A uniform ladder `5.0 m` long rests against a frictionless, vertical wall with its lower end `3.0m` to from the wall. The ladder weighs `160 N`. The coefficient of static friction between the foot of the ladder and the ground is `0.40`. A man weighing `740 N` climbs slowly up the ladder.
How far along the ladder can the man climb before the ladder starts to slip?

Options

A. `3m`
B. `5m`
C. `2.7m`
D. `1.25m`

Explanation

To solve the problem step by step, we will analyze the forces and torques acting on the ladder and find out how far the man can climb before the ladder starts to slip.

### Step 1: Understand the Setup
- The ladder is 5.0 m long and rests against a frictionless wall.
- The base of the ladder is 3.0 m away from the wall.
- The weight of the ladder is 160 N, and the man weighs 740 N.
- The coefficient of static friction between the ladder and the ground is 0.40.

### Step 2: Calculate the Angle of the Ladder
Using the geometry of the situation:
- The height of the wall where the ladder touches can be found using the Pythagorean theorem:
\[ h = \sqrt{(5.0^2) - (3.0^2)} = \sqrt{25 - 9} = \sqrt{16} = 4.0 \text{ m} \]
- The angle \( \theta \) can be calculated using:
\[ \tan(\theta) = \frac{h}{3.0} = \frac{4.0}{3.0} \]
Thus, \( \theta = \tan^{-1}\left(\frac{4}{3}\right) \).

### Step 3: Calculate the Maximum Static Friction Force
The normal force \( N \) at the base of the ladder can be calculated by balancing the vertical forces:
- The total weight acting downwards is the weight of the ladder plus the weight of the man:
\[ N = 160 \, \text{N} + 740 \, \text{N} = 900 \, \text{N} \]
- The maximum static friction force \( f_{\text{max}} \) is given by:
\[ f_{\text{max}} = \mu_s \cdot N = 0.40 \cdot 900 \, \text{N} = 360 \, \text{N} \]

### Step 4: Establish Torque Equilibrium
To find how far the man can climb before the ladder starts to slip, we need to set up a torque balance about the point where the ladder touches the ground.

Assuming the man climbs a distance \( x \) up the ladder:
- The torque due to the man’s weight (740 N) about the base is:
\[ \tau_{\text{man}} = 740 \cdot x \cdot \cos(\theta) \]
- The torque due to the weight of the ladder (acting at its center, which is at 2.5 m) is:
\[ \tau_{\text{ladder}} = 160 \cdot 2.5 \cdot \cos(\theta) \]
- The torque due to the friction force (360 N) is:
\[ \tau_{\text{friction}} = 360 \cdot 3.0 \cdot \sin(\theta) \]

### Step 5: Set Up the Torque Equation
Setting the sum of torques about the base to zero for equilibrium:
\[ 740 \cdot x \cdot \cos(\theta) + 160 \cdot 2.5 \cdot \cos(\theta) = 360 \cdot 3.0 \cdot \sin(\theta) \]

### Step 6: Solve for \( x \)
Rearranging the equation to solve for \( x \):
\[ x = \frac{360 \cdot 3.0 \cdot \sin(\theta) - 160 \cdot 2.5 \cdot \cos(\theta)}{740 \cdot \cos(\theta)} \]

### Step 7: Substitute Values
Using \( \sin(\theta) = \frac{4}{5} \) and \( \cos(\theta) = \frac{3}{5} \):
\[ x = \frac{360 \cdot 3.0 \cdot \frac{4}{5} - 160 \cdot 2.5 \cdot \frac{3}{5}}{740 \cdot \frac{3}{5}} \]

Calculating the values:
\[ x = \frac{360 \cdot 3.0 \cdot 0.8 - 160 \cdot 2.5 \cdot 0.6}{740 \cdot 0.6} \]
\[ x = \frac{864 - 240}{444} \approx 1.41 \, \text{m} \]

### Final Answer
The man can climb approximately **1.41 m** along the ladder before it starts to slip.