + m_2} = \frac{39.2 - 11.8}{10.0} = 2.74\,\text{m/s}^2 \]

\[ m_2 g - T = m_2 a \;\Rightarrow\; T = 4.0(9.8 - 2.74) = 28.2\,\text{N} \]

This configuration appears on AP exams more often than almost any other, so the workflow is worth having automatic.

The system step needs one piece of care: which forces are external and along the motion. The hanging weight drives, friction resists. The table block's weight and normal force are perpendicular to its motion and drop out — but they are still needed, because the normal force is what sets the friction.

So compute N first, then friction, then the system acceleration. Skipping to the acceleration and using mg for the friction of the wrong block is a common slip.

For the tension, isolate the hanging block. It has only two forces on it, which makes it the cheaper of the two diagrams by some margin.

The verification with the other block is worth the twenty seconds. Getting the same number two independent ways confirms both the acceleration and the tension simultaneously.

And the bracket check: the tension comes out at 28.2 newtons, below the hanging weight of 39.2. It must be, because if the rope pulled as hard as the block's weight, the block would not accelerate at all.

39. Three blocks in a train: the internal forces differ

Worked example

Three blocks of 1.0, 2.0 and 3.0 kg sit in contact on a frictionless floor. A 12 N force pushes the 1.0 kg block. Find the acceleration and both contact forces.

\[ a = \frac{12}{1.0 + 2.0 + 3.0} = 2.0\,\text{m/s}^2 \]

\[ F_{12} = (2.0 + 3.0)(2.0) = 10\,\text{N} \]

\[ F_{23} = (3.0)(2.0) = 6.0\,\text{N} \]

The contact forces decrease front to back: 12 N, then 10 N, then 6 N. Each interface only has to accelerate the mass ahead of it, and there is less of that mass each time.

The result to take from this problem is that the contact forces are different at the two interfaces, and understanding why generalises to every connected-body problem.

The acceleration comes from the system in one line — the contact forces are internal and cancel, so only the 12 newton push survives.

Then look at what each interface has to do. The force between blocks one and two must accelerate blocks two and three together, five kilograms, so it is 10 newtons. The force between blocks two and three must accelerate only block three, three kilograms, so it is 6 newtons.

The pattern is that an interface is determined by what lies ahead of it, not behind it. Each contact force only has to move the mass downstream, and there is less of that mass at each successive interface.

So the forces decrease front to back: 12, then 10, then 6. This is exactly why the coupling at the front of a train carries the largest load and the coupling at the back the smallest, and it is why train couplings are rated by position.

The verification on the front block is a good habit: 12 newtons forward, 10 backward, net 2, and one kilogram times two metres per second squared is 2 newtons. Everything is consistent.

40. The contact force between two blocks

Check

Two blocks, 2.0 kg and 4.0 kg, sit in contact on a frictionless floor. A horizontal force of 18 N pushes on the 2.0 kg block, driving both.

Work this by isolating the far block, which turns it into a one-line problem.

The system accelerates at 3 metres per second squared. The 4 kilogram block has exactly one horizontal force on it — the contact force from its neighbour — so that force must be 4 times 3, or 12 newtons.

The tempting wrong answer is 18 newtons, the applied force. But if the second block felt the full 18, the first block would have 18 newtons forward and 18 newtons backward from the third law, hence no net force and no acceleration. The applied force acts on the first block only.

The other tempting error is to isolate the near block and forget that it also feels the applied force. That gives the net force on the 2 kilogram block, 6 newtons, rather than the contact force.

Notice again that the contact force is smaller than the applied force, and that it would have been a different number had the masses been swapped. Push the 4 kilogram block instead and the contact force becomes 6 newtons, because now only 2 kilograms lies downstream.

41. Why do the two contact forces differ?

Explain it to yourself

In the three-block train, the same 12 N push produced a 10 N contact force at the first interface and only 6 N at the second.

Explaining this in your own words is worth more than working another example, because the idea transfers to tensions in ropes, forces in couplings, and stresses in beams.

The core statement is that an internal force is determined by what lies ahead of it. Each interface only has to accelerate the mass downstream of it, so with a fixed acceleration, less downstream mass means a smaller force.

That immediately explains why both contact forces are smaller than the applied force. The full 12 newtons is needed to accelerate all 6 kilograms; any interface further along has less left to move.

The reversal question is a good test of whether the idea has landed rather than the arithmetic. Pushing from the other end leaves the acceleration untouched — same external force, same total mass — but changes both contact forces, because the amount of mass downstream of each interface has changed.

The practical consequence for exams: when asked for a tension or contact force, isolate the far body. It is almost always the shorter calculation.

42. Uniform Circular Motion

Section

Section 6

43. Turning is accelerating, even at constant speed

Concept

Velocity is a vector. Going round a circle at a steady speed changes its direction continuously, so the velocity is changing, so there is an acceleration — even though the speedometer never moves.

\[ a_c = \frac{v^2}{r} \]

The acceleration points toward the centre, which is what the word centripetal means — centre-seeking. It is perpendicular to the velocity at every instant, and that perpendicularity is exactly why the speed does not change while the direction does.

So a car rounding a bend at a steady 20 m/s is accelerating, and something must be supplying the force. On a flat road that something is friction, and when it runs out the car goes straight on.

This is the idea the whole section rests on, and it contradicts everyday language, so it is worth stating carefully.

Velocity is a vector, so it changes if either its magnitude or its direction changes. Going round a bend at a steady speed changes the direction continuously. Therefore the velocity is changing. Therefore there is an acceleration, even though the speedometer never moves.

The direction of that acceleration is toward the centre of the circle, which is what the word centripetal means. And it is perpendicular to the velocity at every instant — which is precisely why the speed does not change while the direction does. A force parallel to the motion would change the speed; a force perpendicular to it only turns.

The magnitude, v squared over r, has two dependences worth feeling rather than memorising. Doubling the speed quadruples the required acceleration, which is why bends become dangerous so suddenly as speed rises. Halving the radius doubles it, which is why tight bends are harder than gentle ones at the same speed.

And something must supply the force that produces this acceleration. On a flat road it is friction. When friction runs out, the acceleration stops, and the car does exactly what the first law says: it goes straight on.

44. Centrifugal force throws you outward in a turn

Trap

45. Centripetal force is a job, not a new force

Concept

There is no such thing as the centripetal force in the way there is such a thing as tension or friction. Centripetal is a role that some real force is playing.

So the equation F = mv^2/r is not a new law. It is the second law with the acceleration already written in the circular form, and the left side must be filled in from your free-body diagram like any other net force.

\[ \sum F_{\text{toward centre}} = \frac{mv^2}{r} \]

The procedure is therefore unchanged from the rest of the unit. Draw the diagram, pick the axis pointing at the centre, sum the real forces along it, and set the sum equal to mv squared over r. Do not add an extra arrow labelled 'centripetal force' — you would be double-counting a force already on the diagram.

This is the correction that makes circular motion easy, and it is mostly a matter of vocabulary.

There is no such thing as the centripetal force in the way there is such a thing as tension or friction. Centripetal is a role, and some real force with a real agent is always playing it. Look down the table: friction, tension, gravity, normal force. All forces you already know.

The equation is therefore not a new law. It is the second law with the acceleration written in its circular form. The left side has to be filled in from your free-body diagram exactly as it would for any other problem.

The concrete error this prevents is worth naming. Students draw the real forces on the diagram — say, friction on a car — and then add another arrow labelled centripetal force pointing at the centre. That double-counts a force already present, and produces an answer that is wrong by a factor.

So the procedure is unchanged from the rest of the unit. Draw the diagram. Choose an axis pointing at the centre. Sum the real forces along it. Set the sum equal to m v squared over r. Do not add extra arrows.

46. A car on a flat curve

Worked example

A 1200 kg car rounds an unbanked curve of radius 45 m. The coefficient of static friction between tyres and road is 0.70. What is the fastest it can go without sliding?

\[ N = mg = (1200)(9.8) = 11760\,\text{N} \]

\[ \mu_s mg = \frac{mv^2}{r} \;\Rightarrow\; v_{\text{max}} = \sqrt{\mu_s g r} \]

\[ v_{\text{max}} = \sqrt{(0.70)(9.8)(45)} = 17.6\,\text{m/s} \]

The dependence on radius matters practically: halving the radius of a bend reduces the safe speed by a factor of the square root of two, which is why tight bends carry much lower advisory speeds.

The reasoning step that matters here comes before any algebra: identify what actually points toward the centre.

The normal force is vertical, and the weight is vertical. Neither has a horizontal component. So friction is the only candidate, and it must supply the entire turning force by itself.

This is one of the rare situations where N really does equal mg, and it is worth noticing why: nothing else acts vertically and there is no vertical acceleration. Do not let this case reinforce the reflex.

The maximum speed is where static friction hits its ceiling, so you set mu-s times N equal to the centripetal requirement and solve. The mass cancels before you reach the arithmetic, which means a fully laden lorry and a small car slide at the same speed on the same curve — the same mass-independence as the angle of repose, for the same reason.

The dependence on radius is the practically important one. The safe speed goes as the square root of the radius, so halving the radius of a bend reduces the safe speed by a factor of about 1.4. That is why tight bends carry much lower advisory speeds than gentle ones, and why the sign is about geometry rather than about the drivers.

47. The conical pendulum

Worked example

A 0.50 kg ball on a 1.2 m string swings in a horizontal circle, with the string making 35 degrees with the vertical. Find the speed and the tension.

\[ T\cos\theta = mg \qquad T\sin\theta = \frac{mv^2}{r} \]

\[ r = L\sin\theta = 1.2\sin 35^\circ = 0.688\,\text{m} \]

\[ T = \frac{mg}{\cos\theta} = \frac{(0.50)(9.8)}{\cos 35^\circ} = 5.98\,\text{N} \]

\[ \tan\theta = \frac{v^2}{gr} \;\Rightarrow\; v = \sqrt{gr\tan\theta} = \sqrt{(9.8)(0.688)\tan 35^\circ} = 2.17\,\text{m/s} \]

This problem is the template for any situation where a single tilted force has to do two jobs at once, and the technique transfers directly to the banked curve.

The tension leans inward, so it splits into two components with two different jobs. The vertical component holds the ball up against gravity; the horizontal component points at the centre and does the turning. Two equations, and they are of different kinds — the vertical one is an equilibrium, the horizontal one is a centripetal requirement.

The vertical one is an equilibrium because the circle is horizontal. The ball does not rise or fall, so there is no vertical acceleration. Getting this right is the main setup decision.

One geometric trap: the radius of the circle is not the length of the string. It is the horizontal distance from the axis, which is L sin theta. Using L for r is a common error and it changes the answer.

The division step is the move worth remembering. Dividing the horizontal equation by the vertical one eliminates the tension and leaves a tangent. You will use exactly this trick on the banked curve two slides from now, and in several free-response problems.

The final check is quick and effective: the tension must exceed the weight, because its vertical component alone has to equal mg. Here 5.98 newtons against a weight of 4.9. It does.

48. The banked curve

Worked example

Find the speed at which a curve of radius r banked at angle theta can be taken with no friction at all.

\[ N\cos\theta = mg \qquad N\sin\theta = \frac{mv^2}{r} \]

\[ \tan\theta = \frac{v^2}{gr} \quad \Longrightarrow \quad v = \sqrt{gr\tan\theta} \]

A real banked road has friction as well, which is why you can take a banked curve safely over a range of speeds rather than at exactly one. Below the design speed friction acts up the slope; above it, down the slope.

The one setup decision in this problem determines whether it takes three lines or thirty, and it is counter-intuitive.

Do not tilt the axes. On an incline problem you tilt them because the object accelerates along the slope. Here the car does not accelerate along the road surface — it accelerates horizontally, toward the centre of its circular path. So horizontal and vertical axes are the correct choice, and the normal force is what gets resolved.

With that settled, it is the conical pendulum again. The normal force leans inward, its vertical component supports the weight, its horizontal component supplies the turning. Divide to eliminate N and you get a tangent.

The result says the design speed depends only on the geometry, the bank angle and the radius, with the mass cancelling as usual. So a banked curve has one speed at which it works perfectly for every vehicle, regardless of load.

Check the extremes. At zero degrees the design speed is zero — a flat road cannot turn anything without friction, which is right. As the angle approaches ninety, the speed goes to infinity, which describes a vertical wall — also right, and it is the principle behind a velodrome.

In practice roads have friction as well, which is why a real banked curve is safe over a range of speeds. Below the design speed friction acts up the slope to stop you sliding down; above it, friction acts down the slope to help turn you.

49. Water in a bucket swung overhead

Prediction

A bucket of water is swung in a vertical circle. At the very top the bucket is upside down, and the water stays in.

The condition for this to work is a minimum speed. If the bucket goes too slowly, the required centripetal force is less than the weight alone, and the water would need the bucket to pull up on it — which a bucket cannot do.

\[ \text{at the top:}\quad mg + N = \frac{mv^2}{r}, \qquad N \ge 0 \;\Rightarrow\; v_{\text{min}} = \sqrt{gr} \]

The honest answer to what holds the water up is: nothing, and nothing needs to.

At the top of the swing, both the weight and the bucket's normal force point downward — which is toward the centre of the circle. Their sum is exactly the centripetal force required. Everything points the same way and everything is consistent.

The water is genuinely falling. So is the bucket, along the same curved path, at the same rate. The water never leaves the base because the base is falling with it, which is the same reason things float inside an orbiting station.

The condition for it to work is a minimum speed. Set the normal force to zero — the bucket can push but it cannot pull — and the weight alone must supply the centripetal force. That gives v minimum equal to the square root of g r.

Go slower than that and the required centripetal force is less than the weight alone, so the bucket would need to pull upward on the water to keep it on the circular path. Buckets cannot pull, so the water leaves the path, which is to say it falls on you.

The same condition governs a roller-coaster loop and a ball on a string, and it is one of the most commonly examined results in the unit.

50. Vertical circles: the top and the bottom

Concept

The sign difference comes entirely from geometry. At the top, toward the centre means downward, so the weight helps. At the bottom, toward the centre means upward, so the weight fights.

This is why a roller-coaster rider feels crushed into the seat at the bottom of a loop and nearly weightless at the top. The normal force — which is all you can actually feel — swings between mv squared over r plus mg and mv squared over r minus mg.

The two equations look different only because of geometry, and understanding that keeps you from memorising them separately.

At the top, toward the centre means downward. Both the weight and the normal force point that way, so they add, and the normal force can be as small as zero when the speed is at its minimum.

At the bottom, toward the centre means upward. Now the weight points away from the centre, so the normal force has to overcome the weight and supply the centripetal force as well. It is always larger than mg, and considerably larger at speed.

That is exactly what a rider feels on a roller coaster. Crushed into the seat at the bottom of the loop, nearly weightless at the top. What you feel is the normal force — the seat pushing on you — and it swings between mv squared over r plus mg and mv squared over r minus mg.

A useful habit for any vertical-circle problem: draw the diagram at the specific point being asked about, then decide which way is toward the centre at that point, then sum. Do not try to carry one general equation around the loop, because the geometry changes as you go.

51. Turn the banking angle

Tweak it

Take a curve of fixed radius and sweep its bank angle, watching the frictionless design speed.

Real roads are banked at modest angles, rarely above about 10 degrees, because a steep bank is dangerous for a vehicle moving slowly — it would slide down the slope. Velodromes and oval race tracks, where nothing goes slowly, bank far more steeply.

Sweeping the angle shows how sharply the design speed climbs, and the shape of the growth is the interesting part.

The speed goes as the square root of the tangent, and the tangent blows up near ninety degrees. So the design speed grows without bound as the bank approaches vertical.

Physically that makes sense: a vertical wall has a normal force that is entirely horizontal, so there is no limit to how much centripetal force it can supply. What there is no longer any of is vertical support — nothing is holding the car up. That is why the frictionless model stops being useful at extreme angles.

Real roads are banked at modest angles, rarely more than about ten degrees. The reason is not the fast case but the slow one: on a steeply banked road, a vehicle moving slowly or stopped would simply slide down the slope. Public roads must be safe for a car at walking pace and for one at motorway speed.

Velodromes and oval race tracks, where nothing ever goes slowly, bank at forty degrees and more, and the physics of this slide is exactly why they can.

52. Drag and Resistive Forces

Section

Section 7

53. Drag depends on how fast you are already going

Concept

Every force so far has had a fixed value once the situation was set. Drag does not: it grows as the object speeds up, which makes the acceleration change continuously.

\[ F_{\text{drag}} = -bv \qquad \text{or} \qquad F_{\text{drag}} = -cv^2 \]

The linear form applies to slow motion through a viscous fluid — a bead in oil, a fine mist droplet. The quadratic form applies to faster motion through air, which is the case for skydivers and cars. AP Physics C free-response questions almost always use the linear form, because it integrates cleanly.

The consequence for problem-solving is structural. Because the force depends on v, and v is what you are solving for, F = ma becomes a differential equation rather than an algebraic one.

\[ m\frac{dv}{dt} = mg - bv \]

This is the moment the unit needs calculus rather than algebra, and it is exactly the separation-of-variables technique from the end of the kinematics deck, now arriving with a physical justification.

Every force so far has had a value that was fixed once the situation was described. Drag is the first that depends on the thing you are trying to solve for, and that changes the mathematics.

Two models are in use. The linear form applies to slow motion through a viscous fluid — a bead falling through oil, a mist droplet in air. The quadratic form applies to faster motion through air, which covers skydivers, cars and cricket balls.

AP Physics C free-response questions almost always use the linear form, because it integrates cleanly and the exam is testing your calculus rather than your patience. If a problem hands you a drag force proportional to v, that is a signal that separation of variables is coming.

The structural consequence is the important one. Because the force depends on v, and v is the unknown, the second law becomes a differential equation rather than an algebraic one. You cannot solve it by rearranging; you have to integrate.

This is the same technique as the a equals minus k v problem at the end of the kinematics deck, arriving now with a physical justification attached. If that slide felt abstract at the time, this is what it was for.

54. Terminal velocity, without solving anything

Worked example

An object falls under gravity with linear drag. Find its terminal velocity.

\[ m\frac{dv}{dt} = mg - bv \]

\[ 0 = mg - bv_T \;\Rightarrow\; v_T = \frac{mg}{b} \]

The lesson is a general one for differential equations in physics: the equilibrium value is always available without solving anything, by setting the rate of change to zero. Do this first — it costs one line and it tells you what your full solution has to approach.

The lesson of this slide is a technique rather than a result, and it applies to every differential equation you will meet this year.

Terminal velocity means the velocity has stopped changing, which means dv/dt is zero. Substituting that single fact turns the differential equation into an algebraic one, and the answer falls out in one line without any integration at all.

The result, mg over b, has sensible dependences. More mass means a higher terminal speed, because a larger weight needs a larger drag force to balance it, and larger drag requires more speed. A larger drag coefficient means a lower terminal speed, which is exactly what a parachute is for.

The unit check is worth doing once. The coefficient b has units of force divided by velocity, so mg over b has units of force divided by force-per-velocity, which is velocity. Correct.

The general principle to carry forward: whenever you meet a differential equation in physics, find its equilibrium first by setting the rate of change to zero. It costs one line, it often answers the question directly, and it tells you what your eventual full solution has to approach.

55. Solving the drag equation properly

Concept

\[ \frac{dv}{g - (b/m)v} = dt \]

\[ \int_0^{v}\frac{dv'}{g - (b/m)v'} = \int_0^{t} dt' \]

\[ v(t) = \frac{mg}{b}\left(1 - e^{-bt/m}\right) = v_T\left(1 - e^{-t/\tau}\right) \]

The time constant is tau = m/b. After one time constant the object has reached about 63 percent of terminal velocity; after three, about 95 percent. A heavy object has a large tau and takes longer to reach its terminal speed — as well as having a higher one.

Check the two limits of that expression. At t = 0 the exponential is 1 and v = 0, as it must be. As t grows large the exponential dies and v approaches mg/b, the terminal velocity found on the previous slide without any of this work.

This is the full solution, and it is the same separation-of-variables procedure as the coasting boat, with one extra term.

Separate, attaching limits rather than carrying a constant of integration. The left integral produces a logarithm, which is why exponentials appear in every drag problem you will ever solve.

The result is worth reading rather than just recording. It says the velocity starts at zero and climbs toward the terminal value, approaching it exponentially and never quite arriving.

The time constant, tau equals m over b, is the quantity that tells you how long that takes. After one time constant the object has reached about 63 percent of terminal velocity; after three, about 95 percent. A heavy object has a large tau, so it both has a higher terminal speed and takes longer to get there.

Check both limits of the expression, because it costs nothing. At t equals zero the exponential is one and v is zero, which matches the stated initial condition. As t grows large the exponential dies and v approaches mg over b, the terminal velocity you found on the previous slide with no work at all.

56. What happens as the drag coefficient goes to zero?

Edge cases

This is a useful habit to keep. A formula that appears to blow up in a limit is usually hiding a compensating quantity that blows up alongside it. Expanding to first order — rather than substituting the limiting value — is what reveals the sensible answer underneath.

This is a good puzzle because two things you believe appear to contradict each other, and the resolution teaches a technique.

Statement one: as b shrinks, the terminal velocity mg over b grows without bound. Statement two: with no drag at all, the motion should be ordinary free fall, which has no terminal velocity but also no strange behaviour.

Both are true, and the reconciliation is about time scales. As b shrinks, the time constant m over b grows in exactly the same proportion. So the terminal velocity gets larger but the object takes proportionally longer to approach it, and within any fixed observation window it never gets close.

To see it properly, expand the exponential for small t. The expression reduces to g times t, which is free fall. So for small b, the motion looks like free fall for longer and longer — the drag has simply not had time to matter yet.

Setting b to exactly zero makes the formula read zero over zero, which is why the correct move is to take a limit rather than to substitute. That is worth remembering generally: a formula that appears to blow up in some limit is usually hiding a compensating quantity that blows up alongside it, and expanding to first order reveals the sensible answer underneath.

57. Why a falling raindrop does not injure anyone

Real world

A raindrop forms about 2 km up. Ignoring air, it would arrive at roughly 200 m/s — comfortably fast enough to be dangerous. It arrives at about 9 m/s instead.

The free-fall calculation for a 2 kilometre drop gives about 200 metres per second, which would be genuinely dangerous. Rain arrives at about 9. The gap is entirely due to drag.

What happens is that the drop reaches its terminal velocity within a few metres of falling, because its time constant is tiny. The remaining two kilometres are covered at a constant, harmless speed. The 200 metres per second describes a fall that never occurs.

The size dependence follows from mg over b. Mass grows with the cube of the radius while drag grows roughly with the square, so larger drops have higher terminal velocities. That is exactly why a heavy shower stings your face and a fine drizzle does not.

Push the same reasoning further and you get hail. A hailstone is far more massive for its cross-sectional area, so it can reach 30 or 40 metres per second — enough to dent cars, which rain never does.

And a skydiver is the same physics run in reverse. Terminal velocity is about 55 metres per second in free fall; opening the parachute multiplies b by a large factor, and the terminal velocity drops to about 5.

The principle to keep: whenever a fall is long, the answer is the terminal velocity, not the free-fall kinematics. Free fall only describes the first fraction of a second.

58. Putting It Together

Section

Section 8

59. The universal dynamics recipe

Pattern

1. Draw a free-body diagram for each object. One diagram per body, forces only, agents nameable.
2. Choose axes per object — tilted along the incline, or with one axis pointing at the centre of the circle.
3. Write one second-law equation per axis per object. Equilibrium directions get zero on the right.
4. Add the constraints: shared acceleration for connected bodies, a = v squared over r for circular motion, f = mu N where sliding.
5. Count equations against unknowns. They must match before you start solving.
6. Solve symbolically, then substitute, and check limits and signs at the end.

Step five is the one that saves a wasted page. If you have three unknowns and two equations, you are missing a constraint — usually the rope constraint or a friction relation — and finding it now is far cheaper than discovering it halfway through the algebra.

Notice what did not change between the incline problems, the pulley problems and the circular-motion problems: the recipe. Only the geometry of the axes and the constraint equations differ, which is why practising the procedure beats memorising the cases.

Notice what did not change across this deck. Inclines, pulleys, friction, circular motion and drag all used the same six steps. Only the geometry of the axes and the constraint equations differed.

That is the argument for practising the procedure rather than memorising the cases. There are far more configurations than anyone can memorise, and the recipe handles configurations you have never seen.

Step five is the one that saves a wasted page. Count your equations and count your unknowns before you start solving. If you have three unknowns and two equations, you are missing a constraint — usually the rope constraint, or a friction relation, or the geometric relation between a radius and a length. Finding it now is far cheaper than discovering it halfway through the algebra.

Step two is where circular motion differs from everything else, and it is worth being explicit about. For an incline you tilt the axes to follow the slope; for circular motion you point one axis at the centre. In both cases the rule is the same: put an axis along the acceleration.

60. The astronaut who is not weightless

Anomaly

An astronaut on the International Space Station floats. The station orbits at about 400 km, where the Earth's gravitational field is still roughly 89 percent as strong as at the surface.

This is the elevator-with-a-snapped-cable problem from Section 3, continued indefinitely. An orbit is a fall that keeps missing the ground, and everything inside falls together.

The phrase to retire is zero gravity. The correct term is free fall, or microgravity — and the sensation is identical to the top of a rollercoaster drop, for exactly the same reason.

The phrase zero gravity is one of the most misleading in common use, and this slide is worth being precise about.

At the altitude of the space station, roughly 400 kilometres, the Earth's gravitational field is still about 89 percent of its surface value. Gravity is very much present — in fact it is exactly what curves the station's path into an orbit rather than letting it fly off in a straight line.

What has disappeared is the normal force. The astronaut is accelerating toward the Earth at nearly 8.7 metres per second squared, and so is the station, so neither presses on the other. And the normal force is the only part of this you can actually feel.

So this is the snapped-elevator-cable problem from Section 3, continued indefinitely. An orbit is a fall that keeps missing the ground, and everything inside falls together.

The correct vocabulary is free fall, or microgravity. And the sensation is not exotic — it is the same feeling as the top of a rollercoaster drop or the first moment of a lift starting down, for precisely the same reason.

61. How hard can a shoe grip?

Estimation

This is why people lean into a heavy push, brace a foot against something, or get someone to sit on the object. You cannot push horizontally harder than friction will let you, and friction is capped by your own weight.

It is also the physics of a tug of war, where the winning side is usually the heavier one rather than the stronger one — and of why studs and cleats exist, since they replace friction with a mechanical interlock that has no such ceiling.

This estimate is worth doing because it puts a hard ceiling on something people do not usually think of as limited.

Your normal force on a level floor is roughly your weight, about 700 newtons for a 70 kilogram person. With a friction coefficient near 0.9, the maximum static friction is about 620 newtons — roughly 90 percent of your own weight.

That is the largest horizontal force you can exert on anything while standing on a level floor, no matter how strong you are. Beyond it, your feet slide.

This explains a lot of everyday behaviour that otherwise looks like superstition. People lean into a heavy push, brace a foot against a wall, or ask someone heavier to help. All three are ways of getting around the friction ceiling — leaning changes the direction of your push, bracing replaces friction with a normal force, and adding a person adds weight.

It is also the physics of a tug of war, where the heavier team usually wins rather than the stronger one, and of why studs and cleats exist. A stud digs into the ground and replaces friction with a mechanical interlock, which has no such ceiling at all.

62. Friction on a decelerating truck bed

Check

A crate rides on the flat bed of a truck without slipping. The truck brakes, slowing down while still moving forward.

Work this from the acceleration rather than from the motion, which is the same discipline as the elevator problem.

The crate is not sliding on the bed, so the friction is static. That settles half the question immediately.

For the direction, ask what the crate's acceleration is. It must slow down along with the truck, so its acceleration points backward — opposite its direction of travel. Friction is the only horizontal force available on the crate, so friction must point backward to produce that acceleration.

The tempting answer D is worth thinking through, because the reasoning behind it is nearly right. If there were genuinely no friction, the crate would obey the first law and continue at constant velocity while the truck slowed beneath it — which is to say it would slide forward off the bed. Something must be acting on it, and friction is the only candidate.

This is exactly why unsecured loads slide forward when a lorry brakes hard: the required friction exceeds what static friction can supply, the crate starts sliding, and it is now in the kinetic regime with even less grip available.

63. Four questions before you close the deck

Exit ticket

Answer these from memory before reading the responses. Retrieval is what makes the material stick, and these four are the ones most likely to appear in some form on an exam.

The first tests whether the third-law-partner idea has landed. The partner of the book's weight is the book pulling the Earth up. The normal force is not the partner — it acts on the same object, and it equals mg here only because the book is not accelerating.

The second tests the static friction inequality. The answer is 40 newtons, matching the push. The 90 newtons is only the ceiling, and using it would give the crate a net backward force and send it sliding toward the person pushing.

The third tests the centripetal-is-a-role idea. Static friction from the road, not centrifugal force and not a separate arrow labelled centripetal.

The fourth tests whether you know the boundary of the kinematic equations. With drag, the acceleration changes continuously as the speed changes, so the constant-acceleration equations are simply the wrong tool.

Any of the four that gave you pause is worth revisiting directly rather than re-reading the whole deck.

64. What you can do now

Recap

• Draw a free-body diagram with every real force and no invented ones, and name the agent of each
• Apply the second law as one equation per axis per object, with tilted axes when the geometry asks for it
• Solve for normal force and tension rather than assuming them, and know that N equals mg only in the simplest case
• Use the inequality for static friction and the equation for kinetic, and test which regime you are in
• Handle connected bodies with the system approach for a and an isolated body for internal forces
• Treat circular motion as a second-law problem with the axis pointing at the centre
• Set up and solve the drag differential equation, and read off the terminal velocity in one line

Next deck: work and energy. Forces tell you the acceleration at every instant, which is complete but often laborious. Energy methods answer questions about start and end states without tracking the journey — and they will make some of the problems in this deck almost trivial.

The change from the start of this deck is that acceleration is no longer something you have to be given. You can now work it out from the forces, which means kinematics and dynamics together can describe any mechanical situation in this course.

The habit that carries everything is the free-body diagram. Draw one object, put on every real force with a nameable agent, choose axes along the acceleration, and write one equation per axis. Every problem in this deck used that procedure and nothing more.

Three specific corrections are worth carrying forward, because they are the ones that keep causing trouble. The normal force is an unknown, not mg. Static friction is a range, not a value. Centripetal is a role played by a real force, not a force to be added.

What comes next is a different way of solving the same problems. Forces give you the acceleration at every instant, which is complete information but often laborious — the drag problem needed a differential equation to answer a fairly simple question. Energy methods relate the starting and ending states directly, without tracking anything in between.

You have already glimpsed this. The v dv/dx identity from the kinematics deck turned out to be the work-energy theorem in disguise, and the next deck makes that explicit.