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3.2 Dynamics

Key Takeaways

  • Constant-acceleration kinematics: v = v₀ + at, s = s₀ + v₀t + ½at², and v² = v₀² + 2a(s − s₀).
  • Newton's second law for a particle is ΣF = ma; for rigid-body rotation about the mass center ΣM = Iα.
  • Work–energy: the net work done equals the change in kinetic energy, W = ΔKE = ½m(v₂² − v₁²).
  • Linear impulse–momentum: ΣF·Δt = m(v₂ − v₁); momentum is conserved when no external impulse acts.
  • Projectile range on level ground is R = v₀² sin(2θ)/g, maximized at a launch angle of 45°.
Last updated: May 2026

Picking the Right Method

Dynamics is worth 4–6 FE Civil questions. The exam rewards method selection: use kinematics when motion is asked, Newton's second law when forces and acceleration link, work–energy when speed relates to distance, and impulse–momentum when speed relates to time or collisions.

Kinematics

For rectilinear motion with constant acceleration a:

EquationSolves for
v = v₀ + atvelocity from time
s = s₀ + v₀t + ½at²position from time
v² = v₀² + 2a(s − s₀)velocity from distance

If acceleration varies, integrate: v = ∫a dt and s = ∫v dt. For curvilinear motion, use normal/tangential components: tangential aₜ = dv/dt and normal aₙ = v²/ρ, where ρ is the radius of curvature.

Projectile Motion

Horizontal velocity is constant; vertical motion has a = −g (9.81 m/s² or 32.2 ft/s²). On level ground:

R=v02sin2θgR = \frac{v_0^2 \sin 2\theta}{g}

Maximum range occurs at θ = 45°. Time of flight is t = 2v₀sinθ/g.

Newton's Laws

Newton's second law for a particle is ΣF = ma. Build an FBD exactly as in statics, but the net force is no longer zero — it equals mass times acceleration. For connected bodies (pulleys, blocks), write ΣF = ma for each mass and link them through a common acceleration.

For rigid-body plane motion: ΣF = ma_G at the mass center and ΣM_G = I_G α for rotation, where I_G is the mass moment of inertia and α is angular acceleration.

Work and Energy

The work–energy principle avoids solving for acceleration:

Wnet=ΔKE=12mv2212mv12W_{net} = \Delta KE = \tfrac{1}{2}m v_2^2 - \tfrac{1}{2}m v_1^2

  • Work of a constant force: W = F·d·cosθ
  • Work of gravity: W = −mgΔh (potential energy change)
  • Work of a spring: W = ½k(x₁² − x₂²)

When only conservative forces act, mechanical energy is conserved: KE₁ + PE₁ = KE₂ + PE₂. Friction removes energy as heat.

Impulse and Momentum

Linear impulse–momentum links force and time:

FΔt=mv2mv1\sum F \, \Delta t = m v_2 - m v_1

When the net external impulse is zero, linear momentum is conserved — the workhorse for collision problems. For a perfectly elastic collision, kinetic energy is also conserved; for a perfectly plastic (inelastic) collision the bodies move together afterward. The coefficient of restitution e relates separation to approach velocity, with e = 1 (elastic) and e = 0 (plastic).

Vibrations Basics

An undamped simple harmonic oscillator (mass m, spring k) has natural angular frequency:

ωn=km,f=ωn2π,T=1f\omega_n = \sqrt{\frac{k}{m}}, \qquad f = \frac{\omega_n}{2\pi}, \qquad T = \frac{1}{f}

A simple pendulum of length L has ωₙ = √(g/L). These free-vibration relations appear in the Dynamics section of the NCEES FE Reference Handbook — locate them before exam day so you only verify, not derive.

Test Your Knowledge

A car accelerates uniformly from rest and reaches 30 m/s after traveling 150 m. What is its acceleration?

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Test Your Knowledge

A 1,500 kg car traveling at 20 m/s collides with and sticks to a stationary 1,000 kg car. What is their common velocity immediately after impact?

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