When several forces act on a rigid body at once, their turning effects simply add — clockwise moments one way, anticlockwise the other. Summing them, with a consistent sign for each sense, tells you the net turning effect of the whole system.
12 min read4 worked examples
Adding moments together
When several forces act on a rigid body at once, each has its own moment about a chosen point. The resultant moment is simply their sum, taking a consistent sign for each sense of turning — say, anticlockwise positive, clockwise negative. If the resultant moment about a point is zero, the body has no tendency to rotate about that point; if it isn't zero, the body will begin to turn in whichever sense the resultant acts.
The method for several forces on a rod or lamina
(1) Choose a point to take moments about, and a positive sense (usually anticlockwise). (2) Find the moment of each force separately, using $F \times$ (perpendicular distance), giving each the correct sign. (3) Add them all together to get the resultant moment.
Forces that aren't perpendicular to the body
A system often mixes forces acting straight on (perpendicular to the rod or edge they act on) with forces acting at an angle. For an angled force, use the result from the previous lesson: the moment is $Fd\sin\theta$, where $d$ is the distance from the pivot to the point of application and $\theta$ is the angle between the force and the object. Every force in the system — perpendicular or angled — contributes one signed term to the same sum.
Worked example 1
Several forces on a rod
A rod $AB$ of length $5\ \text{m}$ has a force of $8\ \text{N}$ acting perpendicular to it, $2\ \text{m}$ from $A$, turning it anticlockwise, and a force of $6\ \text{N}$ acting perpendicular to it, $4\ \text{m}$ from $A$, turning it clockwise. Find the resultant moment about $A$.
MethodSum the individual moments, with sign for sense
1
Taking anticlockwise as positive: moment of the first force $= +8 \times 2 = +16\ \text{N m}$.
Already anticlockwise, so positive.
2
Moment of the second force $= -6 \times 4 = -24\ \text{N m}$.
Clockwise, so negative in this convention.
3
Resultant moment $= 16 + (-24) = -8\ \text{N m}$, i.e. $8\ \text{N m}$ clockwise.
Sum the signed moments.
▪ solved
Worked example 2
An angled force in the system
A rod $AB$ of length $6\ \text{m}$ is pivoted at $A$. A force of $10\ \text{N}$ acts perpendicular to the rod, $2\ \text{m}$ from $A$, turning it anticlockwise. A second force of $18\ \text{N}$ acts at $35°$ to the rod, $5\ \text{m}$ from $A$, turning it clockwise. Find the resultant moment about $A$.
MethodFind each moment separately, using Fd sin θ for the angled force
1
Taking anticlockwise as positive: moment of the first force $=+10\times 2=+20\ \text{N m}$.
Already perpendicular to the rod, and anticlockwise.
2
Moment of the second force: $Fd\sin\theta = 18\times 5\times\sin 35° = 90\times 0.5736=51.6\ \text{N m}$ (3 s.f.), clockwise, so $-51.6\ \text{N m}$.
Only the component perpendicular to the rod contributes.
3
Resultant moment $= 20+(-51.6)=-31.6\ \text{N m}$, i.e. $31.6\ \text{N m}$ clockwise (3 s.f.).
Sum the two signed moments.
▪ solved
Worked example 3
A system of forces on a rectangular lamina
A rectangular lamina $ABCD$ has $AB=6\ \text{m}$ and $AD=4\ \text{m}$, and is pivoted at $A$. A vertical force of $15\ \text{N}$ acts downward at $B$, turning the lamina clockwise. A horizontal force of $10\ \text{N}$ acts at $D$, turning it anticlockwise. A vertical force of $8\ \text{N}$ acts downward at $C$ (diagonally opposite $A$), turning it clockwise. Find the resultant moment about $A$.
MethodFor each force, the moment arm is whichever side length is perpendicular to it
1
The force at $B$ is vertical, so its moment arm about $A$ is the horizontal distance $AB=6\ \text{m}$. Moment $=-15\times 6=-90\ \text{N m}$ (clockwise).
A vertical force's turning effect depends only on horizontal distance from the pivot.
2
The force at $D$ is horizontal, so its moment arm about $A$ is the vertical distance $AD=4\ \text{m}$. Moment $=+10\times 4=+40\ \text{N m}$ (anticlockwise).
A horizontal force's turning effect depends only on vertical distance from the pivot.
3
The force at $C$ is vertical, so its moment arm about $A$ is again the horizontal distance, which is $AB=6\ \text{m}$ (since $C$ is directly above $B$). Moment $=-8\times 6=-48\ \text{N m}$ (clockwise).
$C$'s horizontal offset from $A$ equals $B$'s, because $BC$ is vertical.
4
Resultant moment $=-90+40-48=-98\ \text{N m}$, i.e. $98\ \text{N m}$ clockwise.
Sum all three signed moments.
▪ solved
No need for coordinates or components
Because the lamina is rectangular, the perpendicular distance for a vertical force is always a horizontal side length, and for a horizontal force it's always a vertical side length — read straight off the shape, without setting up any formal coordinate system.
Worked example 4
Finding an unknown distance for equilibriumChallenge
A rod $AB$ of length $6\ \text{m}$ is pivoted at $A$. A force of $30\ \text{N}$ acts perpendicular to the rod, $1.5\ \text{m}$ from $A$, turning it clockwise. A force of $25\ \text{N}$ acts at $50°$ to the rod, at a distance $x$ metres from $A$, turning it anticlockwise. Given that the rod is in equilibrium, find $x$.
MethodSet the resultant moment to zero and solve for x
1
Moment of the first force (clockwise): $-30\times 1.5=-45\ \text{N m}$.
Perpendicular force, direct substitution.
2
Moment of the second force (anticlockwise): $+25\sin 50°\times x = 19.15x\ \text{N m}$ (4 s.f.).
Only the component of the $25\ \text{N}$ force perpendicular to the rod contributes.
3
Equilibrium means the resultant moment is zero: $-45+19.15x=0$.
Rotational equilibrium: the two moments must exactly cancel.
4
$x = \dfrac{45}{19.15}=2.35\ \text{m}$ (3 s.f.).
Solve for $x$.
▪ solved
The same idea, run in reverse
Every earlier example summed known moments to find a resultant. Here the resultant is given (zero, since the rod is in equilibrium), and one piece of information — the position of the second force — is unknown instead. Setting up exactly the same equation and solving for the unknown is the same technique, just rearranged.
Common mistake
Losing track of sign conventions when several forces are involved — mixing clockwise-positive for one force and anticlockwise-positive for another in the same calculation. Fix one sense as positive at the very start of a question, write it down, and apply it consistently to every single force in the system, including any acting at an angle.
Check yourself
A rod $AB$ of length $4\ \text{m}$ is pivoted at $A$. A force of $14\ \text{N}$ acts perpendicular to the rod, $1\ \text{m}$ from $A$, turning it anticlockwise. A force of $9\ \text{N}$ acts perpendicular to the rod at $B$ (4 m from $A$), turning it clockwise. Find the resultant moment about $A$, stating its sense.