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Newton's Laws in Dynamics

Specialist Mathematics
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Newton's Laws in Dynamics

Specialist Mathematics
01 May 2026

Newton’s Laws of Motion

The Three Laws

First Law (Law of Inertia):
An object at rest stays at rest, and an object in motion continues with constant velocity, unless acted on by a net external force.

Second Law:
$$\mathbf{F}_{\text{net}} = m\mathbf{a} = m\frac{d\mathbf{v}}{dt} = \frac{d(m\mathbf{v})}{dt}$$

In component form: $F_x = ma_x$, $F_y = ma_y$, $F_z = ma_z$.

Third Law (Action-Reaction):
If $A$ exerts force $\mathbf{F}$ on $B$, then $B$ exerts $-\mathbf{F}$ on $A$.

Common Forces

Force Symbol Direction Magnitude
Weight $W = mg$ Downward $mg$
Normal force $N$ Perpendicular to surface Determined by equilibrium
Tension $T$ Along string toward anchor Determined by equilibrium
Friction $f$ Opposing relative motion $\leq \mu N$ (kinetic: $= \mu_k N$)
Spring force $F = -kx$ Opposing extension $k

Standard Problem Types

Inclined plane: Block of mass $m$ on smooth incline at angle $\theta$.

Along the incline (down positive): $F = mg\sin\theta = ma \Rightarrow a = g\sin\theta$.

Normal to incline: $N = mg\cos\theta$.

Connected particles (Atwood machine): Masses $m_1 > m_2$ over a frictionless pulley.

System equation: $(m_1 - m_2)g = (m_1+m_2)a \Rightarrow a = \dfrac{(m_1-m_2)g}{m_1+m_2}$.

Tension: $T = \dfrac{2m_1 m_2 g}{m_1+m_2}$.

Example: $m_1 = 5$ kg, $m_2 = 3$ kg: $a = \dfrac{2 \times 9.8}{8} = 2.45$ m/s$^2$, $T = \dfrac{2\times 5\times3\times9.8}{8} = 36.75$ N.

Particle on incline with friction: $\mu_k = 0.3$, $\theta = 30^\circ$, $m = 2$ kg, moving down.

Friction acts up the incline:
$ma = mg\sin30^\circ - \mu_k mg\cos30^\circ = mg(\sin30^\circ - 0.3\cos30^\circ)$
$a = 9.8(0.5 - 0.3\times0.866) = 9.8(0.5-0.260) = 9.8\times0.240 \approx 2.35$ m/s$^2$.

Variable Force and ODE

When force is not constant, apply Newton’s second law as a differential equation:
$$m\frac{dv}{dt} = F(t, v, x)$$

Example: Particle of mass $1$ kg, resistive force $F = -2v$. Initial speed $10$ m/s.

$$\frac{dv}{dt} = -2v \Rightarrow v = 10e^{-2t}$$

$$x = \int_0^t 10e^{-2\tau}\,d\tau = 5(1-e^{-2t})$$

As $t\to\infty$, $v\to0$ and $x\to5$ m (particle travels a finite distance).

KEY TAKEAWAY: Newton’s second law $F = ma$ is the bridge between dynamics (forces) and kinematics (motion). In 2D/3D, apply it componentwise.

EXAM TIP: Always draw a free-body diagram before writing any equation. Label every force with its magnitude and direction. This prevents sign errors.

COMMON MISTAKE: Including reaction forces (from Newton’s third law) in the free-body diagram of a single object. Third-law pairs act on different objects.

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