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Step-by-Step Solution
Step 1: Identify All Forces Acting on the Ladder
1. The ladder of length $ \sqrt{34} \text{ m}$ (which numerically is around $5.83\text{ m}$) is placed so that its foot is $3\text{ m}$ away from the wall and it leans frictionlessly against the wall.
2. The weight of the ladder: $W = mg = 10 \,\text{kg} \times 10 \,\text{m/s}^2 = 100 \,\text{N}.$ This force acts at the midpoint of the ladder.
3. The wall exerts a horizontal normal reaction force $F_w$ on the ladder at the top contact point.
4. The floor exerts two components on the ladder foot: the normal reaction $N$ (vertically upward) and the frictional force $f_r$ (horizontally, opposing any potential sliding). Because the wall is frictionless, the only horizontal force at the top is $F_w$, and the floor's corresponding horizontal reaction is $f_r$.
Step 2: Determine the Geometry for Torque Calculations
- Since the foot of the ladder is $3\text{ m}$ from the wall, the top of the ladder makes contact at a vertical height of $5\text{ m}$ (by the Pythagorean theorem: $ \sqrt{(\sqrt{34})^2 - 3^2} = \sqrt{34 - 9} = \sqrt{25} = 5 $).
- The weight (100 N) acts at the midpoint of the ladder, i.e., at $2.5\text{ m}$ from the point of contact of the wall, along the ladder. Vertically, it is halfway up, so effectively at $2.5\,\text{m}$ from the foot along the length, but we will use torque with perpendicular distances.
Step 3: Apply Torque Equilibrium about the Foot of the Ladder
Let the point where the ladder touches the floor be point B. Taking moments about point B (to eliminate forces exerted by the floor from our torque equations):
The counterclockwise torque is produced by the horizontal reaction of the wall $F_w$. Its moment arm is $5\text{ m}$ (the vertical height of the contact with the wall).
The clockwise torque is produced by the weight of the ladder (100 N). Since the ladder is uniform, the weight acts at its midpoint, which is horizontally $3/2 = 1.5\text{ m}$ from the wall and vertically $5/2=2.5\text{ m}$ from the floor. The perpendicular distance from B to the line of action of the weight is $2.5\text{ m}$ (vertically), but more directly using geometry, the torque arm is half the ladder length projected perpendicular to the foot. Here, an easier approach is to notice that the entire length to the top is $5\text{ m}$, so the midpoint is $2.5\text{ m}$ from the top, hence half of the vertical distance is $2.5 \text{ m}.$
Hence, the torque balance equation about point B becomes:
$$ F_w \times 5 = W \times \frac{5}{2} \times \frac{3}{5}, $$
but to simplify, the direct approach (often derived from geometry) is given by:
$$ F_w \times 5 = \left(\frac{3}{2} \right) (mg). $$
The factor $ \frac{3}{2} $ appears because the horizontal distance to the center of the ladder from the foot is $1.5\text{ m}$ out of $3\text{ m}$ total. Combining properly yields:
$$ F_w \times 5 = \frac{3}{2} \times 100. $$
Solving for $F_w$:
$$ F_w = \frac{\frac{3}{2} \times 100}{5} = \frac{150}{5} = 30\, \text{N}. $$
Step 4: Determine the Normal and Friction Force at the Floor
- Vertically, the net upward force from the floor $N$ must balance the ladder's weight:
$$ N = mg = 100\,\text{N}. $$
- Horizontally, the frictional force at the floor $f_r$ must balance $F_w$, so:
$$ f_r = F_w = 30\,\text{N}. $$
- The total reaction force from the floor, $F_f$, is the vector sum of $N$ (vertical) and $f_r$ (horizontal):
$$ F_f = \sqrt{N^2 + f_r^2} = \sqrt{100^2 + 30^2} = \sqrt{10000 + 900} = \sqrt{10900} = 10\sqrt{109}\,\text{N}. $$
Step 5: Compute the Required Ratio
We want $ \displaystyle \frac{F_w}{F_f} $. Substituting the values:
$$ F_w = 30\,\text{N}, \quad F_f = 10\sqrt{109}\,\text{N}, $$
so
$$ \frac{F_w}{F_f} = \frac{30}{10\sqrt{109}} = \frac{3}{\sqrt{109}}. $$
Final Answer
The ratio $ \displaystyle \frac{F_w}{F_f} $ is
$$ \frac{3}{\sqrt{109}}. $$
