Kinetics of Particles — Newton's Laws & D'Alembert's Principle — Engineering Mechanics Notes

Last reviewed 16 Sept 2026 · 5 min read

Newton's laws of motion

First law (inertia): a body stays at rest or in uniform motion in a straight line unless acted on by an unbalanced external force.
Second law: the rate of change of momentum is proportional to the applied force and takes place in its direction: $F = \frac{d ( m v )}{d t} = ma$ for constant mass.
Third law: to every action there is an equal and opposite reaction (acting on the other body).

Units: 1 newton (N) is the force that gives 1 kg an acceleration of 1 m/s². Weight $W = m g$ ; mass is constant, weight changes with $g$ .

Equation of motion

\sum F = m a

Draw the free-body diagram, choose the direction of motion as positive, include every force (applied loads, weight, normal reaction, friction, tension), then apply $\sum F = ma$ along the direction of motion and $\sum F = 0$ perpendicular to it.

D'Alembert's principle

Rewrite $\sum F = ma$ as $\sum F - ma = 0$ . The term $- ma$ is an imaginary inertia force acting opposite to the acceleration.

FormulaD'Alembert's principle

A body in accelerated motion can be treated as if it were in equilibrium under the real forces plus an inertia force $m a$ applied opposite to the acceleration:

\sum F + (- ma) = 0

This turns dynamics problems into statics problems (dynamic equilibrium) — useful for connected bodies and rotating systems.

Lift (elevator) problems

A person of mass $m$ stands on a weighing machine in a lift. The reading equals the normal reaction $R$ :

Motion of lift	Reaction (apparent weight)
At rest or uniform velocity	$R = m g$
Accelerating upward (or decelerating while moving down)	$R = m (g + a)$
Accelerating downward (or decelerating while moving up)	$R = m (g - a)$
Free fall ( $a = g$ )	$R = 0$ (weightlessness)

Cable tension for a lift of mass $M$ follows the same pattern: $T = M (g \pm a)$ .

Connected bodies

Two masses over a smooth pulley (Atwood machine)

Masses $m_{1} > m_{2}$ hang over a frictionless, massless pulley:

a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}, T = \frac{2 m _{1} m _{2} g}{m _{1} + m _{2}}

The pulley support carries $2 T$ .

Mass on a table pulling a hanging mass

Mass $m_{1}$ on a horizontal table with friction coefficient $μ$ , connected over a pulley to $m_{2}$ hanging:

a = \frac{( m _{2} - μ m _{1} ) g}{m _{1} + m _{2}}, T = m_{1} (a + μg)

Mass on an inclined plane connected to a hanging mass

With $m_{1}$ on a plane at $α$ (moving up the plane), $m_{2}$ hanging:

a = \frac{[ m _{2} - m _{1} ( sin α + μ cos α ) ] g}{m _{1} + m _{2}}

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