R6, S1MechanicsStretch

Ladders, Contact Forces and Limiting Equilibrium

A ladder can satisfy horizontal and vertical force balance and still rotate. Treat it as a rigid body, draw every contact force, and combine force equations with a moment equation. Friction supplies a bound; it reaches its maximum only at impending slip.

45 min Moments

The big picture

Ladder problems combine three ideas: a uniform body has weight acting at its midpoint, a contact can provide a normal force and sometimes friction, and static equilibrium needs zero resultant force and zero resultant moment. Choosing a pivot through two unknown forces removes both from the moment equation. This leaves a small, interpretable system rather than a formula to memorise.

What you'll be able to do

Draw force diagrams for a uniform ladder on rough ground against a smooth wall
Use perpendicular distances and a convenient pivot
Apply friction as an inequality except at limiting equilibrium
Find a limiting coefficient or the maximum safe position of a person under the model

Start from a physical diagram

For a ladder against a smooth vertical wall, the wall reaction is horizontal. At rough horizontal ground, the normal reaction is vertical and friction is horizontal. A uniform ladder’s weight acts vertically at its midpoint. A person is modelled as a particle with weight acting at their position along the ladder.

If the foot tends to slide away from the wall, ground friction acts towards the wall. A smooth wall has no friction force. A normal contact force must push, not pull: a computed negative reaction signals a loss of contact or an inconsistent assumption.

Model assumptions should explain the force diagram: rigid ladder, uniform mass distribution, smooth wall, rough ground and a static person. The model ignores bending and movement, so its limiting result is not a real-world safety recommendation.

Example 1A ladder leans against a smooth wall and rests on rough horizontal ground. How many support forces belong on its diagram?

1One horizontal normal reaction at the wall.

2One vertical ground normal reaction at the foot.

3One horizontal friction force at the foot.

4Add the ladder weight, and any person’s weight, separately.

AnswerThree support-force components. There is no wall friction in this model.

Choose the foot as the pivot

Let the ladder have length $L$ and angle $\theta$ above the ground. The wall reaction $S$ acts at height $L\sin\theta$ , so its moment about the foot is $SL\sin\theta$ .

The ladder weight $W$ acts at the midpoint, whose horizontal distance from the foot is $(L/2)\cos\theta$ . A person's weight $P$ at distance $d$ along the ladder has lever arm $d\cos\theta$ .

Both ground forces act through the foot, so they have zero moment about it. Taking moments at the foot eliminates two unknowns at once.

SL\sin\theta=W\frac L2\cos\theta+Pd\cos\theta

Moment balance about the foot, for the stated model.

Example 1A uniform 5 m ladder of weight 200 N leans at

60^\circ

against a smooth wall, with no person on it. Find the wall reaction.

1Take moments about the foot:

S(5\sin60^\circ)=200(2.5\cos60^\circ)

S=100\cot60^\circ=100/\sqrt3\approx57.735

3The reaction is positive, consistent with the wall pushing the ladder away.

Answer

S=57.7

N (3 s.f.).

Force balance and the friction bound

Vertical equilibrium gives $R=W+P$ . Horizontal equilibrium gives friction magnitude $F=S$ . Static friction satisfies $F\leq\mu R$ . The equality $F=\mu R$ applies only at limiting equilibrium, when slip is about to occur.

For a known ladder position, the equilibrium equations tell you the friction required. Compare it with the maximum available friction. If the required value exceeds that maximum, static equilibrium is impossible under this model.

Example 1For the unloaded ladder above, find the least coefficient of ground friction that permits equilibrium. Would

\mu=0.4

work?

R=200

N and

F=S=100/\sqrt3

2Require

\mu\geq F/R=(100/\sqrt3)/200=1/(2\sqrt3)\approx0.288675

3If

\mu=0.4

, the maximum is

\mu R=80

N, which exceeds the required

57.735

Answer

\mu_{\min}\approx0.289

. At

\mu=0.4

, actual friction remains

57.7

N, not 80 N.

A person climbs: solve the limiting condition

As a person moves farther along the ladder, their weight has a larger moment about the foot. The required wall reaction and friction increase. At the limiting position use friction equality, then solve the moment equation for distance along the ladder.

State whether the distance is along the ladder, horizontal or vertical. Check that the answer lies between zero and the ladder length. A result above the ladder length means the person can reach its top under the idealised static model, rather than a physical position beyond the top.

Example 1A 5 m uniform ladder weighs 200 N and makes

60^\circ

with rough horizontal ground against a smooth wall. A 600 N person stands d metres along it. If

\mu=0.4

, find the limiting d.

R=200+600=800

N. At impending slip,

F=0.4(800)=320

N, so

S=320

2Moments about the foot:

320(5\sin60^\circ)=200(2.5\cos60^\circ)+600d\cos60^\circ

3Thus

800\sqrt3=250+300d

d=(800\sqrt3-250)/300\approx3.78547

m. This lies on the 5 m ladder.

AnswerThe limiting position is about 3.79 m along the ladder. Under the model, smaller distances require less friction.

Critique and extend the model

If the wall were rough, it could supply vertical friction as well as a horizontal normal reaction. You must draw the additional force and establish its direction from the tendency to slip; the smooth-wall equations would no longer apply.

A non-uniform ladder need not have its centre of mass at the midpoint. If its centre of mass is given at a distance c along the ladder, replace the midpoint lever arm by c cos theta.

Moment equilibrium is not optional because the forces balance. Conversely, balanced moments about one point do not guarantee equilibrium without the force equations. Use all three independent planar conditions.

Tip — Label lever arms on a sketch. Vertical weights use horizontal distances; horizontal reactions use vertical distances.

Think like an examiner

Show weight acting at the midpoint only when the ladder is uniform.
Use perpendicular lever arms and identify the pivot.
Use equality in the friction law only at the limiting position.
Check the location and sign of the result against the physical model.

Common misconceptions

Using the entire ladder length as the lever arm for its weight.

Use the horizontal distance from the foot to its centre of mass.

Setting friction to mu R in every static situation.

Friction adjusts up to its maximum; equality needs impending slip.

Adding vertical wall friction at a smooth wall.

A smooth vertical wall supplies only a horizontal normal reaction.

Key results to remember

R=W+P,\qquad F=S,\qquad F\leq\mu R

Force balance and the static friction bound for a smooth wall.

\sum M=0,\quad\sum F_x=0,\quad\sum F_y=0

Three conditions for planar rigid-body equilibrium.

Stretch yourself

Show that an unloaded uniform ladder against a smooth wall requires $\mu\geq\tfrac12\cot\theta$ . What changes as the ladder becomes steeper?

Hint — Use moments at the foot, then divide the required friction by the ground reaction.

Key takeaways

Draw contacts and weights before writing equations.
Choose a pivot that removes unknown reactions.
Combine moments with horizontal and vertical force balance.
Use friction equality only at limiting equilibrium and check the domain of the result.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

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