This problem involves calculating the frictional force at the base of a ladder in static equilibrium. Let’s solve it step by step.

Given:

1. Mass of the ladder: $m_{\text{ladder}} = 8 \, \text{kg}$
2. Length of the ladder: $L = 4 \, \text{m}$
3. Mass of the woman: $m_{\text{woman}} = 60 \, \text{kg}$
4. Distance of the woman from the top of the ladder: $1 \, \text{m}$, so she stands $3 \, \text{m}$ from the bottom.
5. Angle of the ladder with the ground: $\theta = 60^\circ$
6. The wall is smooth, so there’s no friction at the wall.
7. Gravitational acceleration: $g = 9.8 \, \text{m/s}^2$

We need to find frictional force ($F_{\text{friction}}$) at the base of the ladder.

Free Body Diagram and Forces:

1. Weight of the ladder ($W_{\text{ladder}}$): $W_{\text{ladder}} = m_{\text{ladder}} \cdot g = 8 \cdot 9.8 = 78.4 \, \text{N}$ Acts at the midpoint of the ladder ($2 \, \text{m}$ from the base).

2. Weight of the woman ($W_{\text{woman}}$): $W_{\text{woman}} = m_{\text{woman}} \cdot g = 60 \cdot 9.8 = 588 \, \text{N}$ Acts $3 \, \text{m}$ from the base.

3. Normal force at the base ($N$): Acts vertically upward.

4. Frictional force at the base ($F_{\text{friction}}$): Acts horizontally to prevent the ladder from sliding.

5. Reaction force at the wall ($F_{\text{wall}}$): Acts horizontally.

Equilibrium Conditions:

1. Vertical Force Balance: $N = W_{\text{ladder}} + W_{\text{woman}}$ [ N = 78.4 + 588 =