A ladder rests against a frictionless vertical wall, with its upper end `6 m` above the ground and the lower end `4 m` away from the wall. The weight of the ladder is `500 N` and its `CG` at `1//3^(rd)` distance from the lower end. Wall's reaction will be (in newton)

Options

A. 111
B. 333
C. 222
D. 129

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry of the Ladder
The ladder is resting against a frictionless vertical wall. The height of the ladder's upper end from the ground is 6 m, and the distance of the lower end from the wall is 4 m.

### Step 2: Determine the Length of the Ladder
Using the Pythagorean theorem, we can find the length \( L \) of the ladder:
\[ L = \sqrt{(6^2 + 4^2)} = \sqrt{(36 + 16)} = \sqrt{52} = 2\sqrt{13} \text{ m} \]

### Step 3: Identify the Center of Gravity
The center of gravity (CG) of the ladder is located at \( \frac{1}{3} \) of the length from the lower end. Thus, the distance from the lower end to the CG is:
\[ \text{Distance to CG} = \frac{L}{3} = \frac{2\sqrt{13}}{3} \text{ m} \]

### Step 4: Set Up the Torque Equation
To find the normal reaction force \( N \) at the wall, we will take moments about the base of the ladder (point O). The weight of the ladder acts downward at the CG, and the normal force from the wall acts horizontally at the top of the ladder.

The torque due to the weight of the ladder (500 N) about point O is:
\[ \tau_{\text{weight}} = 500 \times \frac{4}{3} \text{ (perpendicular distance from O)} \]

The torque due to the normal force \( N \) is:
\[ \tau_{\text{normal}} = N \times 6 \text{ (perpendicular distance from O)} \]

### Step 5: Set the Total Torque to Zero
Since the ladder is in equilibrium, the sum of the torques about point O must be zero:
\[ 500 \times \frac{4}{3} = N \times 6 \]

### Step 6: Solve for the Normal Force \( N \)
Rearranging the equation gives:
\[ N = \frac{500 \times \frac{4}{3}}{6} \]
\[ N = \frac{2000}{18} = \frac{1000}{9} \text{ N} \]

### Final Answer
The normal reaction force \( N \) at the wall is approximately:
\[ N \approx 111.11 \text{ N} \]

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