AP Physics C: Mechanics Unit 2 Notes — Newton’s Laws of Motion

Newton's First Law and Inertia

What the First Law says (and what it really means)

Newton’s First Law states that if the net force on an object is zero, then the object’s velocity stays constant. “Constant velocity” includes two possibilities: staying at rest, or moving in a straight line at constant speed.

A clean mathematical way to say this is:

\sum \vec{F} = \vec{0} \Rightarrow \vec{a} = \vec{0}

and since acceleration is the rate of change of velocity, \vec{a} = \vec{0} implies \vec{v} is constant.

The key idea isn’t that “forces make things move.” The key idea is that forces change motion. If an object is already moving, it does not need a force to “keep it moving” (that’s an old Aristotelian intuition that Newton overturned). What forces do is change the velocity vector—change speed, direction, or both.

Inertia: the “resistance to changes in motion”

Inertia is the tendency of an object to maintain its state of motion. In Newtonian mechanics, inertia is quantified by mass: bigger mass means “harder to accelerate” for the same net force.

This is why an empty shopping cart is easy to start/stop/turn, while a loaded cart “wants” to keep doing what it’s doing.

Two clarifications that prevent a lot of confusion:

  • Inertia is not a force. It’s a property of matter.
  • “Constant velocity” does not mean “no forces act.” It means the forces that act add up to zero.

Why the First Law matters: equilibrium as a special case of dynamics

The First Law is the foundation for equilibrium reasoning. In many Newton’s Laws problems, the first step is to decide whether the object is accelerating. If not, you can set:

\sum \vec{F} = \vec{0}

That single statement unlocks lots of everyday situations (a book resting on a table, a mass hanging at rest, an object sliding at constant speed with friction balanced by an applied force). In AP Physics C, you’ll often use the First Law to build the equations that solve for unknown forces.

Inertial frames (the “hidden assumption”)

Newton’s Laws are simplest in an inertial reference frame, meaning a frame that is not accelerating (or at least one where Newton’s laws hold to very good approximation). If you analyze motion from an accelerating frame (like a car that is speeding up), objects appear to accelerate “on their own” unless you introduce fictitious forces. In AP Physics C: Mechanics, you’re usually expected to choose an inertial frame (often the ground) unless a problem explicitly pushes you into non-inertial analysis.

Example 1: Book on a table (static equilibrium)

A book rests on a horizontal table.

Conceptual setup: The book is at rest, so \vec{a} = \vec{0} and the net force must be zero.

Forces on the book:

  • Weight downward: \vec{W} with magnitude W = mg
  • Normal force upward from the table: \vec{N}

Choose upward as positive y.

Newton’s First Law condition:

\sum F_y = 0

So:

N - mg = 0

Therefore:

N = mg

A common misconception is “the normal force always equals mg.” Here it does—but only because there is no vertical acceleration and the only vertical forces are N and mg.

Example 2: Constant-speed towing with friction

A crate is pulled across a floor at constant speed by a horizontal rope tension T. Kinetic friction magnitude is f_k.

Constant speed means \vec{a} = \vec{0}, so:

\sum F_x = 0

The horizontal forces are T (forward) and f_k (backward), so:

T - f_k = 0

Thus:

T = f_k

The important First Law idea: constant speed does not imply “no forces,” it implies “balanced forces.”

Exam Focus
  • Typical question patterns:
    • Objects at rest or moving at constant velocity: set up equilibrium equations \sum F_x = 0 and \sum F_y = 0.
    • Conceptual questions distinguishing “no force” vs “no net force.”
    • Identifying whether a situation is inertial and whether Newton’s laws apply directly.
  • Common mistakes:
    • Assuming “motion implies a force in the direction of motion.” Instead, ask whether velocity is changing.
    • Automatically setting N = mg even when there are other vertical forces or vertical acceleration.
    • Mixing up “zero acceleration” with “zero velocity.”

Newton's Second Law

What the Second Law says (the core of dynamics)

Newton’s Second Law is the quantitative rule that connects forces to acceleration:

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

Here:

  • \sum \vec{F} is the net force vector (the vector sum of all external forces on the object)
  • m is the object’s mass
  • \vec{a} is the object’s acceleration vector

This law is what turns a force description into a motion prediction. If you can model the forces, you can solve for acceleration; if you know acceleration, you can connect to kinematics.

Why “net force” and “vector” matter

Forces have direction, so you must add them as vectors. The net force could be zero even when multiple forces act (like N and mg on a resting book). Also, acceleration points in the direction of the net force, not necessarily the direction of motion.

A helpful way to think about it:

  • Net force determines how velocity changes.
  • Mass tells you how much acceleration you get from a given net force.

Component form (how you actually solve problems)

Most problems become manageable when you choose axes and write the Second Law in components:

\sum F_x = ma_x

\sum F_y = ma_y

In AP Physics C, choosing axes strategically is a major skill. For example, on an incline, it’s often best to choose x along the plane and y perpendicular to it. That choice reduces the number of force components you must compute.

Force models you’ll use with the Second Law

Newton’s Second Law is not a “plug-and-chug” formula unless you have correct force models. The most common forces in Unit 2 contexts include:

  • Weight: gravitational force near Earth’s surface with magnitude mg, direction downward.

\vec{W} = m\vec{g}

  • Normal force \vec{N}: contact force perpendicular to a surface. Its magnitude adjusts to enforce the constraint (no interpenetration), so it is not automatically mg.
  • Tension \vec{T}: pulling force along a rope/string. In idealized problems, ropes are massless and inextensible, so tension is the same throughout a single rope segment.
  • Friction:
    • Static friction magnitude satisfies 0 \le f_s \le \mu_s N and adjusts as needed up to a maximum.

f_s \le \mu_s N

  • Kinetic friction magnitude is f_k = \mu_k N and opposes relative motion.

f_k = \mu_k N

  • Spring force (if present): for an ideal spring, Hooke’s law magnitude is proportional to displacement.

F_s = kx

Direction is opposite the displacement from equilibrium.

These models are where many errors occur: students often treat friction as always \mu N (not true for static friction), or assume tension always equals weight (only in special equilibrium cases).

Example 1: Elevator scale reading (apparent weight)

A person of mass m stands on a scale in an elevator accelerating upward with acceleration magnitude a.

Conceptual goal: The scale reads the normal force N, not mg. If the elevator accelerates, the net force is not zero.

Forces on the person:

  • Upward N from the scale
  • Downward mg

Take upward as positive. Apply Newton’s Second Law:

\sum F_y = ma

So:

N - mg = ma

Solve:

N = m(g + a)

Interpretation: when accelerating upward, the scale reads heavier than mg. If accelerating downward with magnitude a (so a_y = -a), you’d get:

N = m(g - a)

This is a classic place where the First Law intuition fails—your “weight feeling” changes because net force is changing your motion.

Example 2: Block on an incline (choosing axes wisely)

A block of mass m slides down a frictionless incline at angle \theta above horizontal.

Conceptual setup: The forces are weight mg downward and normal force N perpendicular to the plane. Because the plane constrains the motion, the acceleration perpendicular to the plane is zero.

Choose axes:

  • x along the plane (down the plane)
  • y perpendicular to plane (out of the plane)

Resolve weight into components:

W_x = mg\sin\theta

W_y = mg\cos\theta

Newton’s Second Law along y: no acceleration perpendicular to plane, so a_y = 0.

\sum F_y = 0

Forces in y: N outward minus mg\cos\theta into the plane.

N - mg\cos\theta = 0

So:

N = mg\cos\theta

Now along x:

\sum F_x = ma_x

Only mg\sin\theta acts along the plane (frictionless), so:

mg\sin\theta = ma_x

Thus:

a_x = g\sin\theta

A frequent mistake is using mg\cos\theta for the downhill component; it’s actually the perpendicular component.

Exam Focus
  • Typical question patterns:
    • Given forces (or a situation), write component equations and solve for a, N, T, friction, or applied force.
    • Inclined plane setups with axis choice and weight components.
    • “Apparent weight” problems (elevators, vertical circular motion later) where N \ne mg.
  • Common mistakes:
    • Forgetting that \sum \vec{F} is the sum of external forces on the chosen object only.
    • Treating static friction as f_s = \mu_s N automatically rather than as a variable up to a maximum.
    • Writing one equation F = ma without separating into perpendicular directions when constraints imply a_y = 0.

Newton's Third Law

What the Third Law says (forces come in pairs)

Newton’s Third Law states: when object A exerts a force on object B, object B exerts a force on object A that is equal in magnitude and opposite in direction.

If A exerts \vec{F}_{A \to B} on B, then B exerts \vec{F}_{B \to A} on A such that:

\vec{F}_{A \to B} = -\vec{F}_{B \to A}

The forces act on different objects. That single sentence prevents most Third Law errors.

Why it matters: interaction forces and system thinking

The Third Law is what makes forces fundamentally about interactions, not properties an object “has.” It also becomes powerful when you switch from analyzing one object to analyzing a system:

  • If you include both interacting objects in your system, Third Law pairs become internal forces and cancel in the net external force.
  • That is why, for example, you can treat a person pushing a cart as a system and ignore the internal push-pull between person and cart when computing the system’s acceleration due to external forces.

How to correctly identify Third Law pairs

A Third Law pair must satisfy all of these:

  • Same interaction (same “contact” or same “gravitational attraction,” etc.)
  • Equal magnitude, opposite direction
  • Act on different objects
  • Occur simultaneously

Common Third Law interaction types:

  • Contact: normal force pair between two surfaces
  • Tension: rope pulling on object and object pulling on rope
  • Gravity: Earth pulls on object and object pulls on Earth

What Third Law pairs are NOT

Students often think forces that “balance” in equilibrium are Third Law pairs. They are usually not.

Example: a book rests on a table.

  • Upward N on the book and downward mg on the book are not a Third Law pair because they act on the same object.
  • The Third Law pair to N (force of table on book) is the force of book on table.
  • The Third Law pair to mg (Earth on book) is the gravitational force of book on Earth.

This distinction matters because if you accidentally pair forces on the same object, you’ll incorrectly conclude \sum \vec{F} = 0 in situations where the object is actually accelerating.

Example 1: Pushing a wall (why you feel a push back)

You push horizontally on a wall with force magnitude F. The wall pushes back on you with the same magnitude in the opposite direction.

  • Force you exert on wall: \vec{F}_{you \to wall}
  • Force wall exerts on you: \vec{F}_{wall \to you}

These are a Third Law pair:

\vec{F}_{you \to wall} = -\vec{F}_{wall \to you}

If the wall doesn’t move, that doesn’t mean “no forces exist.” It means the wall (and building, and Earth) provide other forces so the wall’s net acceleration is essentially zero.

Example 2: Two blocks in contact (action-reaction inside a system)

Block 1 pushes Block 2 on a frictionless surface. Suppose Block 1 exerts a contact force of magnitude C on Block 2 to the right.

Then Block 2 exerts a contact force of magnitude C on Block 1 to the left. These are Third Law partners.

If you analyze Block 2 alone, that contact force C may be the reason Block 2 accelerates. If you analyze the two-block system, those internal contact forces cancel, and only external forces (like an applied push from outside the system) determine the system acceleration.

This “single object vs system” flexibility is a major AP skill: you choose the viewpoint that makes the algebra simplest.

Exam Focus
  • Typical question patterns:
    • Identify the correct action-reaction pair for a given force (especially normal, tension, and gravity).
    • Multi-object systems where internal Third Law forces cancel when treating a combined system.
    • Conceptual questions contrasting Third Law pairs with balanced forces on one object.
  • Common mistakes:
    • Claiming N and mg on the same object are a Third Law pair.
    • Thinking action-reaction forces “cancel” to make acceleration zero (they don’t, because they’re on different objects).
    • Forgetting to label forces with “agent on object” (which helps you pair them correctly).

Free-Body Diagrams

What a free-body diagram is (and why it’s essential)

A free-body diagram (FBD) is a simplified sketch that shows all external forces acting on a chosen object (or chosen system). “Free” means you conceptually isolate the object from its surroundings and replace interactions with forces.

Free-body diagrams matter because Newton’s Second Law requires the net force on the object, and the only reliable way to get that net force is to:

  1. Choose the object (or system)
  2. List and draw every external force acting on it
  3. Choose axes
  4. Write \sum F_x = ma_x and \sum F_y = ma_y using your diagram

If your FBD is missing a force or includes an extra one, the algebra will faithfully produce the wrong answer.

What to include: external forces only

A good FBD includes:

  • Forces from interactions with other objects (contact forces, tension, friction)
  • Long-range forces (weight)

A good FBD does not include:

  • Forces the object exerts on other things (those belong on other objects’ FBDs)
  • “Motion forces” like “force of velocity” or “force in the direction of motion” (these don’t exist as separate forces)
  • Both members of a Third Law pair on the same object

The most common forces and how to draw them

You’ll see the same few forces repeatedly; learning their direction rules is half the battle.

Weight

Weight acts toward Earth’s center (downward in typical problems). Magnitude near Earth’s surface is mg.

Draw \vec{W} straight down from the object.

Normal force

Normal force is perpendicular to the contact surface and points away from the surface. It adjusts in magnitude based on constraints.

Draw \vec{N} perpendicular to the surface (not necessarily upward if the surface is inclined).

Tension

Tension points along the rope, pulling away from the object. A rope can’t push in the ideal model.

Draw \vec{T} along the rope direction.

Friction

Friction acts along the surface, opposing relative motion (or impending relative motion).

  • If the object is sliding right relative to the surface, kinetic friction on the object points left.
  • If the object is not sliding, static friction points in the direction needed to prevent slipping (up to its maximum).

This is a common misconception: friction does not always oppose the applied force; it opposes the relative motion (or tendency of motion) between surfaces.

Choosing axes: make the math match the physics

Because Newton’s Second Law is easiest in components, you should choose axes to reduce component work:

  • On horizontal surfaces, use x horizontal and y vertical.
  • On inclines, use x along the incline and y perpendicular.

Then only resolve forces that are not aligned with your axes (often just weight on an incline). This turns a messy vector sum into two clean scalar equations.

Systems and “internal forces disappear” (when you choose a system FBD)

Sometimes drawing an FBD for a single object forces you to include internal interaction forces that you don’t actually care about. If you instead choose a system containing multiple objects, forces between objects in the system become internal and cancel.

Example: two blocks connected by a rope pulled by an external force. If you choose both blocks as the system, the rope tension between them is internal and won’t appear on the system FBD. The system acceleration depends only on external forces and the total mass.

This is not a trick; it’s a direct payoff of Newton’s Third Law.

Worked problem: Block pulled at an angle (FBD to equations)

A block of mass m is pulled across a horizontal surface by a force of magnitude F at angle \theta above the horizontal. The coefficient of kinetic friction is \mu_k. Find the acceleration.

Step 1: Draw the FBD (describe it clearly)

Forces on the block:

  • Weight mg downward
  • Normal force N upward
  • Applied force components: horizontal component to the right and vertical component upward
  • Kinetic friction f_k to the left (opposes motion)

Resolve applied force:

F_x = F\cos\theta

F_y = F\sin\theta

Step 2: Use vertical equation to find N

There is no vertical acceleration (block stays on surface), so a_y = 0:

\sum F_y = 0

Upward forces: N and F\sin\theta. Downward: mg.

N + F\sin\theta - mg = 0

So:

N = mg - F\sin\theta

This shows why N is not always mg: the upward pull partially “unloads” the block.

Step 3: Write friction using the correct normal force

Kinetic friction magnitude:

f_k = \mu_k N

So:

f_k = \mu_k(mg - F\sin\theta)

Step 4: Apply Newton’s Second Law horizontally

Horizontal net force produces a_x:

\sum F_x = ma_x

Rightward: F\cos\theta. Leftward: f_k.

F\cos\theta - f_k = ma_x

Substitute friction:

F\cos\theta - \mu_k(mg - F\sin\theta) = ma_x

Solve for acceleration:

a_x = \frac{F\cos\theta - \mu_k(mg - F\sin\theta)}{m}

This problem is a perfect example of why the FBD is not optional: if you incorrectly set N = mg, you get the wrong friction, which gives the wrong acceleration.

Worked problem: Static friction adjusts (don’t set it equal to \mu_s N too early)

A block rests on a horizontal surface. You apply a horizontal force F, but the block does not move. The coefficient of static friction is \mu_s.

Key concept: static friction matches what’s needed to prevent motion, up to its maximum.

Horizontal forces on the block:

  • Applied force F to the right
  • Static friction f_s to the left

Since the block is at rest, a_x = 0:

\sum F_x = 0

So:

F - f_s = 0

Thus:

f_s = F

But this is only possible if static friction is not being exceeded:

f_s \le \mu_s N

On a horizontal surface with no other vertical forces, N = mg, so the no-slip condition is:

F \le \mu_s mg

If F exceeds \mu_s mg, the block starts to slide and you must switch to kinetic friction f_k = \mu_k N.

Exam Focus
  • Typical question patterns:
    • “Draw the free-body diagram” then write component Newton’s Second Law equations.
    • Incline or pulley setups where a correct axis choice and correct force directions determine the whole solution.
    • Friction questions that test the difference between f_s and f_k and the meaning of f_s \le \mu_s N.
  • Common mistakes:
    • Including both action and reaction forces on the same FBD (for example, drawing “force of block on table” on the block’s diagram).
    • Drawing friction in the wrong direction (it opposes relative motion or impending slip, not necessarily the applied force).
    • Setting f_s = \mu_s N automatically instead of first solving for the required static friction and then checking whether it exceeds the maximum.