Kinetics of a Particle:
Force and Acceleration
Kinetics relates the forces acting on a particle to the motion they produce. This chapter develops Newton's second law into the equation of motion and applies it across rectangular, normal–tangential, and cylindrical coordinate systems, closing with central-force motion and the mechanics of orbits.
Chapter Objectives
- Explain the physical distinction among mass, weight, resultant force, and acceleration.
- Construct free-body and kinetic diagrams for a particle or particle system in an inertial frame.
- Select rectangular, normal–tangential, or cylindrical equations to match the known motion.
- Calculate accelerations, reactions, friction forces, and central-force motion from Newton's second law.
- Test contact, tension, and static-friction assumptions for physical admissibility.
- Verify results through units, signs, limiting behavior, and independent kinematic checks.
Newton's Second Law of Motion
Kinetics is the branch of dynamics that connects the change in motion of a body to the forces that cause it. Its foundation is Newton's second law, which describes how an unbalanced force produces acceleration.
When an unbalanced force acts on a particle, the particle accelerates in the direction of that force, with a magnitude proportional to the force. The constant of proportionality is the particle's mass m.
Mass and Inertia
Applying a known unbalanced force F and measuring the resulting acceleration a gives a constant ratio \(m = F/a\). This positive scalar is the mass. Because it stays constant throughout any acceleration, mass measures a particle's resistance to a change in velocity — its inertia.
This is one of the most important relationships in mechanics, and its validity rests on experimental evidence. Newton's law fails near the speed of light (where Einstein's relativity applies) and at atomic scales (where quantum mechanics applies), but those regimes are not encountered in ordinary engineering problems.
Always confirm the force is the net (resultant) force before equating it to ma. A common early mistake is plugging in a single applied force while ignoring weight, friction, or normal reactions. Draw the free-body diagram first, every time.
Newton also postulated a law for the mutual attraction between any two particles.
Any two particles attract each other along the line joining their centers, with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.
| F | force of attraction between the two particles |
| G | universal gravitational constant, \(G = 66.73\times10^{-12}\ \mathrm{m^3/(kg\cdot s^2)}\) |
| m₁, m₂ | mass of each particle |
| r | distance between the particles' centers |
Near the earth's surface, the only gravitational force of significant magnitude is between the earth and the particle. Letting \(m_2 = M_e\) (earth's mass) and defining \(g = GM_e/r^2\), the weight follows directly.
g — Acceleration Due to Gravity
Comparing \(W = mg\) with \(F = ma\) identifies g as the acceleration due to gravity. The standard value \(g = 9.81\ \mathrm{m/s^2}\) is measured at sea level and latitude 45° (the "standard location"). A 1-kg body therefore weighs 9.81 N, a 2-kg body weighs 19.62 N, and so on.
The Equation of Motion
When more than one force acts on a particle, the resultant is the vector sum of all forces, \(\mathbf{F}_R = \Sigma\mathbf{F}\). The equation of motion takes its general form using this resultant.
Free-Body Diagram & Kinetic Diagram
The free-body diagram shows the particle isolated from its supports with every acting force drawn on it. The kinetic diagram shows the single resultant ma vector that those forces produce. Setting the two diagrams equal expresses the equation of motion graphically: \(\Sigma\mathbf{F} = m\mathbf{a}\). If the resultant is zero, the acceleration is zero — the particle stays at rest or moves with constant velocity (static equilibrium, Newton's first law).
Inertial (Newtonian) Reference Frame
The equation of motion is only valid when acceleration is measured relative to a non-rotating frame whose axes are fixed or translate at constant velocity. In such a frame the observer does not accelerate, so the measured acceleration is the same from any frame of this type. For rockets and satellites the frame may be fixed to the stars; for motion near the earth's surface, a frame fixed to the earth suffices, since the earth's rotational accelerations are negligibly small for most applications.
You will sometimes see the equation rearranged as \(\Sigma\mathbf{F} - m\mathbf{a} = 0\), where \(-m\mathbf{a}\) is treated as an "inertia force vector" (D'Alembert's principle). It can be handy for intuition, but in this course we keep forces on one side and ma on the other — it keeps signs and bookkeeping cleaner under exam pressure.
Equation of Motion for a System of Particles
The equation of motion extends to a system of particles enclosed in a region of space. The analysis holds regardless of how the particles are connected, so it applies equally to solids, liquids, and gases.
Internal vs. External Forces
For the i-th particle of mass \(m_i\), the internal force \(\mathbf{f}_i\) is the resultant of forces the other particles exert on it, while the external force \(\mathbf{F}_i\) represents gravitational, electrical, magnetic, or contact forces from bodies outside the system. Applying \(\Sigma\mathbf{F} = m\mathbf{a}\) to this particle gives \(\mathbf{F}_i + \mathbf{f}_i = m_i\mathbf{a}_i\).
Summing over all particles and noting that internal forces cancel in equal-and-opposite collinear pairs, only the external forces survive:
Introducing the center of mass G, defined by \(m\mathbf{r}_G = \Sigma m_i \mathbf{r}_i\) with total mass \(m = \Sigma m_i\), and differentiating twice in time (no mass entering or leaving) yields \(m\mathbf{a}_G = \Sigma m_i \mathbf{a}_i\). Substituting gives the compact result:
The sum of external forces equals the total mass times the acceleration of the center of mass. Because real bodies have finite size, this result justifies treating a body as a single particle.
Important Points
- The equation of motion is based on experimental evidence and is valid only within an inertial frame of reference.
- The equation states that an unbalanced force on a particle causes it to accelerate.
- An inertial frame does not rotate; its axes either translate at constant velocity or are at rest.
- Mass is a property of matter measuring resistance to a change in velocity. It is absolute — it does not change with location.
- Weight is a force caused by the earth's gravitation. It is not absolute; it depends on altitude above the earth's surface.
Equations of Motion: Rectangular Coordinates
For a particle moving relative to an inertial x, y, z frame, both the forces and the acceleration are resolved into i, j, k components. Applying \(\Sigma\mathbf{F} = m\mathbf{a}\) and matching components on each side gives three scalar equations.
If the particle is constrained to the x–y plane, only the first two equations are needed.
Procedure for Analysis
The equations of motion relate the forces on a particle to the accelerated motion they cause.
Free-Body Diagram
- Select the inertial coordinate system — usually rectangular x, y, z for rectilinear motion.
- Draw the free-body diagram showing all forces, so they can be resolved into components.
- Establish the direction and sense of the acceleration. If unknown, assume each component acts along its positive axis.
- Represent the acceleration as the ma vector on the kinetic diagram, and identify the unknowns.
Equations of Motion
- If forces resolve directly from the free-body diagram, apply the scalar component equations.
- For complicated 3-D geometry, use Cartesian vector analysis.
- Friction: on a rough surface use \(F_f = \μk N\); the friction force opposes relative motion. If motion impends, use the static coefficient.
- Spring: for a linear elastic spring \(F_s = k s\), where \(s = l - l_0\) is the deformation from the undeformed length.
Kinematics
- Once a is found from \(\Sigma\mathbf{F} = m\mathbf{a}\), apply kinematic equations to find velocity or position.
- If \(a = f(t)\): use \(a = dv/dt\) and \(v = ds/dt\).
- If \(a = f(s)\): integrate \(a\,ds = v\,dv\).
- If \(a\) is constant: use \(v = v_0 + a_c t,\ s = s_0 + v_0 t + \tfrac12 a_c t^2,\ v^2 = v_0^2 + 2 a_c (s - s_0)\).
- A negative result for an unknown component means it acts opposite to the assumed sense.
In connected-body problems, pick one positive direction for the whole system and commit to it across every equation, including the kinematic constraint. The negative sign that pops out (as on \(a_B\) here) is the math telling you the real direction — trust it instead of re-deriving.
Equations of Motion: Normal and Tangential Coordinates
When a particle travels a known curved path, it is natural to resolve forces along the tangential (t), normal (n), and binormal (b) directions. There is no motion in the binormal direction, since the particle is constrained to the path.
Tangential vs. Normal Components
The tangential component \(a_t = dv/dt\) measures the change in the magnitude of velocity: a tangential force along the motion speeds the particle up, one against it slows the particle down. The normal component \(a_n = v^2/ρ\) measures the change in the direction of velocity. It is produced by \(\Sigma F_n\), which always points toward the path's center of curvature — for this reason it is often called the centripetal force.
Procedure for Analysis
Free-Body Diagram
- Establish the t, n, b coordinate system at the particle and draw its free-body diagram.
- The normal acceleration \(a_n\) always acts in the +n direction (toward the center of curvature).
- If \(a_t\) is unknown, assume it acts in the +t direction. There is no acceleration in the b direction.
Equations of Motion
- Apply Eq. 13–7.
Kinematics
- Formulate \(a_t = dv/dt\) or \(a_t = v\,dv/ds\), and \(a_n = v^2/ρ\).
- For a path \(y = f(x)\), the radius of curvature is \(ρ = \dfrac{[1 + (dy/dx)^2]^{3/2}}{|\,d^2y/dx^2\,|}\).
Equations of Motion: Cylindrical Coordinates
When the forces are resolved along the cylindrical unit directions \(\mathbf{u}_r, \mathbf{u}_θ, \mathbf{u}_z\), the equation of motion produces three scalar equations.
If motion is confined to the r–θ plane, only the first two are used. The acceleration components come from kinematics:
Direction of Normal and Friction Forces
When a particle follows a path \(r = f(θ)\), the normal force N is perpendicular to the tangent of the path, and the friction force F acts along the tangent opposing motion. Their directions relative to the radial line are fixed by the angle \(\psi\), measured between the extended radial line and the tangent to the curve.
A positive \(\psi\) is measured counterclockwise (in the +θ sense) from the extended radial line to the tangent; a negative value is measured clockwise.
Procedure for Analysis
Free-Body Diagram
- Establish the r, θ, z inertial coordinate system and draw the free-body diagram.
- Assume \(a_r, a_θ, a_z\) act in the positive directions if unknown; identify all unknowns.
Equations of Motion
- Apply Eq. 13–8.
Kinematics
- Determine r and the time derivatives \(\dot{r}, \ddot{r}, \dot{θ}, \ddot{θ}, \ddot{z}\), then evaluate \(a_r, a_θ, a_z\).
- A negative acceleration component indicates it acts in its negative coordinate direction.
- When differentiating \(r = f(θ)\), use the chain rule carefully.
Central-Force Motion and Space Mechanics
Central-Force Motion
If a particle moves under a single force whose line of action is always directed toward a fixed point, the motion is called central-force motion. It is commonly produced by electrostatic and gravitational forces, and it governs satellites, moons, and planetary orbits.
Using polar coordinates and writing the equation of motion for a particle acted on only by the central force F (directed toward the origin):
The second equation integrates to a constant of the motion — the basis of the equal-areas property.
Areal Velocity
The area swept by the radius as it turns through \(dθ\) is \(dA = \tfrac12 r^2 dθ\). The areal velocity \(dA/dt = \tfrac12 r^2 \dotθ = h/2\) is therefore constant: a particle in central-force motion sweeps out equal areas in equal times.
Eliminating time with the substitution \(\xi = 1/r\) transforms Eq. 13–10 into a differential equation for the path \(r = f(θ)\):
For gravitational attraction \(F = G M_e m / r^2\), the path equation becomes linear, and its solution is the equation of a conic section in polar form.
Eccentricity and the Directrix
A conic section is the locus of a point whose distance to a focus F bears a constant ratio e (the eccentricity) to its distance from a fixed line, the directrix. Comparing the conic equation with Eq. 13–15 identifies the distance from focus to directrix as \(p = 1/C\) and the eccentricity as \(e = C h^2 / (G M_e)\) (Eq. 13–17).
Measuring \(θ\) from the axis of symmetry sets \(\phi = 0\), so the trajectory reduces to:
The constants h and C follow from the launch conditions \(r_0\) and \(v_0\) at the start of free flight (\(v_0\) parallel to the surface, so it has no radial component):
Trajectory Type by Eccentricity (Eq. 13–22)
- \(e = 0\) — circle
- \(e = 1\) — parabola
- \(e < 1\) — ellipse
- \(e > 1\) — hyperbola
The escape velocity is the launch speed that places a satellite on a parabolic path — "on the border" of never returning. Launch speeds below \(v_c\) cause the satellite to reenter and burn up or crash.
Planets and most satellites follow elliptical orbits, with the earth at one focus. The closest approach (perigee) and farthest point (apogee) define the ellipse.
The elliptical nature of planetary orbits was discovered by Johannes Kepler after twenty years of observation — before Newton formulated his laws — so it later served as important confirmation of those laws. For a planet, the sun's mass \(M_s\) replaces \(M_e\) in the formulas.
Every planet moves in an ellipse with the sun located at one focus. (Stated mathematically by Eq. 13–21.)
The line joining a planet to the sun sweeps out equal areas in equal time intervals, regardless of its length. (Stated mathematically by Eq. 13–12.)
The square of the period of any planet is directly proportional to the cube of the major axis of its orbit.
For orbit problems, decide the trajectory type from e before grinding through algebra. Computing eccentricity first tells you whether to expect a bounded (circle/ellipse) or unbounded (parabola/hyperbola) path, and that sanity-checks every later number. Keep G, \(M_e\), and \(r_0\) in consistent SI units — mixing Mm and km is the most common source of wrong answers here.
Physical meaning
Newton's second law states that the vector imbalance of external forces changes velocity. Coordinate components are bookkeeping choices: normal force balance changes direction, tangential force balance changes speed, and neither acceleration component is an additional physical force.
Problem-solving strategy
- Isolate one particle or a clearly defined particle system and draw only its external forces.
- Select an inertial frame and coordinates aligned with known motion: rectangular, normal–tangential, or cylindrical.
- Obtain acceleration from kinematics or a stated constraint before writing independent components of \(\Sigma\mathbf F=m\mathbf a\).
- Solve symbolically for reactions, friction, or acceleration, retaining the assumed sign directions.
- Check contact reactions, friction inequalities, tension nonnegativity, units, and limiting cases before accepting the motion.
Recognition strategy
Forces and instantaneous acceleration are requested during known sliding, so a particle free-body diagram and rectangular components of Newton's second law are the direct method.
Given information
- Take \(+x\) in the pull direction and \(+y\) upward; \(m=18.0\ \text{kg}\), \(P=95.0\ \text{N}\), \(\theta=25.0^\circ\), and \(\mu_k=0.220\).
- The floor is horizontal, vertical acceleration is zero while contact persists, and kinetic friction opposes positive sliding.
Model and interpretation
Model the crate as a particle with weight, normal force, applied pull, and kinetic friction. The upward pull reduces the normal force, so friction must be computed only after vertical equilibrium.
Verification
Units: Newtons divided by kilograms reduce to m/s², and the vertical equation contains forces only, so both acceleration and reaction carry correct SI units.
Signs and direction: The pull's vertical component lowers the reaction below \(mg=176.58\ \text{N}\), while the positive horizontal resultant confirms acceleration in the assumed direction.
Plausibility: The normal force remains positive, so floor contact is admissible; removing friction would raise acceleration to 4.78 m/s², correctly above the calculated value.
Engineering interpretation
Handle angle changes both useful drive force and friction through the normal reaction. The result assumes steady kinetic friction; wheel compliance, floor variation, and operator force limits would matter in equipment selection.
Recognition strategy
The path curvature and instantaneous speed are known, so normal–tangential coordinates expose the centripetal acceleration and the separate braking acceleration without constructing rectangular motion equations.
Given information
- At the crest, take \(+n\) downward toward the center of curvature and \(+t\) in the direction of travel; \(m=1200\ \text{kg}\), \(v=18.0\ \text{m/s}\), and \(\rho=75.0\ \text{m}\).
- The speed is decreasing, so \(a_t=-1.80\ \text{m/s}^2\); contact requires the upward normal reaction to remain nonnegative.
Model and interpretation
Treat the vehicle as a particle. Weight acts in \(+n\), the road normal acts in \(-n\), and the road's tangential braking force acts opposite \(+t\). Rotation and suspension load distribution are outside this model.
Verification
Units: Each mass–acceleration product has newtons, and the resultant combines orthogonal force components with identical dimensions.
Signs and direction: The negative tangential result agrees with braking, while the upward normal reaction is smaller than weight because part of gravity supplies inward acceleration.
Plausibility: At zero speed the formula returns \(N=mg\); loss of contact would occur at \(\sqrt{g\rho}=27.1\ \text{m/s}\), safely above the stated speed.
Engineering interpretation
The crest unloads the tires while braking demands tangential grip, so available tire force is reduced compared with level travel. Suspension dynamics and axle load transfer require a rigid-body vehicle model and test data.
Recognition strategy
The assumed no-slip motion first requires a static-friction inequality, not automatic equality. If the required friction exceeds capacity, the motion must be reformulated with kinetic friction.
Given information
- Take \(+s\) uphill along the belt and \(+n\) normal away from it; \(m=32.0\ \text{kg}\), \(\beta=18.0^\circ\), and belt acceleration is \(+2.20\ \text{m/s}^2\).
- Coefficients are \(\mu_s=0.380\), \(\mu_k=0.280\); gravity is 9.81 m/s² and normal separation is not expected.
Model and interpretation
Model the package as a particle. Weight has a downhill component, so static friction would need to accelerate the package uphill; after slip begins, kinetic friction acts uphill because the belt moves uphill relative to the package.
Verification
Units: Friction and gravity components are newtons; dividing their resultant by 32.0 kg yields m/s², and each coefficient is dimensionless.
Signs and direction: The negative package acceleration is downhill under the uphill-positive convention; kinetic friction is nevertheless uphill because it opposes relative slipping, not necessarily absolute motion.
Plausibility: With zero friction the package would accelerate downhill at \(g\sin18^\circ=3.03\ \text{m/s}^2\); uphill kinetic friction reduces that magnitude to 0.419 m/s² without reversing the resultant.
Engineering interpretation
The belt command cannot be transmitted through available static friction, so package tracking fails immediately. Conveyor design should include surface variability, contamination, acceleration ramps, package geometry, and guarding review.
Chapter Review
Kinetics & the Equation of Motion
Kinetics relates forces to the accelerations they cause through Newton's second law, \(\Sigma\mathbf{F} = m\mathbf{a}\). Always draw the free-body diagram first to account for every force; graphically it equals the kinetic diagram, which shows the resultant ma vector.
Inertial Coordinate Systems
Acceleration must be measured in an inertial frame — axes that do not rotate and are either fixed or translate at constant velocity. The component forms are:
Central-Force Motion
When a single force acts on a particle — such as gravity during a satellite's free flight — the motion is central-force motion. The orbit depends on the eccentricity e, producing a circular, parabolic, elliptical, or hyperbolic trajectory.
The whole chapter is one idea applied four ways: \(\Sigma\mathbf{F} = m\mathbf{a}\) in rectangular, n–t, cylindrical, and central-force settings. Master the free-body-then-kinematics workflow once, and each coordinate system is just a different way to write the components. When unsure which system to use: rectilinear motion → rectangular; known curved path → n–t; angular-motion data or \(r=f(θ)\) → cylindrical.
Before moving on
- Draw an external-force free-body diagram and state an inertial coordinate system.
- Obtain acceleration kinematically, choose matching components, and solve \(\Sigma\mathbf F=m\mathbf a\).
- Check friction capacity, tension, contact reactions, units, signs, and limiting behavior.