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

Options

A. 25`sqrt 2` N, 203 N
B. 50`sqrt 2` N, 230 N
C. 203 N, 25`sqrt 2` N
D. `230 N, 50sqrt 2` N

Explanation

To solve the problem of the ladder leaning against a frictionless wall, we will follow these steps:

### Step 1: Understand the Setup
We have a ladder of length \( L = 3 \, \text{m} \) leaning against a wall. The base of the ladder is \( 1 \, \text{m} \) away from the wall. The weight of the ladder is \( W = 20 \, \text{kg} \), which we will convert to Newtons using \( W = mg \) where \( g = 9.8 \, \text{m/s}^2 \).

### Step 2: Calculate the Weight of the Ladder
\[ W = 20 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 196 \, \text{N} \]

### Step 3: Use Pythagorean Theorem to Find the Height
Let \( h \) be the height at which the ladder touches the wall. Using the Pythagorean theorem:
\[ L^2 = h^2 + d^2 \]
where \( d = 1 \, \text{m} \) (the distance from the wall to the foot of the ladder).
\[ 3^2 = h^2 + 1^2 \implies 9 = h^2 + 1 \implies h^2 = 8 \implies h = \sqrt{8} = 2\sqrt{2} \, \text{m} \]

### Step 4: Set Up the Equations for Equilibrium
1. **Vertical Forces**: The normal force \( F_2 \) from the ground must balance the weight of the ladder:
\[ F_2 - W = 0 \implies F_2 = W = 196 \, \text{N} \]

2. **Horizontal Forces**: The force \( F_1 \) from the wall must balance the horizontal component of the force from the ground:
\[ F - F_1 = 0 \implies F = F_1 \]

3. **Moment about Point A**: To find \( F_1 \), we take moments about point A (the base of the ladder):
\[ F_1 \cdot h - \frac{W}{2} \cdot d = 0 \]
Here, \( \frac{W}{2} \) is the weight acting at the midpoint of the ladder, which is \( \frac{3}{2} = 1.5 \, \text{m} \) from A.
\[ F_1 \cdot (2\sqrt{2}) - 196 \cdot 0.5 = 0 \]
\[ F_1 \cdot (2\sqrt{2}) = 98 \implies F_1 = \frac{98}{2\sqrt{2}} = \frac{49}{\sqrt{2}} \approx 34.6 \, \text{N} \]

### Step 5: Calculate the Resultant Force from the Ground
The resultant force \( F_2 \) from the ground can be found using:
\[ F_2 = \sqrt{F^2 + N^2} = \sqrt{F_1^2 + F_2^2} \]
Substituting \( F_1 \) and \( F_2 \):
\[ F_2 = \sqrt{(34.6)^2 + (196)^2} \]
Calculating this gives:
\[ F_2 \approx \sqrt{1197.16 + 38416} = \sqrt{39613.16} \approx 199 \, \text{N} \]

### Final Results
- The reaction force from the wall \( F_1 \approx 34.6 \, \text{N} \)
- The reaction force from the ground \( F_2 \approx 199 \, \text{N} \)

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