AP Physics C: Mechanics — Unit 2 Newton’s Laws (Concepts, Methods, and Worked Problems)

Forces and Free-Body Diagrams (FBDs)

What a force is (and what it is not)

A force is an interaction between objects that can change an object’s motion (its velocity) or shape. In AP Physics C: Mechanics, you treat forces as vectors—they have both magnitude and direction—and you model how they combine to produce acceleration.

Two big ideas keep you grounded:

1. Forces always come from interactions with other objects (or fields like gravity). If you cannot identify “what is pushing or pulling,” you probably have not finished the physical story.
2. Motion does not “need a force” to keep going. Forces are linked to changes in motion (acceleration), not motion itself.

Force types: contact vs. long-range

You will mostly encounter:

• Contact forces: require physical contact (normal force, friction, tension, spring force).
• Long-range forces: act without contact through a field (gravity is the main one in Unit 2).

This classification matters because it guides your free-body diagram: contact forces come from surfaces/ropes/springs that touch the object; long-range forces come from fields that reach the object.

The purpose of a free-body diagram

A free-body diagram (FBD) is a simplified sketch that shows only one object (or one chosen system) and all external forces acting on it. The point of an FBD is not artistry—it is bookkeeping. A correct FBD turns a messy real situation into a clean application of Newton’s laws.

A great mental model: an FBD is a “force inventory.” If your inventory is wrong or incomplete, your equations will be wrong even if your algebra is perfect.

How to draw an FBD correctly (a reliable procedure)

When you draw an FBD, follow a consistent process:

1. Choose the object (or system) you are analyzing. Draw it as a dot or simple box.
2. List all interactions with the environment: surfaces, ropes, springs, Earth (gravity), etc.
3. Draw one force vector for each interaction acting on the chosen object.
4. Label forces by type and agent when helpful (for example, “normal from table,” “tension from rope”).
5. Choose coordinate axes that simplify component breakdown (often along/perpendicular to an incline or along acceleration).
6. Only after the diagram is correct: write Newton’s second law in components.

Common “force inventory” mistakes to avoid

• Including forces the object exerts on something else. An FBD includes only forces on the object.
• Drawing “motion force” or “force of velocity.” There is no such force.
• Confusing the normal force with weight. Weight is gravitational; normal is a contact force.
• Assuming friction always equals \mu N. That is only true for kinetic friction, and for static friction it is an upper bound.

Worked example 1: Block resting on a horizontal table

A block sits at rest on a table.

Physical reasoning (before equations): At rest means acceleration is zero. So the net force must be zero. The only interactions are gravity (Earth pulls down) and the table (pushes up).

FBD:

• Weight downward: F_g = mg
• Normal force upward: N

Newton’s second law in the vertical direction:

\sum F_y = ma_y

Since a_y = 0:

N - mg = 0

So:

N = mg

The key idea is not the algebra—it is the logic: the normal force adjusts to satisfy the no-acceleration condition.

Worked example 2: Block on an incline (no friction)

A block of mass m rests on a frictionless incline of angle \theta.

Why axes choice matters: If you keep axes horizontal/vertical, you will split the normal force into components. That is possible, but it is unnecessary work. Instead, align axes parallel and perpendicular to the plane.

FBD:

• Weight mg straight down.
• Normal force N perpendicular to the plane.

Resolve weight into components relative to the incline:

F_{g,\parallel} = mg\sin\theta

F_{g,\perp} = mg\cos\theta

Perpendicular direction (no acceleration into the surface):

\sum F_{\perp} = ma_{\perp}

N - mg\cos\theta = 0

N = mg\cos\theta

Parallel direction (the block accelerates down the plane):

\sum F_{\parallel} = ma_{\parallel}

mg\sin\theta = ma

a = g\sin\theta

Exam Focus

• Typical question patterns:
  • “Draw an FBD for … then write Newton’s second law in components.”
  • “Choose axes to simplify an incline/pulley situation and solve for acceleration/tension.”
  • “Given a motion description, identify which forces must exist and which can be zero.”
• Common mistakes:
  • Mixing forces and components: drawing both mg and mg\sin\theta as separate forces.
  • Putting friction in the wrong direction (friction opposes relative slipping or attempted slipping).
  • Forgetting that an object can have zero acceleration while still having forces (balanced forces).

Newton’s First Law and Inertial Frames

The law and its meaning

Newton’s First Law says that if the net external force on an object is zero, the object’s velocity remains constant. Constant velocity includes the special case of being at rest.

This law is fundamentally about inertial frames: reference frames in which objects with zero net force do not accelerate. In AP Physics C, you almost always work in an approximately inertial frame (the lab frame near Earth is close enough for most problems).

Why it matters

Newton’s First Law teaches you a powerful diagnostic:

• If you observe no acceleration, then the net force must be zero.
• If you observe acceleration, then the net force is not zero.

This is the conceptual backbone behind equilibrium problems, “terminal speed” reasoning in later units, and many multiple-choice conceptual questions.

Equilibrium is a physics statement, not a feeling

When an object is in equilibrium, it means:

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

Equilibrium does not mean “nothing is happening.” An object moving at constant velocity is also in equilibrium.

You will often see two flavors:

• Static equilibrium: velocity is zero.
• Dynamic equilibrium: velocity is constant but not zero.

In both cases, acceleration is zero.

Common misconception: “If it’s moving, there must be a force forward”

If you slide a puck on nearly frictionless ice, it keeps moving even after you stop pushing. The forward push was needed to change its velocity (speed it up), not to maintain it. In everyday life, friction usually provides a backward force, so you often do need a forward force to maintain constant speed—but that is because you are canceling friction, not because motion itself requires force.

Worked example: Constant-speed elevator

An elevator moves upward at constant speed.

Key idea: Constant speed means acceleration is zero.

Forces on the elevator car (treat the car as the object):

• Tension upward: T
• Weight downward: mg

Newton’s second law (or first law logic):

\sum F_y = ma_y

T - mg = 0

So:

T = mg

Even though it is moving upward, the net force is zero.

Exam Focus

• Typical question patterns:
  • “An object moves at constant velocity; what can you conclude about the net force?”
  • “A force is removed; predict qualitatively how motion changes (or does not).”
  • “Identify whether a situation is equilibrium and set up equations accordingly.”
• Common mistakes:
  • Treating constant velocity as if it implies a net force in the direction of motion.
  • Confusing “no net force” with “no forces.”
  • Forgetting that equilibrium can occur while moving.

Newton’s Second Law: Turning Forces into Equations of Motion

The core statement

Newton’s Second Law relates the net external force on an object to its acceleration:

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

Here:

• \sum \vec{F} is the vector sum of all external forces on the object.
• m is the inertial mass (resistance to changes in motion).
• \vec{a} is the acceleration vector.

Why it matters

This is the main engine of Unit 2: once you can translate a physical situation into forces, Newton’s second law gives you the acceleration. From acceleration you can connect to kinematics (Unit 1 tools) when needed.

Newton’s second law is also how you solve most multi-object systems: it provides one equation per object (or per chosen system), and the constraints (like a rope enforcing the same acceleration magnitude) close the system.

How it works in practice: components, not just vectors

Most problems are easiest when you break Newton’s second law into perpendicular directions. For a chosen coordinate system:

\sum F_x = ma_x

\sum F_y = ma_y

A major skill in AP Physics C is choosing axes that make one of these equations simple (for example, aligning an axis with the direction of acceleration).

Strategy for solving Newton’s second law problems

A consistent approach prevents many errors:

1. Identify the object(s) or system.
2. Draw an FBD for each object/system.
3. Choose axes and resolve forces into components.
4. Write \sum F = ma in each relevant direction.
5. Add constraint equations (rope constraints, “no motion perpendicular to surface,” etc.).
6. Solve algebraically, then check whether the direction of friction/tension assumptions were consistent.

Worked example 1: Pulling a block with a force at an angle

A block of mass m is pulled on 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.

Conceptual setup: The applied force has components: one horizontal (helps motion) and one vertical (reduces the normal force). Kinetic friction depends on the normal force, so the vertical component matters even if there is no vertical acceleration.

Resolve applied force:

F_x = F\cos\theta

F_y = F\sin\theta

Vertical direction: no vertical acceleration, so a_y = 0.

Forces in vertical:

• Normal up: N
• Weight down: mg
• Applied vertical component up: F\sin\theta

Newton’s second law vertically:

N + F\sin\theta - mg = 0

So:

N = mg - F\sin\theta

Kinetic friction magnitude:

f_k = \mu_k N

So:

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

Horizontal direction:

\sum F_x = ma

F\cos\theta - f_k = ma

Substitute friction:

F\cos\theta - \mu_k\left(mg - F\sin\theta\right) = ma

Solve for acceleration:

a = \frac{F\cos\theta - \mu_k\left(mg - F\sin\theta\right)}{m}

Sanity check: If \theta increases, the upward component increases, normal decreases, friction decreases, so acceleration should increase. The expression matches that trend.

Worked example 2: Incline with friction (finding acceleration)

A block slides down an incline of angle \theta with coefficient of kinetic friction \mu_k.

Parallel to incline: gravity component down the slope, friction up the slope.

Normal force (perpendicular equilibrium):

N = mg\cos\theta

Kinetic friction:

f_k = \mu_k mg\cos\theta

Newton’s second law along the incline (take downhill as positive):

mg\sin\theta - f_k = ma

Substitute:

mg\sin\theta - \mu_k mg\cos\theta = ma

So:

a = g\left(\sin\theta - \mu_k\cos\theta\right)

This shows how friction “steals” some of the downhill component.

Exam Focus

• Typical question patterns:
  • “Write Newton’s second law in components and solve for acceleration.”
  • “Given acceleration data, solve backward for an unknown force or coefficient of friction.”
  • “Two-dimensional motion: find components of acceleration from net force components.”
• Common mistakes:
  • Writing \sum F = ma with the wrong sign convention (then interpreting a negative result incorrectly).
  • Forgetting that friction depends on the normal force, which may not equal mg.
  • Treating mass as a force (mass is not measured in newtons).

The Main Force Models You Must Master

Newton’s laws are only as useful as your force models. Unit 2 heavily emphasizes knowing what each common force represents physically, when it applies, and how to express it quantitatively.

Weight (gravitational force near Earth)

Weight is the gravitational force exerted by Earth on an object near the surface.

Magnitude:

F_g = mg

Direction: toward the center of Earth (downward in most problems).

Why it matters: weight is one of the most universal forces in mechanics problems, and it is easy to misuse because people casually say “weight” when they mean “mass.” In physics, mass and weight are different.

Normal force

The normal force is a contact force a surface exerts on an object, perpendicular to the surface.

Key properties:

• It is not automatically equal to mg.
• It adjusts based on other forces and constraints.
• It is perpendicular to the surface (not necessarily “up”).

Example idea: If you push down on a block, the normal force increases. If you pull up on it, the normal force decreases. If the block loses contact, the normal force becomes zero.

Tension

Tension is the pulling force transmitted through a rope, string, or cable.

In many AP problems you assume:

• The rope is massless and does not stretch.
• The pulley is massless and frictionless.

Under those assumptions:

• The tension is the same throughout a single continuous rope segment.
• The rope constraint can enforce equal-magnitude accelerations of connected masses.

A subtle but important point: tension is not “whatever you want.” It is an unknown that comes from applying Newton’s laws plus constraints.

Friction: static vs. kinetic

Friction is a contact force parallel to the surface that opposes relative motion (or attempted relative motion) between surfaces.

Kinetic friction

When surfaces are sliding relative to each other, kinetic friction magnitude is:

f_k = \mu_k N

Direction: opposite the direction of relative motion.

Static friction

When surfaces do not slide, static friction adjusts as needed up to a maximum:

f_s \le \mu_s N

The maximum possible static friction is:

f_{s,\max} = \mu_s N

Direction: opposite the direction the object would begin to slip if static friction were absent.

The most common mistake is setting f_s = \mu_s N automatically. That is only true at the threshold of slipping.

Spring force (Hooke’s law)

A spring exerts a restoring force that tries to bring it back to its natural (unstretched) length.

Hooke’s law magnitude:

F_s = kx

where:

• k is the spring constant.
• x is the displacement from equilibrium length (stretch or compression).

Direction: opposite the displacement (restoring). If you stretch to the right, the spring pulls left.

In Unit 2, springs often appear as forces in Newton’s second law rather than energy objects (energy methods are later).

Drag / air resistance (usually qualitative in this unit)

Air resistance depends on speed and shape; common models include proportional to speed or proportional to speed squared. In many AP Newton’s-law setups, drag is introduced qualitatively:

• Drag acts opposite velocity.
• Drag increases as speed increases.
• Terminal speed occurs when drag balances weight (net force zero).

Unless a specific drag law is provided, do not invent one.

Worked example 1: Static friction that is not at its maximum

A block of mass m rests on a horizontal surface. You apply a small horizontal force F, and the block does not move.

Reasoning: No motion means acceleration is zero, so net force is zero. Static friction must match the applied force (up to its maximum).

Horizontal equilibrium:

F - f_s = 0

So:

f_s = F

But static friction can only do this if:

F \le \mu_s N

Since vertically N = mg for this case, the “no slip” condition is:

F \le \mu_s mg

This example captures the key idea: static friction is a responsive force, not a fixed value.

Worked example 2: Spring pulling a block on frictionless surface

A block of mass m attached to a spring is displaced by distance x from equilibrium on a frictionless horizontal surface. What is its acceleration at that instant?

Reasoning: The only horizontal force is the spring restoring force of magnitude kx toward equilibrium.

Newton’s second law (take positive direction away from equilibrium so the restoring force is negative):

\sum F_x = ma

-kx = ma

So:

a = -\frac{k}{m}x

The negative sign encodes the restoring nature: acceleration points back toward equilibrium.

Exam Focus

• Typical question patterns:
  • “Determine normal force on inclines/elevators, then compute friction.”
  • “Static vs kinetic friction: decide which applies from the motion description.”
  • “Use Hooke’s law with Newton’s second law to relate displacement and acceleration.”
• Common mistakes:
  • Assuming N = mg in situations with vertical components of applied force or vertical acceleration.
  • Setting f_s = \mu_s N even when the object is clearly not at impending motion.
  • Giving spring force the wrong direction (forgetting it is restoring).

Multi-Object Systems: Ropes, Pulleys, and the Idea of a “System”

Why multi-object problems feel harder (and how to make them manageable)

Connected-object problems are not fundamentally new physics—they are mostly accounting challenges.

The key additional concept is a constraint: a relationship between motions imposed by a connection like an inextensible rope. Constraints reduce the number of independent accelerations.

Another key concept is choosing whether to analyze each object separately or combine them into a single system.

Internal vs. external forces (and why system choice helps)

If you choose a system consisting of multiple objects, then forces between those objects become internal forces. Internal forces often cancel in the system’s net force accounting, leaving only external forces.

This can simplify problems dramatically—especially when tension forces are otherwise unknown.

A useful rule: if your question asks for the acceleration of the whole connected set, a system approach can eliminate internal tensions. If your question asks for tension, you usually must analyze at least one object separately.

Rope constraints (typical AP assumptions)

Under the common AP idealizations (massless rope, frictionless pulley, no stretching):

• Connected masses often share the same acceleration magnitude.
• Tension is uniform along a single rope segment.

Be careful: if a pulley is movable or there are multiple rope segments supporting a mass, the acceleration and tension relationships can be different. Many AP problems stick to fixed pulleys, but you should still think rather than memorize.

Worked example 1: Atwood machine (two hanging masses)

Two masses m_1 and m_2 hang over a frictionless, massless pulley. Assume m_2 > m_1, so m_2 accelerates down and m_1 accelerates up with magnitude a.

Step 1: FBDs
For m_1 (accelerating up):

• Tension up: T
• Weight down: m_1 g

For m_2 (accelerating down):

• Weight down: m_2 g
• Tension up: T

Step 2: Newton’s second law for each mass
Take up as positive for m_1:

T - m_1 g = m_1 a

Take down as positive for m_2:

m_2 g - T = m_2 a

Step 3: Solve
Add the equations to eliminate T:

m_2 g - m_1 g = m_1 a + m_2 a

Factor:

(m_2 - m_1)g = (m_1 + m_2)a

So:

a = \frac{(m_2 - m_1)g}{m_1 + m_2}

Now solve for tension using T - m_1 g = m_1 a:

T = m_1 g + m_1 a

Substitute expression for a:

T = m_1 g + m_1\left(\frac{(m_2 - m_1)g}{m_1 + m_2}\right)

You can simplify further, but the main physics is already captured.

Concept check: If m_1 = m_2, then a = 0, as expected.

Worked example 2: Block on table connected to hanging mass (with friction)

A block of mass m_1 on a horizontal table is connected by a rope over a pulley to a hanging mass m_2. The coefficient of kinetic friction between m_1 and the table is \mu_k. If the system moves with m_2 downward, find the acceleration.

Step 1: FBD for m_1 (on table)
Horizontal forces:

• Tension pulling toward pulley: T
• Kinetic friction opposing motion: f_k = \mu_k N

Vertical forces:

• Normal N up
• Weight m_1 g down

Since no vertical acceleration for m_1:

N - m_1 g = 0

So:

N = m_1 g

Thus:

f_k = \mu_k m_1 g

Newton’s second law horizontally for m_1 (take direction toward pulley as positive):

T - \mu_k m_1 g = m_1 a

Step 2: FBD for m_2 (hanging)
Forces:

• Weight down: m_2 g
• Tension up: T

Newton’s second law for m_2 (take downward as positive):

m_2 g - T = m_2 a

Step 3: Solve
Add the two equations:

m_2 g - \mu_k m_1 g = m_1 a + m_2 a

So:

a = \frac{m_2 g - \mu_k m_1 g}{m_1 + m_2}

Interpretation: Friction reduces the driving force; if m_2 g is too small compared to \mu_k m_1 g, the computed a would be negative, signaling that your assumed direction of motion is not consistent.

Exam Focus

• Typical question patterns:
  • “Two masses connected by a rope: find acceleration and tension.”
  • “Decide whether the system moves; compare maximum static friction to the driving force.”
  • “Use a system approach to eliminate internal tension, then solve.”
• Common mistakes:
  • Assigning different acceleration magnitudes to masses connected by a single taut rope (under ideal assumptions).
  • Mixing sign conventions between objects (define positive directions clearly for each).
  • Forgetting to check whether motion is possible given static friction limits.

Newton’s Third Law: Action-Reaction Pairs and Contact Interactions

The law and its precise meaning

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

This is about interactions between two different objects. The force pair acts on different bodies.

Why it matters (beyond memorization)

Newton’s third law prevents a very common reasoning trap: thinking that “the stronger object exerts a bigger force.” In reality, during an interaction, the forces are equal and opposite; what differs is how those forces affect each object’s motion because the masses (and thus accelerations) may differ.

This law is also the conceptual foundation for:

• Understanding normal force and friction as interaction forces.
• Explaining propulsion (walking, car tires pushing on the road, rockets pushing exhaust).
• Avoiding the incorrect idea that action-reaction pairs “cancel” in a single object’s FBD.

How to correctly identify third-law pairs

A third-law pair must satisfy all of the following:

• Same interaction type (both contact, or both gravitational, etc.).
• Same magnitude, opposite direction.
• Act on different objects.
• Occur simultaneously.

A helpful naming scheme: “force on B by A” pairs with “force on A by B.”

Common confusion: “If forces are equal and opposite, why does anything accelerate?”

Because the equal-and-opposite forces are on different objects. Acceleration depends on the net force on a single object.

Example: When you push a cart, your hands push the cart forward, and the cart pushes your hands backward with equal magnitude. The cart can still accelerate because the forces on the cart do not necessarily cancel; the cart may have a net forward force if friction is small.

Worked example 1: Block on a table (normal force pair)

A block rests on a table.

Forces on the block:

• Weight: Earth on block: mg downward.
• Normal: table on block: N upward.

Third-law pairs:

• The pair to N (table on block) is “block on table,” an equal and opposite force downward exerted by the block on the table.
• The pair to mg (Earth on block) is “block on Earth,” an equal and opposite gravitational force upward exerted by the block on Earth.

Notice that none of these partner forces belong on the block’s FBD—they act on other objects.

Worked example 2: Two skaters pushing off

Two skaters initially at rest push off each other on frictionless ice. Skater 1 has mass m_1 and skater 2 has mass m_2.

During the push:

• Skater 1 exerts a force on skater 2.
• Skater 2 exerts an equal-magnitude opposite force on skater 1.

So the interaction forces are equal, but accelerations satisfy:

a_1 = \frac{F}{m_1}

a_2 = \frac{F}{m_2}

The lighter skater has the larger acceleration.

Exam Focus

• Typical question patterns:
  • “Identify the Newton’s third-law pair corresponding to a given force.”
  • “Explain why two objects exert equal forces but have different accelerations.”
  • “Distinguish between forces that cancel (on one object) and third-law pairs (on different objects).”
• Common mistakes:
  • Putting both third-law forces on the same FBD.
  • Thinking the heavier object exerts a larger force in an interaction.
  • Confusing balanced forces (net force zero on one object) with action-reaction pairs.

Circular Motion as a Newton’s Laws Problem (Centripetal Force is Not a New Force)

The key idea: acceleration can happen without speeding up

In circular motion at constant speed, velocity changes direction continuously, so there is acceleration even though speed is constant. That acceleration points toward the center of the circle.

The magnitude of centripetal acceleration is:

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

where:

• v is speed.
• r is radius of curvature.

Why it matters

Many students think “centripetal force” is a separate force. In Newton’s-law language, centripetal force is the net inward force required to produce centripetal acceleration.

So you do not add a new arrow labeled “centripetal force” unless you explicitly mean “net radial force.” Instead, you identify real forces (tension, gravity, normal, friction) and set their inward components equal to mv^2/r.

Newton’s second law in the radial direction

Choose a radial axis pointing inward (toward the center). Then:

\sum F_{\text{radial}} = m\frac{v^2}{r}

This single equation is the workhorse for most circular-motion force problems.

Worked example 1: Car rounding a flat curve (friction provides centripetal force)

A car of mass m turns on a flat curve of radius r at speed v. The coefficient of static friction between tires and road is \mu_s. What is the maximum speed before slipping?

Conceptual setup: On a flat curve, the only horizontal force available is static friction, directed toward the center. If the required centripetal force exceeds the maximum static friction, the car slips outward.

Maximum static friction:

f_{s,\max} = \mu_s N

On level ground, vertical equilibrium gives N = mg, so:

f_{s,\max} = \mu_s mg

Required centripetal force:

F_c = m\frac{v^2}{r}

At the threshold of slipping:

m\frac{v^2}{r} = \mu_s mg

Cancel m and solve:

v = \sqrt{\mu_s g r}

Interpretation: Larger radius, higher friction coefficient, and larger gravitational acceleration allow higher speed.

Worked example 2: Vertical circle with a mass on a string

A mass m moves in a vertical circle of radius r at speed v at the top of the circle. What tension is required in the string at the top?

Forces at the top: Both weight and tension point toward the center (downward).

Radial inward direction at the top is downward, so:

T + mg = m\frac{v^2}{r}

Solve for tension:

T = m\frac{v^2}{r} - mg

A physically important consequence: if v is too small, tension would need to be negative (impossible). The string would go slack. The “just taut” condition at the top is T = 0, giving:

m\frac{v^2}{r} = mg

So:

v = \sqrt{gr}

This is a classic Newton’s-law circular-motion result.

Exam Focus

• Typical question patterns:
  • “Set up radial force equation for a car/roller coaster/string and solve for speed or tension.”
  • “Decide which forces point toward the center at the top/bottom of a vertical circle.”
  • “Interpret whether contact is maintained (normal force or tension becoming zero).”
• Common mistakes:
  • Adding a separate “centripetal force” in addition to real forces.
  • Using a_c = v/r instead of v^2/r.
  • Getting the radial direction wrong (especially at the top of a vertical circle).

Apparent Weight, Accelerating Frames, and “g-Forces” (Newton’s Laws in Disguise)

Apparent weight is usually a normal force

In many problems, what a scale reads is the normal force the scale exerts on you. That reading is often called apparent weight.

This matters because when you accelerate, the normal force can differ from mg even though your gravitational weight (in the physics sense) is still mg near Earth.

Elevator reasoning: the cleanest way to think about it

Consider a person of mass m standing on a scale in an elevator. Forces on the person:

• Normal force from scale upward: N
• Weight downward: mg

Newton’s second law in the vertical direction:

N - mg = ma

This single equation handles all elevator cases once you define the sign of a.

• If the elevator accelerates upward, a > 0 and N > mg.
• If the elevator accelerates downward, a < 0 and N < mg.
• If the elevator moves at constant velocity, a = 0 and N = mg.

Worked example: Scale reading in an accelerating elevator

A person of mass m is in an elevator accelerating downward with magnitude a. Find the scale reading.

Choose upward positive, so acceleration is -a.

Newton’s second law:

N - mg = m(-a)

So:

N = m(g - a)

If a = g, then N = 0: the person is in free fall and the scale reads zero.

Pseudo-forces (when they appear and why you must be careful)

If you analyze motion from a non-inertial frame (an accelerating frame), Newton’s second law in the simple form \sum \vec{F} = m\vec{a} does not hold unless you introduce a pseudo-force (also called a fictitious force).

In AP Physics C, you can almost always avoid pseudo-forces by writing equations in an inertial frame. However, you should recognize the idea:

• In an accelerating car frame, objects seem to “want to slide backward.” In an inertial frame, the real explanation is that the object tends to maintain its velocity while the car accelerates forward.

If you do use a pseudo-force approach, it must be consistent: the pseudo-force on mass m is opposite the frame acceleration and has magnitude ma.

Worked example: Block against a wall in a horizontally accelerating truck

A truck accelerates to the right with acceleration magnitude a. A block is pressed against a vertical wall inside the truck. The coefficient of static friction between block and wall is \mu_s. What minimum acceleration prevents the block from sliding down?

Inertial-frame approach (recommended):

• The wall pushes on the block with a normal force N horizontally (to the right) to accelerate it with the truck.
• Static friction acts upward to prevent sliding.

Horizontal direction: the only horizontal force on the block is N, producing acceleration a:

N = ma

Vertical direction: for no vertical motion, acceleration is zero, so friction balances weight:

f_s = mg

But static friction is limited:

f_s \le \mu_s N

So:

mg \le \mu_s N

Substitute N = ma:

mg \le \mu_s ma

Cancel m:

g \le \mu_s a

Thus the minimum acceleration is:

a = \frac{g}{\mu_s}

This problem tests whether you can connect a horizontal Newton’s-law requirement (normal force needed for horizontal acceleration) to a vertical friction condition.

Exam Focus

• Typical question patterns:
  • “Elevator/scale problems: relate apparent weight to acceleration.”
  • “Identify when normal force differs from mg due to vertical acceleration.”
  • “Combine perpendicular directions: one direction sets normal force, another uses friction limit.”
• Common mistakes:
  • Calling mg the “scale reading” (the scale reads N, not mg).
  • Forgetting sign conventions for acceleration in vertical motion.
  • Using kinetic friction when the problem states no slipping (static friction applies).