• Explain smooth and rough surfaces, the normal reaction and the friction force.
• State the properties of static friction, limiting static friction and the coefficient of friction.
• Use the resultant reaction, the angle of friction and the relation ${\mu }_s=\tan \lambda$.
• Solve problems on the equilibrium of a body on a rough horizontal plane, on a rough inclined plane, and of rods and ladders with friction.

🌳 Concept tree

├── 1 Smooth and rough surfaces

│ ├── Where friction comes from

│ ├── The coefficient of friction

│ └── The reaction at a contact

├── 2 Properties of the static friction force

│ ├── Properties

│ └── Coefficient of static friction

├── 3 Kinetic friction force

├── 4 The resultant reaction R′

├── 5 The angle of friction

│ ├── Definition

│ └── Relation with the coefficient of friction

• ├── 6 Equilibrium of a body on a rough horizontal plane

│ ├── Setting

│ └── Equations of equilibrium

• ├── 7 Equilibrium of a body on a rough inclined plane

│ ├── Forces and equations

│ └── The relation between θ and λ

• └── 8 Conditions of equilibrium and rigid bodies with friction
• ├── Equilibrium of a particle and of a rigid body

└── Method for rods and ladders on rough ground

• On a rough surface the resultant reaction $R^{\mathrm{\prime }}$ is the resultant of the normal reaction $N$ and the friction force $F$; it is inclined to the normal.

Its magnitude is R′=N2+F2

• At the limiting state $F=F_s={\mu }_{s}N$, so
  $$R^{\mathrm{\prime }}=\sqrt{N^2+{\mu }_s^2{N}^2}=N\sqrt{1+{\mu }_s^2}$$
• At limiting equilibrium the angle of friction λ is linked to the coefficient of static friction by μs = tan λ.
• The resultant reaction can then be written as R′ = R sec λ, where R is the normal reaction.

Equilibrium of a particle and of a rigid body

• A particle under concurrent forces is in equilibrium when the resultant force is zero: the sum of the components along $x$ is $0$ and along $y$ is $0$.
• A rigid body under a system of coplanar forces is in equilibrium when two conditions hold:
• The resultant vector is zero R⃗=0⃗.
• The sum of the moments about any point is zero M⃗=0⃗.
• In components ∑X=0,∑Y=0,∑M=0.
• These conditions are necessary and sufficient for equilibrium.

Method for rods and ladders on rough ground

• Draw all the forces weights at the centres of uniform bodies, the reaction of the rough ground with components N (vertical) and F (horizontal), and the reaction of a smooth wall or wedge, which is perpendicular to the surface it touches.
• Write $\sum X=0$, $\sum Y=0$ and take moments about the point where most unknown forces meet (usually the foot of the rod).
• If the body is about to slip, use $F={\mu }_{s}N$.

✍️ Worked examples

• Example 5:
• A uniform bar $AB$ of weight $5$ kg.wt and length $30$ cm rests with end $A$ on rough horizontal ground and a point $C$ on a smooth wedge $12.5$ cm above the ground.
• The bar is about to slip when inclined at $30^{\circ }$ to the ground.
• Find the reaction of the wedge and the coefficient of friction at $A$.
• Solution:
• $AC={\displaystyle \frac{12.5}{\sin 30^{\circ }}}=25$ cm.
• Forces weight 5 at the midpoint (15 cm from A), wedge reaction R2 perpendicular to the bar at C, and at A the normal R1 (up) and friction F (horizontal).
• Moments about A 5×15cos⁡30∘−R2×25=0.
• $R_2={\displaystyle \frac{75\times \frac{\sqrt{3}}{2}}{25}}={\displaystyle \frac{3\sqrt{3}}{2}}$ kg.wt.
• Horizontal F=R2sin⁡30∘=334 kg.wt.
• Vertical R1+R2cos⁡30∘=5, so R1=5−332⋅32=5−94=114 kg.wt.
• About to slip μs=FR1=33/411/4=3311.
• Example 6:
• A uniform ladder of weight $20$ kg.wt rests with one end on rough horizontal ground and the other on a smooth vertical wall, inclined at $60^{\circ }$ to the horizontal.
• The coefficient of friction at the ground is ${\displaystyle \frac{1}{2\sqrt{3}}}$.
• Show that a girl of weight $60$ kg.wt can climb at most half the length of the ladder.
• Solution:
• Let the ladder length be $\mathrm{\ell }$ and the girl’s greatest distance from the foot $A$ be $x$; then the ladder is about to slip.
• Vertical R1=20+60=80 kg.wt (ground normal reaction).
• Horizontal the wall reaction R2 equals the friction, so R2=μsR1=8023=403 kg.wt.
• Moments about A R2×ℓsin⁡60∘=20×ℓ2cos⁡60∘+60×xcos⁡60∘.
• Left side 403⋅32 ℓ=20ℓ.
• Right side 5ℓ+30x.
• $20\mathrm{\ell }=5\mathrm{\ell }+30x$, so $30x=15\mathrm{\ell }$ and $x={\displaystyle \frac{\mathrm{\ell }}{2}}$.