AcademyEquilibrium and Materials
Academy
Center of Gravity
Level 1 - Physics topic page in Equilibrium and Materials.
Principle
The center of gravity is the point where a body's total weight can be treated as acting.
Notation
\(x_{\mathrm{cg}}\)
center-of-gravity coordinate
\(\mathrm{m}\)
\(W\)
total weight
\(\mathrm{N}\)
\(M\)
total mass
\(\mathrm{kg}\)
\(g\)
gravitational field strength
\(\mathrm{m\,s^{-2}}\)
\(m_i,dm\)
discrete mass or small mass element
\(\mathrm{kg}\)
\(dW\)
small weight element
\(\mathrm{N}\)
\(x_i\)
mass coordinate
\(\mathrm{m}\)
\(x_{\mathrm{left}},x_{\mathrm{right}}\)
left and right support limits
\(\mathrm{m}\)
Method
Derivation 1: Replace many weights with one equivalent weight
In a uniform gravitational field each small mass element contributes a downward force proportional to its mass, so the total weight is still proportional to total mass.
Element weight
\[dW=g\,dm\]
Total weight
\[W=\int dW=g\int dm=Mg\]
Match moments about the origin
\[Wx_{\mathrm{cg}}=\int x\,dW\]
Substitute the element weight
\[Mgx_{\mathrm{cg}}=g\int x\,dm\]
Continuous result
\[x_{\mathrm{cg}}=\frac{1}{M}\int x\,dm\]
Derivation 2: Reduce the result to discrete masses
For separated masses, the integral becomes a sum of mass-weighted positions.
Total mass
\[M=\sum_i m_i\]
Torque match
\[Mgx_{\mathrm{cg}}=\sum_i m_i g x_i\]
Discrete result
\[x_{\mathrm{cg}}=\frac{\sum_i m_i x_i}{\sum_i m_i}\]
Derivation 3: Test whether the body can remain supported
The support sketch shows the practical condition: the vertical weight line must pass through the support base.
Left edge condition
\[x_{\mathrm{cg}}\ge x_{\mathrm{left}}\]
Right edge condition
\[x_{\mathrm{cg}}\le x_{\mathrm{right}}\]
Support test
\[x_{\mathrm{left}}\le x_{\mathrm{cg}}\le x_{\mathrm{right}}\]
Outside this interval the weight produces an unbalanced tipping moment.
Rules
These are the compact results from the derivations above.
Discrete masses
\[x_{\mathrm{cg}}=\frac{\sum_i m_i x_i}{\sum_i m_i}\]
Continuous body
\[x_{\mathrm{cg}}=\frac{1}{M}\int x\,dm\]
Total weight
\[W=Mg\]
Support condition
\[x_{\mathrm{left}}\le x_{\mathrm{cg}}\le x_{\mathrm{right}}\]
Examples
Question
Masses
\[2\,\mathrm{kg}\]
and \[6\,\mathrm{kg}\]
sit at \[x=0\]
and \[x=4\,\mathrm{m}\]
Find \[x_{\mathrm{cg}}\]
Answer
\[x_{\mathrm{cg}}=\frac{2(0)+6(4)}{8}=3.0\,\mathrm{m}\]
Checks
- Weight positions are mass weighted.
- In uniform gravity, center of gravity and center of mass coincide.
- Symmetry works only with symmetric density.
- Tipping starts outside the support base.