AcademyForces and Newton's Laws
Academy
Equilibrium and Inertia
Level 1 - Physics topic page in Forces and Newton's Laws.
Principle
In an inertial frame, zero resultant force means zero acceleration, not necessarily rest.
Notation
\(\sum\vec{F}\)
resultant external force
\(\mathrm{N}\)
\(\vec{a}\)
acceleration
\(\mathrm{m\,s^{-2}}\)
\(\vec{v}\)
velocity
\(\mathrm{m\,s^{-1}}\)
\(T\)
tension magnitude
\(\mathrm{N}\)
\(N\)
normal reaction magnitude
\(\mathrm{N}\)
\(W\)
weight magnitude
\(\mathrm{N}\)
Method
Equilibrium is the zero-acceleration case of the force model, so each component sum must vanish.
Start from dynamics
\[\sum\vec{F}=m\vec{a}\]
Set acceleration
\[\vec{a}=\vec{0}\]
Rest and constant velocity both satisfy this.
Balance axes
\[\sum F_x=0,\qquad \sum F_y=0\]
Solve support forces
\[F_{\mathrm{unknown}}\ \text{makes the relevant component sum zero}\]
The free-body diagram below shows a body at rest with two nonzero forces whose vertical components cancel.
The diagram is not force-free; it is resultant-free.
Rules
These are the compact equilibrium statements for an inertial frame.
First law
\[\sum\vec{F}=\vec{0}\Rightarrow\vec{a}=\vec{0}\]
Component balance
\[\sum F_x=0,\qquad \sum F_y=0\]
Constant velocity
\[\vec{v}=\text{constant}\]
Rest case
\[\vec{v}=\vec{0}\ \text{is one equilibrium case}\]
Examples
Question
A
\[6.0\,\mathrm{kg}\]
lamp hangs motionless from one vertical cable. Find the cable tension using \[g=9.8\,\mathrm{m\,s^{-2}}\]
Answer
Vertical equilibrium gives
\[\sum F_y=T-mg=0\]
so \[T=mg=6.0(9.8)=58.8\,\mathrm{N}\]
Checks
- Zero resultant force means constant velocity, including rest.
- Balance each component separately.
- Support forces are found from force balance; they are not automatically equal to weight.
- Static friction adjusts to the value required for equilibrium until its limit is reached.