Example

Question:

A 3 m long ladder weighing 20 kg leans on a frictionless wall. Its feet rest on the floor 1 m from the wall as shown in the figure.
Find the reaction forces of the wall and the floor.

Solution:

The ladder \(AB\) is 3 m long, its foot \(A\) is at a distance \(AC = 1\,\mathrm{m}\) from the wall.
From Pythagoras: \(BC = 2\sqrt{2}\,\mathrm{m}\).
The forces on the ladder are:

• weight \(W\) at centre of gravity \(D\)
• reaction forces \(F_1\) (horizontal, at wall \(B\)), \(F_2\) (normal, at floor \(A\)), \(F\) (friction at floor \(A\))

For translational equilibrium (vertical): \[ N - W = 0,\quad N = W \] \(W = 20 \times 9.8 = 196\,\mathrm{N}\) For horizontal equilibrium: \[ F - F_1 = 0,\quad F = F_1 \] For rotational equilibrium about \(A\): \[ 2\sqrt{2} F_1 = (1/2) W \] \[ F_1 = \frac{1/2 \times 196}{2\sqrt{2}} \approx 34.6\,\mathrm{N} \] \[ F = F_1 = 34.6\,\mathrm{N} \] Now, vertical reaction at floor: \[ F_2 = W^2 - F^2 = \sqrt{196^2 - 34.6^2} \approx 199\,\mathrm{N} \] The force \(F_2\) makes an angle \(\alpha\) with the vertical: \[ \tan \alpha = \frac{F}{F_2},\quad \alpha = \tan^{-1} \left( \frac{34.6}{199} \right) \approx 10^\circ \]
Summary:
Reaction force by wall: \(F_1 \approx 34.6\,\mathrm{N}\)
Reaction force by floor: \(F_2 \approx 199\,\mathrm{N}\), at an angle \(\alpha \approx 10^\circ\) to vertical.