1. Geometry of the ladder

The ladder has length $\sqrt{34}\,\text{m}$ and its foot is $3\,\text{m}$ away from the wall.

So the height reached on the wall is $h = \sqrt{34-9} = \sqrt{25} = 5\,\text{m}.$

Thus, the ladder forms a $3$-$5$-$\sqrt{34}$ triangle.

2. Forces acting on the ladder

Since the wall is frictionless:

• At the wall, only a horizontal reaction acts: $F_w$.
• At the floor, the reaction has two components:
  • horizontal friction force $f$,
  • vertical normal reaction $N$.

The resultant reaction of the floor is $F_f = \sqrt{N^2 + f^2}.$

The weight of the ladder is $W = mg = 10 \times 10 = 100\,\text{N},$ acting at the midpoint of the ladder.

3. Force balance

Horizontal equilibrium

The only horizontal forces are:

• wall reaction $F_w$,
• floor friction $f$.

Hence, $f = F_w.$

Vertical equilibrium

The only vertical forces are:

• floor normal $N$ upward,
• weight $100\,\text{N}$ downward.

Hence, $N = 100\,\text{N}.$

4. Torque balance about the foot of the ladder

Take moments about the foot, so the floor reactions produce no torque.

• The wall reaction $F_w$ acts horizontally at the top of the ladder, at height $5\,\text{m}$. So its moment is $F_w \cdot 5.$

• The weight acts at the midpoint of the ladder. Since the foot is at $x=3$ and the top at $x=0$, the midpoint is at horizontal distance $\frac{3}{2}\,\text{m}$ from the foot. So the moment of weight is $100 \cdot \frac{3}{2} = 150.$

For rotational equilibrium, $F_w \cdot 5 = 150.$ So, $F_w = 30\,\text{N}.$

Thus, $f = 30\,\text{N}, \qquad N=100\,\text{N}.$

5. Resultant reaction at floor