Moments and centres of mass of discrete massesEdexcel A-Level Further Maths: Revision notes

The moment of a force about a point is the force multiplied by the perpendicular distance from the point to the line of action of the force: $\text{moment}=F\times d.$ It is measured in N m and has a sense, clockwise or anticlockwise. A body is in equilibrium when (1) the resultant force is zero and (2) the sum of the moments about any point is zero. The weight of a body acts at its centre of mass. For a uniform rod this is the midpoint. Example: a force of 20 N acts at a perpendicular distance of 0.3 m from a pivot, so its moment is $20\times0.3=6$ N m.

The centre of mass of a system is the single point at which its total weight can be taken to act. For particles of masses $m_1,m_2,\dots$ at positions $x_1,x_2,\dots$ on a line, the moment of the total weight about the origin equals the sum of the moments of the separate weights: $(\sum m)g\bar x=\sum mgx$, so $\bar x=\frac{\sum mx}{\sum m}.$ Example: masses 2 kg, 3 kg, 5 kg at $x=1,4,9$ give $\bar x=\frac{2+12+45}{10}=5.9$ m. The centre of mass is nearer to the heavier particles, not at the mean of the positions. Adding or removing a particle: include (or leave out) its $m$ and $mx$ in both sums. A 10 kg particle added at $x=8$ gives $\bar x=\frac{59+80}{20}=6.95$ m.

Treat the $x$ and $y$ coordinates separately with the same formula: $\bar x=\frac{\sum mx}{\sum m},\qquad \bar y=\frac{\sum my}{\sum m}.$ Example: 1 kg, 2 kg, 3 kg, 4 kg at $(0,0)$, $(5,0)$, $(4,3)$, $(1,3)$. Then $\sum m=10$, $\sum mx=26$ and $\sum my=21$, so the centre of mass is $(2.6,\,2.1)$. Distances from the centre of mass to a particle use Pythagoras. The distance to the 2 kg particle is $\sqrt{2.4^2+2.1^2}=3.19$ m.

Problems often add or remove a particle, or give the centre of mass and ask for an unknown mass or position.

• Unknown mass: write $\bar x$ as an expression in $m$ and solve. If $\sum mx=5m$ and the total mass is $10+m$, then $\bar x=2.5$ gives $5m=25+2.5m$, so $m=10$.
• Removing a particle: subtract its $m$, $mx$ and $my$ from the totals.
• Condition on the centre of mass: for example it lies on $y=x$. Write $\bar x$ and $\bar y$ in terms of the unknown, then equate. Example: remove the particle at $C$ (leaving $\sum m=10$, $\sum mx=0$, $\sum my=-1$) and add $M$ kg at $(4,6)$. On $y=x$: $\frac{4M}{10+M}=\frac{6M-1}{10+M}$, so $M=0.5$.

A uniform rod of mass $M$ and length $L$ has its weight at its midpoint. If it rests horizontally on two supports, take moments to find the reactions, then resolve vertically to check. Example: a rod $AB$ of length 6 m and mass 12 kg rests on supports $C$ and $D$ with $AC=1$ m and $DB=2$ m. Moments about $C$: $3R_D=12g\times2$, so $R_D=78.4$ N and $R_C=39.2$ N. A rod is about to tilt about a support when the reaction at the other support becomes zero. With a particle of mass $m$ at $B$ and $R_C=0$, moments about $D$ give $12g\times1=mg\times2$, so $m=6$.