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

Geometry: $h=\sqrt{3^2-1^2}=2\sqrt2\approx2.83$ m; weight $W=20\times9.8=196$ N acts at the midpoint, 0.5 m horizontally from the foot.

Torque about the foot: $F_1\times h = W\times0.5 \Rightarrow F_1=\dfrac{196\times0.5}{2\sqrt2}\approx34.6$ N. …

The figure shows a ladder resting against a wall, drawn as a right triangle. The foot of the ladder is at point A on the floor, the top touches the wall at point B, and the corner where wall meets floor is point C. The ladder itself is the hypotenuse AB, labelled 3 m. The horizontal distance from the foot to the wall, AC, is marked 1 m. The vertical height from the floor to the top, CB, is therefore $2\sqrt{2}$ m — this follows from the Pythagorean theorem: $3^2 = 1^2 + (2\sqrt{2})^2$.

Three forces act on the ladder. At the top end B, the wall exerts a horizontal reaction $F_1$ pushing to the right — the wall is frictionless, so there is no vertical component there. At the foot A, the floor exerts a resultant force $F_2$ that is the vector sum of two separate effects: a normal force $N$ acting vertically upward, and a friction force $F$ acting horizontally to the left (it must oppose the tendency of the ladder to slide outwards). The ladder’s weight $W$ acts vertically downward at its midpoint D, which is 1.5 m from either end along the ladder.

The physical idea this figure teaches is equilibrium of a rigid body under non-concurrent forces. For the ladder to remain stationary, two conditions must hold simultaneously: the net force on it must be zero, and the net torque about any point must also be zero. The figure lets you set up those equations.

The key formulas the textbook develops from this figure are the equilibrium conditions. Taking torques about point A (to eliminate the unknown forces $N$ and $F$ at the foot) gives:

Here $W$ is the weight of the ladder, $\frac{1}{2}$ m is the perpendicular distance from A to the line of action of $W$ (the horizontal distance from A to the midpoint D), and $2\sqrt{2}$ m is the perpendicular distance from A to the line of action of $F_1$ (the vertical height of the wall). This torque balance yields $F_1 = \frac{W}{4\sqrt{2}}$.

The horizontal force balance then gives the friction force at the foot:

And the vertical force balance gives the normal reaction at the foot:

The figure thus illustrates how a single diagram encodes all the geometry needed to write torque and force equations for a classic equilibrium problem. The right-triangle dimensions (1 m, $2\sqrt{2}$ m, 3 m) are chosen so that the numbers work out cleanly, but the method applies to any ladder length and any foot distance.

Geometry

The ladder is 3 m long, with its foot on the floor 1 m from the wall. The height where it touches the wall is …

Concept: Static Equilibrium of a Ladder Against a Frictionless Wall

Step 1: Find the height at which the ladder touches the wall

Ladder length $L=3$ m, foot-to-wall distance $=1$ m:

Step 2: Locate the weight

Weight $W=mg=20\times9.8=196$ N acts at the ladder's midpoint, which is at horizontal distance $\tfrac{1}{2}(1\text{ m})=0.5$ m from the foot.

Step 3: Identify the forces

Wall (frictionless) can only push horizontally: reaction $F_1$ at the top. Floor supplies a vertical normal reaction $N$ and horizontal friction $f$ at the foot.

Step 4: Translational equilibrium

Step 5: Rotational equilibrium about the foot (eliminates $N$ and $f$) …