To solve this problem, we need to apply the concepts of static equilibrium. Specifically, we will use the following equilibrium conditions:

1. Force equilibrium: The sum of all vertical and horizontal forces acting on the ladder must be zero.
2. Torque equilibrium: The sum of all torques acting about any point must also be zero.

Step 1: Define the forces acting on the ladder

• $W_L$ = weight of the ladder, $W_L = m_L \cdot g = 9.3 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 91.14 \, \text{N}$
• $W_P$ = weight of the person, $W_P = m_P \cdot g = 79.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 774.2 \, \text{N}$
• $F_N$ = normal force at the bottom of the ladder (vertical reaction force)
• $F_f$ = frictional force at the bottom of the ladder (horizontal reaction force)
• $F_R$ = reaction force at the top of the ladder (horizontal force only, since the gutter is frictionless)

Step 2: Set up force equilibrium equations

Vertical forces:

The sum of all vertical forces must be zero: $F_N - W_L - W_P = 0$ $F_N = W_L + W_P = 91.14 \, \text{N} + 774.2 \, \text{N} = 865.34 \, \text{N}$

Horizontal forces:

The sum of all horizontal forces must also be zero: $F_f - F_R = 0$ $F_f = F_R$

Step 3: Set up torque equilibrium equation

We can calculate the torque about the point where the ladder touches the ground (bottom of the ladder). To maintain equilibrium, the sum of all torques around this point must be zero. We need to consider the torques due to the weights of the ladder and the person, as well as the horizontal force at the top of the ladder.

Taking clockwise torques as negative and counterclockwise torques as positive:

$\text{Torque due to the ladder’s weight:} \quad -W_L \cdot \left(\frac{2.50}{6.50} \cdot 1.90 \right)$ [ \text{Torque due to the person’s weight:} \quad +3.25(A)³= `

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