Dynamics · Chapter 3 of 10 · Intermediate
Kinetics of Particles: Newton's Second Law
Kinematics described the motion. Kinetics explains it: draw the forces, set them equal to mass times acceleration, and the motion follows.
Readiness check
This chapter joins the forces of statics to the accelerations of Chapter 2. Tick only what you can do closed-notes.
- Draw a complete free-body diagram.
- Resolve forces along chosen axes.
- Compute friction as μN.
- Find an = v²/ρ for curved motion.
- Keep mass and weight distinct.
The core idea
Draw every force on the particle, sum them, and set the result equal to mass times acceleration; the equation of motion is statics with an ma on the right.
ΣF = maΣFn = m v²/ρΣFt = m atThe free-body diagram is unchanged from statics; the only difference is that the forces need not balance. Whatever they fail to balance by, divided by the mass, is the acceleration. Choosing axes that line up with the motion (along a line, or normal and tangential to a curve) turns the vector equation into simple scalar ones.
The skills, taught in order
Kinetics is one equation applied with discipline. Five skills cover the equation of motion, the two coordinate forms, the diagram procedure, and connected bodies.
3.1 The equation of motion
Newton's second law, ΣF = ma, is a vector statement: the resultant of all forces equals mass times the acceleration vector. In practice you write it as scalar component equations along chosen axes. The mass is constant (variable mass waits for Chapter 6).
3.2 Rectilinear motion
When motion is along a straight line, align one axis with it. The forces along that axis give ma; forces perpendicular sum to zero (no acceleration that way). Inclines and friction are the standard cases: weight is resolved into components along and across the surface, and friction opposes the motion as μN.
3.3 Curvilinear motion in n-t coordinates
For motion along a curve, use normal and tangential axes. The tangential equation ΣFt = m at governs speeding up or slowing down; the normal equation ΣFn = m v²/ρ supplies the inward force that bends the path. The normal force is what tyres, rails, and strings provide.
| Form | Equations | Best for |
|---|---|---|
| Rectangular | ΣFx = max, ΣFy = may | straight or projectile motion |
| Normal-tangential | ΣFn = mv²/ρ, ΣFt = mat | motion along a known curve |
3.4 The free-body and kinetic diagrams
Draw the free-body diagram (all forces) and, beside it, the kinetic diagram (the ma vector). Setting the two equal, component by component, is a bookkeeping method that prevents lost or invented forces. It is the single most reliable habit in kinetics.
3.5 Connected particles
When bodies are linked by cables over pulleys, their accelerations are related by the constraint that the cable length is fixed. Write ΣF = ma for each body separately, then use the kinematic constraint to close the system. The internal cable tension appears in both equations.
Engineering connection: traction and braking, cornering limits, hoist and elevator design, and the force inputs that Chapter 4 integrates into energy.
Worked example 1: crate sliding down an incline
A 50 kg crate is released from rest on a 20° incline with kinetic friction coefficient μk = 0.25. Find its acceleration down the slope and the normal force from the surface.
- ProblemFind the down-slope acceleration and the normal force for the crate in Figure 1.
- Given / findm = 50 kg, θ = 20°, μk = 0.25, from rest. Find a and N.
- AssumptionsParticle on a rigid plane, kinetic friction (it is sliding), g = 9.81 m/s².
- ModelAxes along and perpendicular to the incline. Perpendicular: ΣF = 0 gives N. Along: ΣF = ma gives a, with friction opposing motion.
- EquationsN = mg cosθ mg sinθ − μkN = ma a = g(sinθ − μkcosθ)
- SolveN = 50 × 9.81 × cos20° = 461 N. Friction = 0.25 × 461 = 115 N. a = 9.81(sin20° − 0.25cos20°) = 9.81(0.342 − 0.235) = 1.05 m/s² down the slope.
- CheckThe driving component mg sinθ = 168 N exceeds friction 115 N, so the crate accelerates, consistent with a positive a. If μk were raised to tan20° = 0.364, the acceleration would vanish, the impending-motion limit.
- ConclusionResolving weight into slope-aligned components is the key move. The perpendicular equation is pure statics; only the along-slope equation carries the ma.
Worked example 2: the skid limit on a flat curve
A 1200 kg car rounds a flat (unbanked) curve of radius 80 m. The tyres can supply a friction coefficient μs = 0.7. Find the maximum speed before the car slides outward.
- ProblemFind the maximum speed before the car in Figure 2 slides off the flat curve.
- Given / findm = 1200 kg, ρ = 80 m, μs = 0.7, g = 9.81 m/s². Find vmax.
- AssumptionsLevel road, friction is the only horizontal force, the limit is impending slip (friction at μsN).
- ModelVertical: N = mg. Normal (horizontal, toward centre): friction provides ΣFn = m v²/ρ, capped at μsN.
- EquationsN = mg μsmg = m vmax²/ρ vmax = √(μsgρ)
- SolveThe mass cancels: vmax = √(0.7 × 9.81 × 80) = √549 = 23.4 m/s ≈ 84 km/h. The friction at that point is μsmg = 0.7 × 1200 × 9.81 = 8240 N, exactly the centripetal force needed.
- CheckMass cancelling is the famous result: a heavy and a light car skid at the same speed on the same curve. The answer scales with √ρ, so a curve four times tighter cuts the safe speed in half.
- ConclusionOn a flat curve, grip alone holds the car. Banking the road (the physics of the banked curve) adds a component of the normal force inward, raising the safe speed beyond this friction-only limit.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Adding a centrifugal force | An outward force drawn on the free body | "Is this force from a real contact or field?" | No outward force acts; the inward net force equals m v²/ρ. |
| Treating ma as a force | ma placed on the free-body diagram | "Is ma a force or the response?" | ma belongs on the right side; keep it off the free body. |
| Wrong friction direction | Friction drawn along the motion | "Which way does slip tend?" | Friction opposes relative sliding, so it points against the motion. |
| Forgetting the constraint | Connected bodies solved independently | "How are the accelerations related?" | Use the cable-length constraint to link the accelerations. |
Practice ladder
A 10 kg block on a frictionless floor is pulled by a horizontal 40 N force. Find its acceleration.
Show answer
a = ΣF/m = 40/10 = 4 m/s². With no friction and level ground, the applied force is the entire net force.
For the Worked Example 1 crate, what coefficient of friction would let it slide at constant velocity once pushed?
Show answer
Constant velocity means a = 0, so sinθ = μkcosθ, giving μk = tan20° = 0.364. At exactly this value the driving and resisting forces balance, the boundary between accelerating and decelerating.
A 0.5 kg ball on a 0.8 m string is whirled in a horizontal circle. If the string breaks at 50 N tension, find the maximum speed.
Show answer
The tension is the centripetal force: T = m v²/ρ, so v = √(Tρ/m) = √(50 × 0.8/0.5) = √80 = 8.9 m/s. Beyond this speed the required force exceeds what the string can carry.
Choose a real accelerating system (a lift starting up, a car braking, a bag on a turning bus). Draw the free body, write ΣF = ma in suitable axes, and estimate the acceleration or a limiting condition.
What good work looks like
A complete free-body diagram with no invented forces, axes matched to the motion, ΣF = ma written in components, and a result checked for sign and magnitude.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Take one real system through free-body diagram, kinetic diagram, and ΣF = ma in the right axes, ending with an acceleration or a limiting speed and a check.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. State the equation of motion and what each side means.
ΣF = ma: the resultant of the real forces equals mass times the acceleration vector.
2. Write the n-t component equations.
ΣFn = m v²/ρ (toward the centre) and ΣFt = m at (along the path).
3. Where does ma go, and where does it not?
On the right side of the equation (the kinetic diagram), never on the free-body diagram.
4. Which way does friction point?
Opposite the relative sliding (or impending sliding) of the surfaces.
5. How are connected particles linked?
By a kinematic constraint (fixed cable length) relating their accelerations.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Meriam and Kraige, Engineering Mechanics: Dynamics (7th ed), Chapter 3, Section A (Force, Mass, and Acceleration) |
| Cross-reference | Hibbeler, Dynamics, Ch. 13 · Beer and Johnston, Ch. 12 |
| Core topics | 3.1 Equation of motion · 3.2 Rectilinear · 3.3 Curvilinear (n-t) · 3.4 Free-body and kinetic diagrams · 3.5 Connected particles |
| Engineering connection | Traction, braking, cornering limits, and hoist design. |
| Read next | Chapter 4: Work and Energy. |