Ladders, Pegs and Hinges

Analysing forces on ladders in limiting equilibrium using moments and friction

SPEC 9.1 • MOMENTS

The Ladder Model

Ladder problems are a classic application of moments and equilibrium. We model a ladder as a uniform rod resting against a wall.

Standard Setup

A ladder of length

l

and weight

W

leans against a vertical wall at angle

θ

to the horizontal ground.

Key Assumptions

The ladder is a uniform rod (weight acts at midpoint)

The wall is usually smooth (no friction at wall)

The ground is usually rough (friction prevents slipping)

The system is in equilibrium (or limiting equilibrium)

Forces on a Ladder

Identify all forces acting on the ladder before applying equilibrium conditions.

W = Weight (at midpoint)

R = Normal reaction (ground)

F = Friction (ground)

S = Normal reaction (wall)

Equilibrium Conditions

Apply the two conditions for equilibrium to solve ladder problems:

Condition 1

\sum F = 0

(Forces balance)R(→):

S - F = 0

R(↑):

R - W = 0

Condition 2

\sum M = 0

(Moments balance)Take moments about a convenient point (usually the base of the ladder)

Strategy

Taking moments about the base eliminates

R

and

F

from the moment equation (both pass through this point), leaving only

W

and

S

.

Taking Moments About the Base

For a ladder of length

l

at angle

θ

to the horizontal:

Moment Equation

M(A): Taking moments about the base

Clockwise moment (W):

W cos θ \times \frac{l}{2}

(component ⊥ to ladder × distance along ladder)

Anticlockwise moment (S):

S sin θ \times l

(component ⊥ to ladder × distance along ladder)

S sin θ \cdot l - W cos θ \cdot \frac{l}{2} = 0

Simplifying

Cancelling

l

:

S = \frac{W c o s θ}{2 s i n θ} = \frac{W}{2 t a n θ}

Limiting Equilibrium

When a ladder is about to slip, friction reaches its maximum value:

Limiting Friction

F = μ R

At the point of slipping, friction

F

equals the coefficient of friction

μ

times the normal reaction

R

.

Important

In limiting equilibrium:

F = μ R

(friction at maximum)

R(→):

S - F = 0

R(↑):

R - W = 0

Therefore:

μ R = S

, which gives

μ W = S

Finding Minimum μ

To find the minimum coefficient of friction needed to prevent slipping:

μ_{m i n} = \frac{S}{R} = \frac{S}{W}

01Worked Example

A uniform ladder of length 6 m and weight 200 N rests against a smooth vertical wall at an angle of 60° to the horizontal ground. The ground is rough. Find the normal reaction at the wall and the friction at the ground.

1

R(↑):

R - W = 0

R = 200

N

2

M(A):

S sin 60° \times 6 - W cos 60° \times 3 = 0

S \times 6 \times \frac{3}{2} = 200 \times 3 \times 0.5

S = \frac{300}{3 3} = \frac{100 3}{3} \approx 57.7

N

3

R(→):

S - F = 0

F = \frac{100 3}{3} \approx 57.7

N

Answer

S = \frac{100 3}{3}

N ≈ 57.7 N,

F = \frac{100 3}{3}

N ≈ 57.7 N

02Worked Example

A uniform ladder of weight 150 N rests against a smooth wall at 65° to the horizontal. Find the minimum coefficient of friction at the ground required to prevent the ladder from slipping.

1

R(↑):

R - W = 0

R = 150

N

2

M(A):

S sin 65° \times l - W cos 65° \times \frac{l}{2} = 0

S = \frac{W c o s 65°}{2 s i n 65°} = \frac{150 \times 0.4226}{2 \times 0.9063} = 34.97

N

3

R(→):

S - F = 0

F = 34.97

N

4

F = μR:

μ = \frac{F}{R} = \frac{34.97}{150} = 0.233

Answer

μ_{m i n} = 0.233

(or

\frac{c o t 65°}{2}

)

03Worked Example

A uniform ladder of length $l$ and weight $W$ rests against a smooth wall at angle $θ$ to the horizontal. Show that the ladder will slip when $tan θ < \frac{1}{2 μ}$ , where $μ$ is the coefficient of friction at the ground.

1

R(↑):

R - W = 0

R = W

2

M(A):

S sin θ \cdot l - W cos θ \cdot \frac{l}{2} = 0

S = \frac{W c o s θ}{2 s i n θ} = \frac{W}{2 t a n θ}

3

R(→):

S - F = 0

F = \frac{W}{2 t a n θ}

4

F ≤ μR:

\frac{W}{2 t a n θ} \leq μ W

tan θ \geq \frac{1}{2 μ}

AnswerLadder slips when

tan θ < \frac{1}{2 μ}

Additional Load on a Ladder

When an additional load is placed on the ladder, we add an extra weight force:

Modified Equations

R(↑):

R - W - P = 0

→

R = W + P

R(→):

S - F = 0

→

F = S

M(A):

S sin θ \cdot l - W cos θ \cdot \frac{l}{2} - P cos θ \cdot x = 0

Key Point

The additional load

P

at distance

x

from the base adds an extra clockwise moment:

P cos θ \cdot x

04Worked Example

A uniform ladder of length 8 m and weight 180 N rests against a smooth wall at 55° to the horizontal. The coefficient of friction at the ground is 0.4. A person of weight 720 N starts climbing from the bottom. How far up the ladder can they climb before it starts to slip?

1

R(↑):

R - 180 - 720 = 0

R = 900

N

2

F = μR:

F = 0.4 \times 900 = 360

N

3

R(→):

S - F = 0

S = 360

N

4

M(A):

S sin 55° \times 8 - 180 cos 55° \times 4 - 720 cos 55° \times x = 0

360 \times 0.819 \times 8 = 180 \times 0.574 \times 4 + 720 \times 0.574 \times x

2358.7 = 413.3 + 413.3 x

x = 4.71

m

AnswerThe person can climb 4.71 m up the ladder.

Rough Wall and Ground

Some problems have friction at both the wall and the ground:

Modified Equilibrium

Horizontal:

F = S

Vertical:

R + G = W

Where

G

is the friction force at the wall (acts upward to oppose the ladder's tendency to slide down).

Two Friction Relationships

At limiting equilibrium:

At ground:

F = μ_{1} R

At wall:

G = μ_{2} S

05Worked Example

A non-uniform rod AB of length 5 m and weight 200 N rests against a rough wall ( $μ = 0.4$ ) and rough ground ( $μ = 0.5$ ) at 60° to the horizontal. In limiting equilibrium, the friction at the ground is 80 N. Find the distance of the centre of mass from end A.

1

R(→):

S - F = 0

S = 80

N

2

At wall (limiting):

G = 0.4 S = 0.4 \times 80 = 32

N

3

R(↑):

R + G - 200 = 0

R = 200 - 32 = 168

N

4

M(A): Let centre of mass be at distance

d

from A.

S sin 60° \times 5 + G cos 60° \times 5 - 200 cos 60° \times d = 0

80 \times \frac{3}{2} \times 5 + 32 \times 0.5 \times 5 = 200 \times 0.5 \times d

2003 + 80 = 100 d

d = 23 + 0.8 = 4.26

m

AnswerCentre of mass is 4.26 m from A

Rod Against a Peg

When a rod rests against a smooth peg (not a wall), the reaction is perpendicular to the rod, not horizontal.

Key Difference

At a smooth peg, the reaction

N

is perpendicular to the rod, not horizontal. You must resolve

N

into horizontal and vertical components.

Resolving the Peg Reaction

If the rod makes angle

θ

with horizontal:

Horizontal component:

N sin θ

Vertical component:

N cos θ

06Worked Example

A uniform rod AB of mass 40 kg and length 10 m rests with end A on rough horizontal ground. The rod rests against a smooth peg C where AC = 8 m. The rod is in limiting equilibrium at 15° to the horizontal. Find the reaction at C and the coefficient of friction at the ground.

1

M(A): (N is ⊥ to rod, so moment = N × distance)

N \times 8 - 40 g cos 15° \times 5 = 0

N = \frac{40 g c o s 15° \times 5}{8} = 25 g cos 15° = 236.7

N

2

R(→):

N sin 15° - F = 0

F = 236.7 \times 0.259 = 61.3

N

3

R(↑):

R + N cos 15° - 40 g = 0

R = 392 - 228.6 = 163.4

N

4

F = μR:

μ = \frac{F}{R} = \frac{61.3}{163.4} = 0.375

Answer

N = 240

N,

μ = 0.38

07Worked Example

A uniform rod AB of weight 100 N and length 4 m has end A on rough ground ( $μ = 0.3$ ). The rod rests against two smooth pegs C and D at distances 1 m and 3 m from A. The rod makes 40° with the horizontal. Find the reactions at C and D.

1

R(→):

N_{1} sin 40° + N_{2} sin 40° - F = 0

F = (N_{1} + N_{2}) sin 40°

2

R(↑):

R + N_{1} cos 40° + N_{2} cos 40° - 100 = 0

R = 100 - (N_{1} + N_{2}) cos 40°

3

M(A): (N₁, N₂ are ⊥ to rod)

N_{1} \times 1 + N_{2} \times 3 - 100 cos 40° \times 2 = 0

N_{1} + 3 N_{2} = 153.2

... (1)

4

M(C):

N_{2} \times 2 - 100 cos 40° \times 1 + R cos 40° \times 1 - F sin 40° \times 1 = 0

Solving with equilibrium:

N_{1} = 38.3

N,

N_{2} = 38.3

N

Answer

N_{1} = 38.3

N,

N_{2} = 38.3

N

Ladders with Hinges

Some ladder problems involve a hinge at the base instead of resting freely on the ground:

Hinge Reactions

A hinge can exert a reaction force in any direction. We resolve into components:

X = Horizontal component of hinge reaction

Y = Vertical component of hinge reaction

Key Difference from Ladder Problems

With a hinge, there is no friction. The hinge provides whatever force is needed for equilibrium. The rod is typically held by a string or support at the other end.

Solving Hinged Rod Problems

For a uniform rod hinged at one end and held by a string:

Equilibrium Equations

R(→):

X - T cos θ = 0

R(↑):

Y + T sin θ - W = 0

M(A):

T sin θ \times l - W \times \frac{l}{2} = 0

(about hinge)

Strategy

Taking moments about the hinge eliminates both

X

and

Y

, allowing you to find

T

directly. Then use force equilibrium to find the hinge components.

Finding the Total Hinge Reaction

After finding

X

and

Y

, the magnitude of the hinge reaction is:

H = X^{2} + Y^{2}

The direction:

tan α = \frac{Y}{X}

08Worked Example

A uniform rod AB of length 4 m and weight 120 N is hinged at A to a vertical wall. The rod is held horizontal by a string attached to B and to a point C on the wall above A. The string makes an angle of 30° with the rod. Find the tension in the string and the magnitude of the reaction at the hinge.

1

M(A):

T sin 30° \times 4 - 120 \times 2 = 0

T \times 0.5 \times 4 = 240

T = 120

N

2

R(→):

T cos 30° - X = 0

X = 120 \times \frac{3}{2} = 603 \approx 103.9

N

3

R(↑):

Y + T sin 30° - 120 = 0

Y = 120 - 120 \times 0.5 = 60

N

4

Resultant:

H = X^{2} + Y^{2} = (603)^{2} + 6 0^{2} = 120

N

Answer

T = 120

N,

H = 120

N

Practice

Question 1: A uniform ladder of length 5 m and weight 200 N rests against a smooth vertical wall at 55° to the horizontal ground. Find the friction force at the ground.

1

R(↑):

R - W = 0

R = 200

N

2

M(A):

S sin 55° \times 5 - 200 cos 55° \times 2.5 = 0

S = 70.0

N

3

R(→):

S - F = 0

F = 70.0

N

Answer

F = 70.0

N

Practice

Question 2: A uniform ladder of weight 160 N rests against a smooth wall at 70° to the ground. The coefficient of friction at the ground is 0.25. A person of weight 640 N stands on the ladder. What is the maximum distance from the base they can stand without the ladder slipping?

1

R(↑):

R - 160 - 640 = 0

R = 800

N

2

F = μR:

F = 0.25 \times 800 = 200

N

3

R(→):

S - F = 0

S = 200

N

4

M(A):

S sin 70° \times l - 160 cos 70° \times \frac{l}{2} - 640 cos 70° \times x = 0

200 sin 70° \times l = 80 cos 70° \times l + 640 cos 70° \times x

x = 0.734 l

AnswerMaximum distance is

0.734 l

(about 73% up the ladder)

Practice

Question 3: A uniform ladder of length 6 m and weight 240 N rests against a rough wall (

μ = 0.2

) at 65° to rough ground (

μ = 0.4

). Find all four forces acting on the ladder when in limiting equilibrium.

1

Limiting equilibrium:

F = 0.4 R

,

G = 0.2 S

2

R(→):

S - F = 0

S = 0.4 R

... (1)

3

R(↑):

R + G - 240 = 0

R + 0.2 S = 240

... (2)

4

Solve: Sub (1) into (2):

R + 0.08 R = 240

1.08 R = 240

R = 222

N,

S = 88.9

N,

F = 88.9

N,

G = 17.8

N

Answer

R = 222

N,

F = 88.9

N,

S = 88.9

N,

G = 17.8

N

Summary

Key Concepts

ConceptKey Formula/Approach

Forces on ladder

W

(weight),

R

(ground normal),

F

(friction),

S

(wall normal)

Force equilibriumR(→):

S - F = 0

, R(↑):

R - W = 0

Moment equationM(A):

S sin θ \cdot l - W cos θ \cdot \frac{l}{2} = 0

Limiting equilibrium

F = μ R

, gives

μ = \frac{S}{W} = \frac{1}{2 t a n θ}

Person on ladderAdd

P cos θ \cdot x

to clockwise moments

Rod against pegReaction

N

perpendicular to rod; resolve into

N sin θ

and

N cos θ

Moments about topEliminates

S

; useful for finding

μ = F / R

directly

Hinged ladderHinge gives

X

(horizontal) and

Y

(vertical);

H = X^{2} + Y^{2}

Exam Strategy

Always draw a clear force diagram first

Take moments about the base to eliminate

R

and

F

Use perpendicular distances:

\frac{l}{2} cos θ

for

W

,

l sin θ

for

S

For "about to slip" questions, use

F = μ R