AcademyEquilibrium and Materials
Academy
Conditions for Static Equilibrium
Level 1 - Physics topic page in Equilibrium and Materials.
Principle
A rigid body is static only when net force and net torque both vanish.
Notation
\(\sum\vec{F}\)
resultant external force
\(\mathrm{N}\)
\(\sum\vec{\tau}\)
resultant external torque
\(\mathrm{N\,m}\)
\(\vec{r}\)
position vector from pivot
\(\mathrm{m}\)
\(\vec{F}\)
applied force
\(\mathrm{N}\)
\(\theta\)
angle between lever arm and force
rad or deg
\(m,\vec{a}_{\mathrm{cm}}\)
body mass and center-of-mass acceleration
\(\mathrm{kg,\;m\,s^{-2}}\)
Method
Derivation 1: Remove translational acceleration
Static equilibrium first requires the body's center of mass to have no acceleration.
Newton condition
\[\sum\vec{F}=m\vec{a}_{\mathrm{cm}}\]
Static case
\[\vec{a}_{\mathrm{cm}}=\vec{0}\]
Force balance
\[\sum\vec{F}=\vec{0}\]
Derivation 2: Build the torque equation about a chosen pivot
Rotation depends on where the force acts, so the moment arm must be included.
Torque vector
\[\vec{\tau}=\vec{r}\times\vec{F}\]
Torque size
\[\tau=rF\sin\theta\]
Only the component perpendicular to the lever arm contributes.
The diagram uses the left support as the pivot. Forces through that point still affect force balance, but they create no torque about that point.
Derivation 3: Remove rotational acceleration
Torque balance
\[\sum\vec{\tau}=\vec{0}\]
Use one sign convention for clockwise and counterclockwise moments.
Rules
These are the compact results from the derivations above.
Force balance
\[\sum\vec{F}=\vec{0}\]
Torque balance
\[\sum\vec{\tau}=\vec{0}\]
Torque vector
\[\vec{\tau}=\vec{r}\times\vec{F}\]
Torque magnitude
\[\tau=rF\sin\theta\]
Examples
Question
A
\[20\,\mathrm{N}\]
force acts perpendicular to a \[0.50\,\mathrm{m}\]
handle. Find the torque magnitude.Answer
\[\tau=rF=0.50(20)=10\,\mathrm{N\,m}\]
Checks
- Zero net force does not prevent rotation.
- A force through the pivot has zero torque.
- Torque signs need one convention.
- Equilibrium is independent of pivot choice.