The problem is about calculating the forces acting on a ladder leaning against a wall at an angle, taking into account the weight of the ladder, friction on the floor, and the minimum static friction coefficient required to maintain equilibrium.

Problem Analysis:

Given:

1. Weight of the ladder $W = 400 \, \text{N}$.
2. Length of the ladder $L = 5 \, \text{m}$.
3. Angle with the horizontal $\theta = 53^\circ$.
4. Friction is only present on the floor; the wall is frictionless.

We are required to:

1. Find all reactions (forces) acting on the ladder to maintain equilibrium.
2. Calculate the minimum static friction coefficient ($\mu_s$) on the floor to prevent slipping.

Solution Steps:

1. Identify the Forces:

• Weight of the Ladder (W): Acts vertically downward at the ladder's center of gravity, which is located at half its length, $\frac{L}{2}$.
• Normal Force on the Floor (N): Acts perpendicular to the floor, at the base of the ladder.
• Frictional Force on the Floor (f): Acts parallel to the floor, preventing the ladder from slipping. Given by $f = \mu_s N$.
• Normal Force from the Wall (F_w): Acts horizontally, perpendicular to the wall, at the point where the ladder contacts the wall.

2. Set Up Equilibrium Equations:

For the ladder to be in static equilibrium, both the sum of forces and the sum of moments (torques) about any point should be zero.

Sum of Forces in the Horizontal Direction:

$F_w = f$

Sum of Forces in the Vertical Direction:

$N = W$

Sum of Moments about the Base (Point A):

Taking moments around the base of the ladder to eliminate $f$ and $N$: $F_w \cdot L \cdot \cos(\theta) = W \cdot \frac{L}{2} \cdot \sin(\theta)$

Simplifying: $F_w = \frac{W \cdot \sin(\theta)}{2 \cdot \cos(\theta)}$

Calculate $F_w$:

Substitute $W = 400 \, \text{N}$, $\theta = 53^\circ$: $F_w = \frac{400 \cdot \sin(53^\circ)}{2 \cdot \cos(53^\circ)}$ Using approximate values, $\sin(53^\circ) \approx 0.7986$ and $\cos(53^\circ) \approx 0.6018$: $F_w = \frac{400 \cdot 0.7986}{2 \cdot 0.6018} \approx 264.4 \, \text{N}$

3. Find the Minimum Static Friction Coefficient $\mu_s$:

Since $f = \mu_s N$ and $f = F_w$: $\mu_s = \frac{F_w}{N} = \frac{264.4}{400} \approx 0.661$

Summary:

1. Normal force from the floor $N = 400 \, \text{N}$.
2. Horizontal force from the wall $F_w \approx 264.4 \, \text{N}$.
3. Minimum static friction coefficient $\mu_s \approx 0.661$.

Would you like additional details on any step, or have further questions?

Related Questions:

1. How would the reaction forces change if the angle of the ladder with the horizontal were decreased?
2. What would happen to the minimum static friction coefficient if the ladder's weight were increased?
3. How does the length of the ladder affect the stability in terms of friction required?
4. How would adding a person climbing up the ladder affect the equilibrium conditions?
5. Why does the wall being frictionless simplify the problem?

Tip:

In ladder problems, taking moments around one end often simplifies calculations, as it eliminates forces acting at that point.