A weightless ladder 20 ft long rests against a frictionless wall at an angle of `60^(@)` from the horizontal. A 150 pound man is 4 ft from the top of the ladder. A horizontal force is needed to keep it from slipping. Choose the correct magnitude from the following.

Options

A. 175 1b
B. 100 1b
C. 120 1b
D. 17.3 1b

Answer

To solve the problem of the weightless ladder resting against a frictionless wall, we will use the principles of equilibrium and torque. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the Setup
We have a weightless ladder that is 20 ft long, leaning against a frictionless wall at an angle of 60 degrees from the horizontal. A 150-pound man is positioned 4 ft from the top of the ladder.

### Step 2: Identify Forces Acting on the Ladder
1. The weight of the man acting downward at a distance of 4 ft from the top of the ladder.
2. The normal force exerted by the ground acting upward at the base of the ladder.
3. A horizontal force \( F \) that we need to find, acting at the base of the ladder to prevent slipping.
4. The normal force exerted by the wall acting horizontally at the top of the ladder.

### Step 3: Set Up the Torque Equation
To prevent the ladder from slipping, we need to ensure that the net torque about any point is zero. We will take torques about the base of the ladder (where it touches the ground).

1. **Torque due to the man's weight**:
- The distance from the base to where the man is located is \( 20 \cos(60^\circ) - 4 \).
- The torque due to the man's weight (150 lbs) is:
\[ \tau_{\text{man}} = 150 \times (20 \cos(60^\circ) - 4) = 150 \times (10 - 4) = 150 \times 6 = 900 \text{ ft-lbs} \]

2. **Torque due to the horizontal force \( F \)**:
- The distance from the base to the top of the ladder is \( 20 \sin(60^\circ) \).
- The torque due to the horizontal force \( F \) is:
\[ \tau_F = F \times 20 \sin(60^\circ) = F \times 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} F \]

### Step 4: Set the Net Torque to Zero
For the ladder to be in equilibrium:
\[ \tau_F - \tau_{\text{man}} = 0 \]
This gives us:
\[ 10\sqrt{3} F = 900 \]

### Step 5: Solve for \( F \)
Now, we can solve for \( F \):
\[ F = \frac{900}{10\sqrt{3}} = \frac{90}{\sqrt{3}} = 30\sqrt{3} \text{ lbs} \]
Calculating \( 30\sqrt{3} \):
\[ F \approx 30 \times 1.732 \approx 51.96 \text{ lbs} \]

### Final Answer
The magnitude of the horizontal force needed to keep the ladder from slipping is approximately **52 lbs**.