Physics for ME · Module 4 of 16
Newton's Laws and Force Models
F = ma is the engine; the force models are the fuel. Gravity, normal force, friction, tension, drag, and springs cover nearly every machine.
Readiness check
From Modules 2 and 3. Tick only what you can do closed-notes.
- Resolve forces into components along declared axes.
- Use the constant-acceleration equations once a is known.
- Convert mass to weight (W = mg) without hesitation.
- Work slope components W sin θ and W cos θ.
- Keep signs consistent through a multi-equation solution.
The core idea
The net force sets the acceleration. Each physical contact contributes one modeled force.
ΣF = maF ≤ μN · F = ks · Fdrag ∝ v²First law: zero net force means constant velocity. Second law: the quantitative engine. Third law: every force has an equal-opposite partner on the other body. The force models give each touch (surface, rope, spring, fluid) its equation.
The skills, taught in order
Newton's three laws say when and how motion changes; the force models say what is pushing. Learn the laws first, then collect the six models, and almost every machine problem becomes the same short procedure.
4.1 Newton's first law
A body with zero net force keeps moving exactly as it was: at rest, or in a straight line at constant velocity. Motion needs no force to continue, only to change. Such a body is in equilibrium, ΣF = 0, and the law holds only in an inertial (non-accelerating) frame of reference.
4.2 Newton's second law
When the net force is not zero, the body accelerates in the direction of that force: ΣF = ma, applied one axis at a time as ΣFx = max and ΣFy = may. Only external forces count, the mass must be constant, and the frame must be inertial. The product ma is the result of the forces, never itself a force to add to the diagram.
4.3 Newton's third law
If A pushes B, then B pushes A with equal magnitude and opposite direction. The two forces of the pair always act on different bodies, so they never appear together on one free-body diagram and never cancel each other out.
4.4 Gravity and weight
Weight is the gravitational pull on a body, W = mg, directed down. Mass (kg) measures inertia and never changes; weight (N) is a force and follows g. Near Earth g runs from about 9.78 to 9.82 m/s², and this course uses g = 9.81 m/s². On the Moon g = 1.62 m/s², so the same mass weighs about a sixth as much.
Micro-example. A 10 kg part weighs 10 × 9.81 = 98.1 N on Earth and 16.2 N on the Moon, while its mass stays 10 kg in both places.
4.5 Normal force
A surface pushes on whatever touches it, perpendicular to the surface. The normal force n takes whatever value balances the perpendicular forces; it is not automatically equal to the weight. Press down on a box and n grows, pull up on it and n shrinks.
Micro-example. A 20 kg box on a level floor has n = 196 N at rest, but an added 50 N downward push raises it to 246 N, and a 50 N upward rope lowers it to 146 N.
4.6 Friction
Friction acts along a surface, opposing relative sliding, in two regimes. Static friction adjusts itself to prevent motion, up to a limit: fs ≤ μsn. Once sliding starts, kinetic friction is roughly constant: fk = μkn. For a given pair of surfaces μk is a little less than μs, which is why a load is harder to start than to keep moving. Both coefficients are pure numbers.
| Surfaces | μs | μk |
|---|---|---|
| Steel on steel | 0.74 | 0.57 |
| Aluminium on steel | 0.61 | 0.47 |
| Rubber on dry concrete | 1.0 | 0.8 |
| Rubber on wet concrete | 0.30 | 0.25 |
| Teflon on steel | 0.04 | 0.04 |
Representative values (Young and Freedman, Table 5.1). Real surfaces vary, so treat them as approximate.
4.7 Tension
A rope, cable, or chain pulls along its own length and never pushes. For a light (massless) cord running over a frictionless pulley, the tension is the same throughout, which is what lets one rope tie two bodies into a single problem.
4.8 Springs
An ideal spring resists being stretched or compressed with a restoring force proportional to the displacement, F = ks (Hooke's law), where k is the stiffness in N/m and s the stretch. The force always points back toward the natural length.
Micro-example. A spring of stiffness k = 200 N/m stretched 50 mm pulls back with F = 200 × 0.05 = 10 N.
4.9 Drag and fluid resistance
A fluid resists a body moving through it, directed opposite the velocity. At low speed the resistance is roughly f = kv; at the higher speeds of vehicles and falling bodies it grows with the square, f = Dv². When drag rises to equal the weight, the net force is zero and the body falls at a constant terminal speed.
| Force | Model | Direction |
|---|---|---|
| Weight | W = mg | down, toward Earth |
| Normal | solve from the perpendicular balance | perpendicular to the surface |
| Friction | fs ≤ μsn, then fk = μkn | along the surface, opposing slip |
| Tension | solve as an unknown | along the rope, pulling |
| Spring | F = ks | back toward natural length |
| Drag | f = Dv² (or kv) | opposite the velocity |
Engineering connection: the physics core behind Statics and Dynamics. Friction returns in full in Statics Module 8, springs in Module 10, and drag in the worked example of Module 1.
Worked example 1: the dragged sled
A 20 kg sled is pulled across a floor by an 80 N rope at 25° above horizontal. The kinetic friction coefficient is μk = 0.15. Find the sled's acceleration.
- ProblemFind the acceleration of the sled in Figure 1.
- Given / findm = 20 kg, P = 80 N at 25°, μk = 0.15. Find a.
- AssumptionsRigid sled, steady sliding (kinetic friction), level floor.
- ModelFigure 2: W = 196.2 N down, N up, friction backward, pull split into 80 cos 25° = 72.5 N and 80 sin 25° = 33.8 N.
- EquationsΣFy = 0: N + 33.8 − 196.2 = 0 ΣFx = ma: 72.5 − μkN = 20a
- SolveN = 162.4 N. Friction = 0.15 × 162.4 = 24.4 N. Net force = 72.5 − 24.4 = 48.1 N. a = 48.1/20 = 2.41 m/s².
- CheckN < W because the rope lifts a little: 162.4 = 196.2 − 33.8. Limits: with μ = 0 the answer would be 72.5/20 = 3.63 m/s²; friction took a believable bite.
- ConclusionThe y-equation (no acceleration) fed N into the x-equation (acceleration): that two-step is the standard choreography of every Newton's-law problem, and the FBD made it mechanical.
Worked example 2: will the crate slide?
A 60 kg crate rests on a ramp. The coefficients between crate and ramp are μs = 0.35 and μk = 0.27. (a) At a 15° slope, does it stay put, and how much friction acts? (b) At what angle does it begin to slide? (c) Once sliding on a 25° slope, what is its acceleration?
- ProblemFor the crate in Figure 3: (a) at 15° does it stay and what friction acts; (b) at what angle does it slide; (c) what is its acceleration sliding at 25°?
- Given / findm = 60 kg, μs = 0.35, μk = 0.27, g = 9.81 m/s². Find the friction at 15°, the slipping angle, and the acceleration at 25°.
- AssumptionsRigid crate treated as a particle, uniform ramp, axes taken along and perpendicular to the surface.
- ModelWeight W = mg splits into W sin θ down the slope and W cos θ into it. The perpendicular balance gives n = W cos θ; friction opposes impending or actual sliding.
- Equationsn = mg cos θ fs ≤ μsn (static) a = g(sin θ − μk cos θ) (sliding)
- SolveW = 588.6 N. (a) At 15°: the down-slope pull is W sin 15° = 152 N, while the static limit is μsn = 0.35 × W cos 15° = 199 N. Since 152 < 199, the crate holds and friction supplies exactly 152 N, not 199 N. (b) Sliding starts when W sin θ = μsW cos θ, that is tan θ = μs, so θ = arctan 0.35 = 19.3°. (c) At 25°: a = 9.81(sin 25° − 0.27 cos 25°) = 1.75 m/s².
- CheckStatic friction is an inequality: below 19.3° it returns exactly the pull it must (152 N at 15°), never more. The slip angle depends only on μs, not on mass, which is why the angle of repose is a clean way to measure a coefficient.
- ConclusionStatic friction is a range, not a fixed value, so the FBD must solve for what it actually supplies. The same incline analysis sizes brakes, chutes, and the angle of repose of stockpiles, and it returns in full in Statics Module 8.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| A "force of motion" invented | An extra forward arrow on a coasting body | "What is physically touching the body to push it?" | Motion needs no force to continue (first law); only changes of motion need force. |
| Third-law pairs on one body | N and W called an action-reaction pair | "Are these two forces on different bodies?" | W's partner is the body pulling Earth up; N's partner is the body pressing the floor. Pairs never cancel on one FBD. |
| N assumed equal to W | Friction wrong whenever pulls have vertical parts | "Does anything else push or pull vertically?" | Solve ΣFy for N every time. The worked example's rope stole 33.8 N of it. |
| Mass and weight swapped in F = ma | Answers off by a factor of 9.81 | "Did I put kilograms or newtons next to a?" | F = ma takes mass in kg. Weight is one of the forces, not the m. |
Practice ladder
A 1200 kg car accelerates at 3 m/s². What net force acts on it, and what does that force become if the same engine pushes a 2400 kg van?
Show answer
F = 3600 N. Same force on twice the mass: a = 1.5 m/s². Second law both directions.
An 8 kg lamp hangs from the ceiling of an elevator accelerating upward at 2 m/s². Find the cable tension.
Show answer
T − mg = ma: T = 8(9.81 + 2) = 94.5 N, versus 78.5 N at rest. Scales in elevators lie for exactly this reason.
A 50 g ball reaches terminal velocity falling through air with drag F = 0.002 v² (newtons, v in m/s). Find the terminal velocity.
Show answer
At terminal speed drag balances weight: 0.002 v² = 0.4905, so v = √245.3 = 15.7 m/s. Past this speed the net force, and hence acceleration, is zero: first law in action mid-fall.
Instrument a real acceleration: use a phone accelerometer log in an elevator, car, or train start. Extract the peak acceleration, estimate the net force on your own body, and identify which physical contact supplied it.
What good work looks like
The logged trace, a peak a with the computed F = ma on your mass, and the correct supplier named (floor normal force or seat friction), with the third-law partner stated.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Create a one-page "Force Models Card": all six models (gravity, normal, friction, tension, spring, drag) with equation, direction rule, one worked mini-example each, and the limit where the model fails.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. State Newton's three laws in engineering words.
1: zero net force, constant velocity. 2: ΣF = ma, componentwise. 3: forces come in equal-opposite pairs on different bodies.
2. Give the six force models and their equations or rules.
Gravity W = mg down; normal N perpendicular to contact (solve for it); friction F ≤ μN opposing slip; tension along the rope, pulling; spring F = ks restoring; drag growing with speed, opposing velocity.
3. Why is N not always equal to mg?
N comes from the vertical equilibrium of all forces; angled pulls, pushes, or accelerations change it.
4. What is terminal velocity, in force language?
The speed at which drag equals weight, making net force and acceleration zero.
5. Why do third-law pairs never cancel?
They act on different bodies, so they never meet on the same free-body diagram.