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Module 2: Motion in Two and Three Dimensions

 

PHYS-2325 M2L3 Newton's Laws and Forces

"If I have seen further it is by standing on the shoulders of giants."
— Isaac Newton


Newton's three laws of motion are the governing rules of classical mechanics — they explain why objects move the way they do. While Modules 1 and 2 answered the what (describing motion with position, velocity, and acceleration vectors), this lesson answers the why: forces cause acceleration. You will learn to identify every force acting on an object, organize them in a free-body diagram, and apply Newton's second law, Fnet = ma, along each coordinate axis independently. You will then extend this framework to three real-world force types — friction, drag, and the centripetal force needed to maintain circular motion — gaining the tools to solve the full range of dynamics problems you will encounter throughout this course and in your engineering career.

A free-body diagram showing multiple force vectors acting on a block on an inclined ramp, with the net force arrow and coordinate axes labeled.
The free-body diagram — the physicist's essential tool for applying Newton's second law.

Course Competencies and Learning Objectives

A ★ indicates that this page contains content related to that LO.

CC2.1 Analyze the components of motion and any applied forces

★ LO2.3.1 State Newton's First, Second, and Third Laws and identify the conditions under which each applies

★ LO2.3.2 Draw a correct free-body diagram for an object subject to multiple forces in two dimensions

★ LO2.3.3 Apply Newton's second law component-by-component to solve for unknown forces or accelerations

★ LO2.3.4 Calculate static and kinetic friction forces using the friction model f = μN

★ LO2.3.5 Determine terminal speed from the drag force equation and interpret the forces on a falling object

★ LO2.3.6 Apply Newton's second law to uniform circular motion, identifying the net centripetal force

Required Reading

Click the blue buttons to go to the OpenStax reading assignments. Complete all six sections before watching the videos.

5.1 Newton's First and Second Laws 5.2 Some Particular Forces 5.3 Applying Newton's Laws 6.1 Friction 6.2 The Drag Force and Terminal Speed 6.3 Uniform Circular Motion

 

Optional Reading

Explore More

Want additional practice with free-body diagrams, friction, or circular motion? These free resources complement the reading and offer interactive tools.

Media

Work through the pre-lecture videos for the reading sections first, then watch the mini-lectures for detailed worked examples on each topic.

Video 1: Pre-Lecture — Newton's First and Second Laws

Pre-Lecture: Newton's First and Second Laws (5.1)

This video introduces Newton's First Law (the law of inertia) and Newton's Second Law (Fnet = ma) — the two foundational principles that relate force to the motion of an object. You will also see Newton's Third Law stated and illustrated with examples.

  • Newton's First Law: an object at rest stays at rest; an object in motion stays in motion unless acted on by a net external force
  • Inertia and mass — resistance to changes in motion
  • Newton's Second Law: Fnet = ma — vector equation, applies along each axis independently
  • Newton's Third Law: forces always come in equal-and-opposite action-reaction pairs

Time: Video pending upload to Canvas

Video 2: Pre-Lecture — Some Particular Forces

Pre-Lecture: Common Forces — Weight, Normal, Tension, Spring (5.2)

This video catalogs the most frequently encountered forces in mechanics: gravitational weight, normal force from a surface, tension in a rope or cable, and the spring force. Understanding each force's direction and magnitude is essential before drawing any free-body diagram.

  • Weight: w = mg, directed straight downward; magnitude w = mg
  • Normal force: perpendicular to the contact surface; adjusts to maintain equilibrium or produce a given acceleration
  • Tension: along a rope or cable, pulling toward the attachment point; same magnitude throughout a massless rope
  • Spring force: Fs = −kx (Hooke's Law) — opposes displacement from equilibrium

Time: Video pending upload to Canvas

Video 3: Pre-Lecture — Applying Newton's Laws

Pre-Lecture: Free-Body Diagrams and Newton's Second Law Strategy (5.3)

This video presents the systematic problem-solving strategy for Newton's second law: identify all forces, draw a free-body diagram, choose a coordinate system, write ΣF = ma along each axis, and solve. This workflow applies to every dynamics problem in the course.

  • Step 1: Identify the object (system) and all external forces acting on it
  • Step 2: Draw the free-body diagram — one arrow per force, labeled with symbol and direction
  • Step 3: Choose a coordinate system aligned with the acceleration when possible
  • Step 4: Write ΣFx = max and ΣFy = may; solve the system of equations

Time: Video pending upload to Canvas

Video 4: Pre-Lecture — Friction

Pre-Lecture: Static and Kinetic Friction (6.1)

This video explains the microscopic origin of friction and the two-regime model used in problem-solving: static friction (which prevents motion up to a maximum value) and kinetic friction (which opposes sliding motion with a constant magnitude).

  • Static friction: fs ≤ μsN — adjusts to prevent motion; maximum value fs,max = μsN
  • Kinetic friction: fk = μkN — constant once sliding; direction opposes relative motion
  • Coefficients of friction μs and μk: μk < μs in all practical cases
  • Friction on inclines — component approach; condition for sliding vs. staying

Time: Video pending upload to Canvas

Video 5: Pre-Lecture — Drag Force and Terminal Speed

Pre-Lecture: The Drag Force and Terminal Speed (6.2)

This video introduces the drag force as a velocity-dependent resistive force that opposes motion through a fluid. As a falling object speeds up, drag increases until it exactly balances gravity — at that point the object reaches terminal speed and stops accelerating.

  • Drag force model: FD = ½CρAv2 — depends on speed squared, fluid density, cross-section, and drag coefficient
  • Direction: always opposite to the velocity of the object through the fluid
  • Terminal speed: vT = √(2mg / CρA) — when FD = mg, acceleration = 0
  • Why heavier or more streamlined objects have higher terminal speeds

Time: Video pending upload to Canvas

Video 6: Pre-Lecture — Uniform Circular Motion

Pre-Lecture: Centripetal Force in Uniform Circular Motion (6.3)

From M2L2b you know that circular motion requires a centripetal acceleration directed toward the center. This video applies Newton's second law to that acceleration: whatever force (or combination of forces) provides the centripetal acceleration is the net centripetal force, Fc = mv2/r.

  • Centripetal force is not a new type of force — it is the net force directed toward the center
  • Fnet, centripetal = mac = mv2/r = m(4π2r)/T2
  • Examples: tension in a string, gravity for orbits, normal force for loop-the-loop, friction for a car on a curve
  • Banked curves — how banking angle reduces the friction required to maintain circular motion

Time: Video pending upload to Canvas

Video 7: Mini-Lecture — Newton's 2nd Law: Horizontal, Vertical, and Tilted

Mini-Lecture: Applying Fnet = ma in Three Orientations

This video works through three versions of the same core problem — a single object with multiple forces — demonstrating how the strategy changes when the surface is horizontal, vertical (elevator), and tilted (inclined ramp). The key insight: always align the coordinate system with the acceleration direction.

  • Horizontal surface: N = mg; net force is horizontal; solve for a or F
  • Vertical (elevator): N − mg = ma; interpret N > mg (accelerating up) vs. N < mg (accelerating down)
  • Tilted (ramp, frictionless): rotate axes so x is along the ramp; component of gravity along ramp = mg sin θ
  • Systematic strategy: draw FBD, choose axes, write ΣF = ma, solve

Time: Video pending — to be recorded by instructor

Video 8: Mini-Lecture — Block on a Frictionless Ramp

Mini-Lecture: Newton's Second Law on an Incline (No Friction)

A detailed worked example: a block slides down a frictionless inclined surface. The video shows how to decompose the gravitational force into components along and perpendicular to the ramp surface, apply Newton's second law along each axis, and find the acceleration and normal force.

  • Choose axes: x along the ramp (positive downslope), y perpendicular to ramp
  • Along ramp: mg sin θ = ma → a = g sin θ
  • Perpendicular to ramp: N − mg cos θ = 0 → N = mg cos θ
  • Note: a = g sin θ is independent of mass — all objects on the same frictionless ramp accelerate identically

Time: Video pending — to be recorded by instructor

Video 9: Mini-Lecture — Reading a Graph: Acceleration Along a Ramp

Mini-Lecture: Extracting Forces from an Acceleration-Time Graph

This video demonstrates how to read a graph of acceleration vs. time for an object on a ramp and use Newton's second law to calculate the forces involved — including identifying when friction is present versus absent from the shape of the graph.

  • Constant acceleration on graph → constant net force → constant friction (or no friction)
  • Using a = g sin θ − μkg cos θ to back-calculate μk from measured acceleration
  • Reading the transition point where a changes (e.g., block reaches bottom of ramp)
  • Connecting graphical analysis (M1L6) to Newton's law dynamics

Time: Video pending — to be recorded by instructor

Video 10: Mini-Lecture — Several Objects Pushed or Pulled as One

Mini-Lecture: Multi-Body Systems — One Acceleration, Multiple FBDs

When two or more objects are connected (by a rope, rod, or contact) and move together, they share the same acceleration. This video shows how to treat the system as a whole to find the acceleration and then isolate each object to find the internal forces (tension, contact forces) between them.

  • Treat the connected system as a single object: Fapplied = (m1 + m2 + ...)a
  • Isolate one object to find tension or contact force between objects
  • Key check: Newton's Third Law — tension pulls equally on both objects at the junction
  • Worked example: two blocks on a surface connected by a string, one pulled by an applied force

Time: Video pending — to be recorded by instructor

Video 11: Mini-Lecture — Static and Kinetic Friction: Block Pushed Along a Floor

Mini-Lecture: The Two Friction Regimes — When Does the Block Start Moving?

This video walks through the classic block-on-floor friction problem in full detail: determining whether the block moves at all (static regime), finding the applied force at which it just starts to slide, and computing the acceleration once kinetic friction takes over.

  • Static regime: fs = Fapplied as long as Fapplied ≤ μsmg → a = 0
  • Threshold for motion: Fapplied = μsmg = fs,max
  • Kinetic regime: fk = μkmg (constant); net force = Fapplied − fk = ma
  • Why μk < μs: it takes more force to start sliding than to keep sliding

Time: Video pending — to be recorded by instructor

Video 12: Mini-Lecture — Friction on a Ramp

Mini-Lecture: Kinetic and Static Friction on an Inclined Surface

Combining the ramp geometry from Video 8 with friction: this video solves for the acceleration of a block sliding down a rough inclined surface and determines the minimum angle at which the block will start to slide from rest.

  • Along ramp: mg sin θ − fk = ma → a = g(sin θ − μk cos θ)
  • Perpendicular to ramp: N = mg cos θ (unchanged by friction)
  • Angle for impending motion: tan θc = μs → θc = tan−1(μs)
  • Going up vs. coming down: friction always opposes direction of motion, so the two cases differ

Time: Video pending — to be recorded by instructor

Video 13: Mini-Lecture — Drag Force and Terminal Speed

Mini-Lecture: Calculating Terminal Speed and Drag-Limited Acceleration

This worked-example video applies the drag force model quantitatively: starting from rest, a falling object accelerates under gravity while drag grows with v2, until the net force approaches zero at terminal speed. The video derives and applies the terminal speed formula.

  • Net force on falling object: Fnet = mg − FD = mg − ½CρAv2
  • At terminal speed: Fnet = 0 → mg = ½CρAvT2
  • Solving for vT = √(2mg / CρA) — units check and numerical example
  • Why a heavier object of the same shape has a higher terminal speed

Time: Video pending — to be recorded by instructor

Video 14: Mini-Lecture — Centripetal Force and Circular Motion

Mini-Lecture: Identifying and Calculating the Net Centripetal Force

This video applies Newton's second law directly to circular motion: the net force toward the center of the circular path equals mv2/r. Multiple physical setups are shown to illustrate that different forces (tension, gravity, normal, friction) can all serve as the centripetal force depending on the context.

  • ΣFcentripetal = mv2/r — always directed toward center, never "outward"
  • Ball on a string (horizontal circle): T = mv2/r
  • Satellite in orbit: Fgravity = mv2/r
  • Car on a flat curve: fs = mv2/r — maximum speed before sliding = √(μsgr)

Time: Video pending — to be recorded by instructor

Practice and Apply - Conceptual

Steps to Solve a Newton's Second Law Problem

Arrange the following steps in the correct order for solving a Newton's second law dynamics problem using free-body diagrams.

Order Items

Arrange the following items in the correct order.

  1. Identify the object (system) whose motion you will analyze.
  2. List all external forces acting on the object with their directions.
  3. Draw the free-body diagram with one labeled arrow per force.
  4. Choose a coordinate system — align x with the acceleration direction when possible.
  5. Write Newton's second law along each axis: ΣFx = max and ΣFy = may.
  6. Solve the resulting equations for the unknown quantity.
  7. Check units, significant figures, and whether the answer is physically reasonable.

Which Objects Have a Non-Zero Net Force?

From the scenarios below, select all in which the net force on the object is not zero.

Select All

Select all scenarios in which the net force is non-zero.

  • A block accelerating down a frictionless ramp.
  • A car speeding up from a stoplight.
  • A ball on a string swinging in a horizontal circle at constant speed.
  • A skydiver before reaching terminal speed.
  • A rocket decelerating as it approaches a landing pad.
  • A book resting on a table with no applied force.
  • A car traveling at constant velocity on a straight highway.
  • A skydiver falling at terminal speed.
  • A block sitting stationary on a rough inclined surface (not sliding).

Match the Force to Its Formula

Match each force on the left to its correct formula or description on the right.

Match Items

Select each set of two matching items.

Items to Match

Weight
w = mg, directed straight downward toward Earth's center.
Kinetic friction
fk = μkN — constant magnitude, opposes direction of sliding.
Static friction (maximum)
fs,max = μsN — upper limit that prevents the onset of sliding.
Drag force
FD = ½CρAv2 — opposes velocity through a fluid; grows with v squared.
Net centripetal force
Fc = mv2/r — directed toward the center of the circular path.
Normal force
N — perpendicular to the contact surface; prevents interpenetration.

Key Concept Check

Click each card to flip it and test your understanding of Newton's Laws, friction, drag, and circular motion.

State Newton's First Law. What does it imply about objects at rest and objects in motion?
Answer

An object at rest remains at rest, and an object in motion continues in motion at constant velocity, unless acted upon by a net external force. This means that no force is needed to maintain motion — only to change it. The tendency to resist change is called inertia, and mass quantifies it.

 

Why is the centripetal force NOT listed as a separate force on a free-body diagram?
Answer

Because centripetal force is not a new or separate type of force — it is simply the net force in the inward radial direction. It is provided by one of the forces already on the FBD (tension, gravity, normal force, friction, etc.). Writing "centripetal force" as an additional arrow would count it twice.

 

A block is pushed across a floor at constant velocity. Is the net force zero? How do you know?
Answer

Yes, the net force is zero. Constant velocity means zero acceleration. By Newton's second law, ΣF = ma = 0 when a = 0. The applied push exactly balances the kinetic friction force in the horizontal direction, and the normal force exactly balances the weight in the vertical direction.

 

What is the difference between static and kinetic friction? Which is larger?
Answer

Static friction acts on a stationary object to prevent it from sliding; it adjusts in magnitude (0 ≤ fs ≤ μsN). Kinetic friction acts on a sliding object and has constant magnitude fk = μkN. Static friction is larger: μs > μk, which is why it takes more force to start an object sliding than to keep it sliding.

 

What happens to the acceleration of a skydiver as they fall from rest? Why does it eventually reach zero?
Answer

Initially, acceleration = g (only gravity acts). As speed increases, the drag force FD = ½CρAv2 grows. The net downward force (mg − FD) decreases, so acceleration decreases. When FD = mg, the net force = 0 and acceleration = 0. The skydiver now falls at constant terminal speed vT = √(2mg/CρA).

 

What is the maximum speed a car can take a flat (unbanked) circular curve of radius r with coefficient of static friction μs?
Answer

Static friction provides the centripetal force: fs = mv2/r. The maximum static friction is fs,max = μsmg = μsN (since N = mg on a flat road). Setting them equal: μsmg = mv2/r → vmax = √(μsgr). Note: mass cancels — all cars have the same speed limit on a given curve.

 

Practice and Apply - Computational

Important: Key Equations and Strategy Reminders

Newton's Second Law (apply along each axis): ΣFx = max   |   ΣFy = may

Friction: fk = μkN (kinetic)   |   fs ≤ μsN; max = μsN (static)

Drag and terminal speed: FD = ½CρAv2   |   vT = √(2mg / CρA)

Centripetal: ΣFcentripetal = mv2/r   |   vmax, flat curve = √(μsgr)

  • Always draw the FBD first — identify every force before writing any equations.
  • Align axes with acceleration — along the ramp, toward the center of a circle, etc.
  • Normal force ≠ weight unless the surface is horizontal and ay = 0.
  • Report forces in N, acceleration in m/s2, speed in m/s. Three significant figures unless stated otherwise.

Sample Problem Videos

Watch each sample problem video, then attempt the corresponding written practice problem below.

Sample Problem 1: Car in a Banked Circular Turn

Sample Problem: Banked Curve — Finding the Ideal Banking Angle and Speed

A complete worked solution for a car navigating a banked circular turn. The video finds the banking angle at which no friction is needed to maintain circular motion, and the corresponding ideal speed. It then extends to finding the maximum speed with friction.

  • FBD for a car on a banked curve: weight downward, normal force perpendicular to banked surface
  • Horizontal (centripetal): N sin θ = mv2/r
  • Vertical (no vertical acceleration): N cos θ − mg = 0 → N = mg / cos θ
  • Ideal angle (no friction needed): tan θ = v2 / (rg)

Time: Video pending — to be recorded by instructor

Sample Problem 2: Friction with Applied Force at an Angle

Sample Problem: Block Pushed at an Angle — Normal Force, Friction, and Acceleration

A complete worked solution for a block on a horizontal surface pushed by a force at an angle below the horizontal. The downward component of the applied force increases the normal force (and therefore friction), which changes the block's acceleration compared to a purely horizontal push.

  • FBD: weight (down), normal (up), applied force at angle (down-right), kinetic friction (left)
  • Vertical: N − mg − F sin φ = 0 → N = mg + F sin φ
  • Horizontal: F cos φ − μkN = ma → solve for a
  • Key insight: pushing down increases N and therefore increases friction — pulling up would decrease it

Time: Video pending — to be recorded by instructor

Sample Problem 3: Terminal Speed of a Falling Raindrop

Sample Problem: Terminal Speed — Applying the Drag Force Model

A complete worked solution finding the terminal speed of a spherical raindrop. Given the drop's radius, the density of water and air, and the drag coefficient, the video applies the terminal speed formula and discusses what physical factors control the speed of rainfall.

  • Calculate mass from density and radius: m = ρwater × (4/3)πr3
  • Calculate cross-sectional area: A = πr2
  • Apply: vT = √(2mg / CρairA) with C ≈ 0.45 for a sphere
  • Interpret: larger drops fall faster; smaller drops (mist) fall nearly at rest

Time: Video pending — to be recorded by instructor

Practice Problem 1 — Newton's Second Law on a Ramp with Friction

A 5.00-kg block is placed on a rough inclined surface tilted at θ = 30.0° above the horizontal. The coefficients of friction are μs = 0.400 and μk = 0.300. (a) Does the block slide? (b) If it does slide, find its acceleration down the ramp. (c) Find the normal force on the block.

Show Solution

Given: m = 5.00 kg, θ = 30.0°, μs = 0.400, μk = 0.300, g = 9.80 m/s2

Part (a) — Does the block slide?

Angle for impending motion: tan θc = μs = 0.400 → θc = tan−1(0.400) = 21.8°

Since θ = 30.0° > θc = 21.8°, the gravitational component along the ramp exceeds maximum static friction.

Yes, the block slides.

Part (b) — Acceleration down the ramp:

Axes: x along ramp (positive down-slope), y perpendicular to ramp.

y-axis: N − mg cos θ = 0 → N = (5.00)(9.80) cos 30.0° = 49.0 × 0.866 = 42.4 N

fk = μkN = (0.300)(42.4) = 12.7 N (directed up the ramp)

x-axis: mg sin θ − fk = ma

(5.00)(9.80) sin 30.0° − 12.7 = (5.00)a

24.5 − 12.7 = 5.00a → a = 11.8 / 5.00 = 2.36 m/s2 down the ramp

Part (c) — Normal force:

N = mg cos θ = (5.00)(9.80)(0.866) = 42.4 N

Note: N < mg = 49.0 N because the incline supports only the component of weight perpendicular to the surface.

Practice Problem 2 — Multi-Body System with Tension

Two blocks are connected by a light string. Block A (mA = 3.00 kg) sits on a frictionless horizontal table and is connected via a string over a frictionless pulley to Block B (mB = 2.00 kg) hanging vertically. (a) Find the acceleration of the system. (b) Find the tension in the string.

Show Solution

Given: mA = 3.00 kg (on table), mB = 2.00 kg (hanging), frictionless table and pulley, g = 9.80 m/s2

Part (a) — System acceleration:

Both blocks have the same magnitude of acceleration a (string is inextensible).

Block A (horizontal): T = mAa   →    T = 3.00a    ...(1)

Block B (vertical, positive direction downward): mBg − T = mBa   →    2.00(9.80) − T = 2.00a    ...(2)

Add equations (1) and (2):

mBg = (mA + mB)a → a = mBg / (mA + mB) = (2.00)(9.80) / (3.00 + 2.00)

a = 19.6 / 5.00 = 3.92 m/s2

Part (b) — Tension:

T = mAa = (3.00)(3.92) = 11.8 N

Check with Block B: mBg − T = mBa → 19.6 − 11.8 = 7.84 = (2.00)(3.92) = 7.84 N ✓

Note: T < mBg = 19.6 N because Block B is accelerating downward — if it were in equilibrium, T would equal its weight.

Practice Problem 3 — Banked Circular Turn

A highway curve has a radius of r = 120 m and is banked at an angle of θ = 18.0°. (a) Find the ideal speed (no friction needed) for this curve. (b) If the coefficient of static friction is μs = 0.350, find the maximum speed at which a car can safely navigate this curve.

Show Solution

Given: r = 120 m, θ = 18.0°, μs = 0.350, g = 9.80 m/s2

Part (a) — Ideal speed (no friction):

FBD: weight mg down, normal force N perpendicular to banked surface.

Horizontal (centripetal): N sin θ = mv2/r

Vertical: N cos θ = mg → N = mg / cos θ

Substitute N: (mg / cos θ) sin θ = mv2/r → g tan θ = v2/r

videal = √(rg tan θ) = √(120 × 9.80 × tan 18.0°) = √(120 × 9.80 × 0.3249)

videal = √(381.9) = 19.5 m/s (about 70 km/h)

Part (b) — Maximum speed with friction:

Above the ideal speed, friction acts down the bank (inward) to supplement the inward component of N.

vmax = √[rg(μs + tan θ) / (1 − μs tan θ)]

Numerator factor: μs + tan θ = 0.350 + 0.3249 = 0.6749

Denominator factor: 1 − μs tan θ = 1 − (0.350)(0.3249) = 1 − 0.1137 = 0.8863

vmax = √[120 × 9.80 × 0.6749 / 0.8863] = √[1176 × 0.7615] = √[896] = 29.9 m/s (about 108 km/h)

Ready to Move On?

Lesson Checklist

Before moving on to the next lesson, confirm you can do each of the following:

  • ☐ State Newton's three laws and give a real-world example illustrating each.
  • ☐ Draw a correct free-body diagram for a block on a flat surface, on an incline, and hanging from a string.
  • ☐ Apply ΣFx = max and ΣFy = may to find unknown forces or accelerations in 2D problems.
  • ☐ Determine whether an object on a rough incline will slide using tan θc = μs.
  • ☐ Calculate acceleration on a rough incline: a = g(sin θ − μk cos θ).
  • ☐ Calculate terminal speed using vT = √(2mg / CρA) and explain why it is reached.
  • ☐ Apply Newton's second law to circular motion: identify the centripetal force and calculate mv2/r.
  • ☐ Solve a banked-curve problem for ideal speed (tan θ = v2/rg).

Newton's Laws are the foundation of the rest of the course. Every subsequent topic — energy, momentum, rotation — builds directly on the force-and-acceleration framework you've built here.