A body on the point of toppling about an edge is a moments problem in disguise: at the exact instant it would tip, the reaction at every other support has dropped to zero.
12 min read4 worked examples
What tilting means
A body resting on more than one support can fail to stay flat in two different ways: it can slide, or it can tilt — rotate about one support as it lifts away from the others. Just as sliding is on the point of happening when friction reaches its maximum, tilting is on the point of happening when the reaction at the support the body is lifting away from drops to exactly zero.
The key idea
At the moment a body is about to tilt about a particular point, the normal reaction at every other support is zero — all of the body's weight is momentarily supported at the tilting point alone. Setting that reaction to zero and taking moments about the tilting point turns "on the point of tilting" into an ordinary moments equation.
The method
Step by step
(1) Identify the point the body would rotate about if it tilted — this is the support closest to the force or load causing the tilt. (2) Set the reaction at every other support to zero. (3) Take moments about the tilting point, using the remaining forces (weight, applied load) — the reaction you set to zero simply drops out, since it now contributes nothing. (4) Solve for whatever is asked: a critical distance, force, or mass.
Worked example 1
A load moving along a beam on two supports
A uniform beam $AB$ of length $8\ \text{m}$ and weight $60\ \text{N}$ rests horizontally on two supports, at $A$ and at a point $C$, $2\ \text{m}$ from $B$. A load of $90\ \text{N}$ is placed on the beam and moved slowly from $A$ toward $B$. Find the distance from $A$ at which the beam is on the point of tilting.
MethodTilting happens about C, the support nearer to B; set the reaction at A to zero
1
As the load moves toward $B$, the beam would tip up at $A$ and rotate about $C$ (the support closer to where the load ends up), so $C$ is the tilting point.
The support furthest from the load stays in contact longest; tilting occurs about the nearer one.
2
$C$ is at $8-2=6\ \text{m}$ from $A$. On the point of tilting, the reaction at $A$ is zero.
All support is momentarily at $C$ alone.
3
Take moments about $C$. Let $x$ be the load's distance from $A$ at the critical point, so its distance from $C$ is $6-x$. The beam's own weight acts at its midpoint, $4\ \text{m}$ from $A$, i.e. $4-6=-2\ \text{m}$ from $C$ (on the $A$ side).
Distances are measured from the new pivot, $C$.
4
Moments about $C$: the load ($90\ \text{N}$, on the $B$ side of $C$) turns the beam one way; the beam's weight ($60\ \text{N}$, on the $A$ side of $C$) turns it the other way. Balance: $90(6-x) = 60(2)$.
Principle of moments, at the critical (limiting) position.
5
$540-90x=120 \Rightarrow 90x=420 \Rightarrow x=4.67\ \text{m}$ (3 s.f.) from $A$.
Solve for $x$.
▪ solved
Worked example 2
Walking along a cantilevered plank
A uniform plank $AB$ of length $6\ \text{m}$ and weight $80\ \text{N}$ rests on a support at $A$ and on the edge of a wall at $C$, $1\ \text{m}$ from $A$, with the rest of the plank overhanging beyond $C$. A person of weight $650\ \text{N}$ walks from $A$ toward $B$. Find how far beyond $C$ the person can walk before the plank tilts.
MethodTilting is about C; set the reaction at A to zero
1
As the person walks past $C$ onto the overhang, the plank would tip up at $A$ and rotate about $C$. On the point of tilting, the reaction at $A$ is zero.
$C$ is the only support left bearing weight at the critical instant.
2
The beam's weight acts at its midpoint, $3\ \text{m}$ from $A$, i.e. $3-1=2\ \text{m}$ from $C$, on the $A$ side.
Midpoint of a $6\ \text{m}$ plank measured from $A$, then re-measured from $C$.
3
Let $d$ be the person's distance beyond $C$ at the critical point. Taking moments about $C$: the person's weight (beyond $C$) balances the plank's weight (on the $A$ side of $C$): $650d = 80(2)$.
Principle of moments at the tilting point.
4
$d = \dfrac{160}{650}=0.246\ \text{m}$ (3 s.f.).
Solve for $d$.
▪ solved
A useful sanity check
A relatively heavy person only needs to step a small distance past $C$ before the light plank tips — $0.246\ \text{m}$, less than a quarter of a metre. That makes physical sense: the person's weight ($650\ \text{N}$) is much larger than the plank's ($80\ \text{N}$), so it takes only a small moment arm on the person's side to outweigh a much larger moment arm on the plank's own, lighter weight.
Worked example 3
The force needed to tip a table
A uniform table can be modelled as a rod $PQ$ of length $1.2\ \text{m}$ and weight $200\ \text{N}$, standing on two legs at $P$ and $Q$. The tabletop overhangs $0.3\ \text{m}$ beyond $Q$, to a point $R$. A downward force $F$ is applied at $R$. Find the value of $F$ at which the table is on the point of tilting about $Q$.
MethodSet the reaction at P to zero, then take moments about Q, the pivot still in contact
1
A downward force beyond $Q$ tends to rotate the table about $Q$, lifting $P$ off the ground. On the point of tilting, the reaction at $P$ is exactly zero.
$P$ is the leg losing contact, so its reaction vanishes at the critical instant.
2
Take moments about $Q$ — the leg that stays in contact, and so the true pivot. The weight ($200\ \text{N}$) acts at the rod's midpoint, $0.6\ \text{m}$ from $Q$, on the $P$ side.
Moments are always taken about the point the body actually rotates around.
3
$F$ acts at $R$, a distance $0.3\ \text{m}$ from $Q$, on the opposite side. At the point of tilting, $F$'s moment about $Q$ exactly balances the weight's: $F(0.3) = 200(0.6)$.
Principle of moments about $Q$.
4
$F = \dfrac{120}{0.3} = 400\ \text{N}$.
Solve for $F$.
▪ solved
Always take moments about the point still on the ground
The pivot for a tilting problem is always the support the body is rotating around, not the one losing contact — here, $Q$, not $P$. It's a point genuinely still touching the ground, with a real (if momentarily unknown) reaction of its own, which is exactly why taking moments about it is safe: that unknown reaction, acting exactly at the pivot, automatically drops out of the equation.
Worked example 4
A range of positions that avoid tiltingChallenge
A uniform beam $AB$ of length $10\ \text{m}$ and weight $120\ \text{N}$ rests on two supports, at $C$ ($2\ \text{m}$ from $A$) and $D$ ($3\ \text{m}$ from $B$). A load of $200\ \text{N}$ is placed on the beam at a distance $x$ metres from $A$. Find the range of values of $x$ for which the beam does not tilt.
MethodFind the two critical positions — tilting about C, and tilting about D — and combine
1
$C$ is at $2\ \text{m}$ from $A$; $D$ is at $10-3=7\ \text{m}$ from $A$. The beam's weight acts at its midpoint, $5\ \text{m}$ from $A$.
Locate every relevant point from a common origin, $A$.
2
Tilting about C (load too far toward $A$, lifting $D$): reaction at $D$ is zero. Taking moments about $C$: the beam's weight is $5-2=3\ \text{m}$ from $C$ on the $B$ side; the load is $(2-x)$ from $C$ on the $A$ side, for $x<2$.
Set up the pivot at $C$ for this case.
3
Balance: $200(2-x) = 120(3) \Rightarrow 2-x = 1.8 \Rightarrow x=0.2\ \text{m}$. So the beam tilts about $C$ if $x<0.2$.
Solve the critical (limiting) case.
4
Tilting about D (load too far toward $B$, lifting $C$): reaction at $C$ is zero. Taking moments about $D$ (at $7\ \text{m}$ from $A$): the beam's weight is $7-5=2\ \text{m}$ from $D$ on the $A$ side; the load is $(x-7)$ from $D$ on the $B$ side, for $x>7$.
Set up the pivot at $D$ for this case.
5
Balance: $200(x-7) = 120(2) \Rightarrow x-7=1.2 \Rightarrow x=8.2\ \text{m}$. So the beam tilts about $D$ if $x>8.2$.
Solve the second critical case.
6
Combining both conditions, the beam does not tilt for $0.2 \le x \le 8.2$.
Between the two critical positions, both reactions stay non-negative.
▪ solved
Two separate tilting conditions
Whenever a body rests on two supports and a load can move freely along it, there are always two ways it can tilt — about whichever support the load moves away from. Each gives its own limiting equation, and the safe range for the load is always the interval between the two critical positions found, exactly as in a compound inequality.
Common mistake
Taking moments about the wrong point — the support that's losing contact, rather than the one still touching the ground, which is the true pivot. Setting a reaction to zero only tells you which force to drop from your equation; the point you take moments about is always the support the body is rotating around, which still has a (possibly unknown, but non-zero) reaction of its own.
Check yourself
A uniform beam $AB$ of length $5\ \text{m}$ and weight $40\ \text{N}$ rests on two supports, at $A$ and at $C$, $1\ \text{m}$ from $B$. A load of $70\ \text{N}$ is moved from $A$ toward $B$. Find the distance from $A$ at which the beam is on the point of tilting.
$C$ is $4\ \text{m}$ from $A$. Tilting occurs about $C$, with the reaction at $A$ zero. Beam's weight acts $2.5\ \text{m}$ from $A$, i.e. $4-2.5=1.5\ \text{m}$ from $C$.
Taking moments about $C$: $70(4-x) = 40(1.5) \Rightarrow 4-x=0.857 \Rightarrow x=3.14\ \text{m}$ (3 s.f.) from $A$.