Rigid-body equilibrium and ladders

What you’ll learn

• Combine force balance with moment balance for non-parallel forces.
• Determine limiting equilibrium of ladders and rods.

Understand the idea

A rigid body can rotate as well as translate, so equilibrium requires zero resultant force and zero resultant moment. Draw all contact reactions, friction forces, applied forces and weights. Choose the moment centre to remove as many unknown forces as possible.

For a ladder against a smooth vertical wall, the wall reaction is horizontal. At a rough horizontal floor, the normal reaction is vertical and friction is horizontal. When a uniform ladder makes angle $\theta$ with the floor, its weight acts at the midpoint and has horizontal moment arm $(L/2)\cos\theta$ about the foot.

Key formulae

$$\sum F_x=0,\quad\sum F_y=0,\quad\sum M_O=0$$

$$M=FL\sin\phi$$

$$F\le\mu R$$

$$H L\sin\theta=W\frac L2\cos\theta\quad\text{(uniform ladder, smooth wall)}$$

Worked example

A uniform ladder of weight $200\,\mathrm N$ rests at $60^\circ$ to a rough horizontal floor against a smooth wall. Find the wall reaction.

1. Call the horizontal wall reaction $H$ and ladder length $L$.
2. Taking moments about the foot gives $HL\sin60^\circ=200(L/2)\cos60^\circ$.
3. Cancel $L$ and solve $H=100\cot60^\circ$.

Answer: $H=100/\sqrt3\approx57.7\,\mathrm N$.