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.
What is the actual frictional force when the man has climbed `1.0 m` along the ladder?

Options

A. A. `360N`
B. B. `171N`
C. C. `900N`
D. D. `740N`

Explanation

To solve the problem, we will analyze the forces acting on the ladder and apply the principles of static equilibrium. Here are the steps to find the actual frictional force when the man has climbed 1.0 m along the ladder:

### Step 1: Understand the Setup
- The ladder is 5.0 m long and rests against a frictionless wall.
- The foot of the ladder is 3.0 m away from the wall.
- The weight of the ladder (W1) is 160 N.
- The weight of the man (W2) is 740 N.
- The man climbs 1.0 m up the ladder.

### Step 2: Determine the Geometry
- The height (h) at which the ladder touches the wall can be found using the Pythagorean theorem:
\[ h = \sqrt{(5.0)^2 - (3.0)^2} = \sqrt{25 - 9} = \sqrt{16} = 4.0 \text{ m} \]

### Step 3: Calculate the Forces
- The total weight acting downwards when the man is 1.0 m up the ladder is:
\[ W_{\text{total}} = W1 + W2 = 160 \text{ N} + 740 \text{ N} = 900 \text{ N} \]

### Step 4: Analyze Forces in Vertical and Horizontal Directions
- In the vertical direction, the normal force (N1) at the base of the ladder must balance the total weight:
\[ N1 = W_{\text{total}} = 900 \text{ N} \]

- In the horizontal direction, the frictional force (F) at the base of the ladder must balance the normal force exerted by the wall (N2):
\[ F = N2 \]

### Step 5: Calculate Torque
- Taking moments about the base of the ladder (point A), we can set up the equation for torque:
- The torque due to the weight of the man (W2) and the ladder (W1) acts clockwise.
- The torque due to the normal force at the wall (N2) acts counterclockwise.

- The distance from point A to the center of mass of the ladder (which is at 2.5 m along the ladder) is:
\[ \text{Distance for W1} = 2.5 \times \frac{3}{5} = 1.5 \text{ m} \]
\[ \text{Distance for W2} = (1.0 \text{ m}) \times \frac{3}{5} = 0.6 \text{ m} \]

- The total torque equation about point A:
\[ W1 \cdot 1.5 + W2 \cdot 0.6 = N2 \cdot 4.0 \]
Substituting the values:
\[ 160 \cdot 1.5 + 740 \cdot 0.6 = N2 \cdot 4.0 \]
\[ 240 + 444 = N2 \cdot 4.0 \]
\[ 684 = N2 \cdot 4.0 \]
\[ N2 = \frac{684}{4} = 171 \text{ N} \]

### Step 6: Find the Frictional Force
- From the horizontal force balance, we know:
\[ F = N2 = 171 \text{ N} \]

Thus, the actual frictional force when the man has climbed 1.0 m along the ladder is **171 N**.

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