Question

A 17.7 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface where the coefficient of static friction is 0.45. The angle between the horizontal and the ladder is θ. The person wants to climb up the ladder a distance of 1.4 m along the ladder from the ladder’s foot. What is the minimum angle θmin (between the horizontal and the ladder) so that the person can reach a distance of 1.4 m without having the ladder slip? The acceleration of gravity is 9.8 m/s^2. Answer in units of degrees.

Answer 1

To solve this problem, we need to analyze the forces acting on the ladder and determine the minimum angle \( \theta_{\text{min}} \) such that the ladder does not slip.

1. **Identifying Forces**:
The forces acting on the ladder are:
- The weight of the person (\( W_p = m_p g \)), where \( m_p = 17.7 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \).
- The normal force (\( N \)) at the bottom of the ladder from the floor.
- The frictional force (\( F_f \)) at the bottom of the ladder opposing the sliding motion, which can be expressed as \( F_f = \mu_s N \), where \( \mu_s = 0.45 \) is the coefficient of static friction.
- The normal force (\( N_w \)) at the top of the ladder from the frictionless wall.

2. **Weight Calculation**:
\[
W_p = m_p g = 17.7 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 173.46 \, \text{N}
\]

3. **Setting Up the Moment Equation**:
- Taking moments about the base of the ladder, we have:
\[
W_p (d \cos(\theta)) = N_w (d \sin(\theta))
\]
where \( d \) is the length of the ladder. The distance the person climbs (1.4 m along the ladder) corresponds with a component of weight acting at a distance from the base.

4. **Frictional Condition**:
The frictional force must be sufficient to prevent slipping:
\[
N_w \leq \mu_s N
\]
Due to the equilibrium of vertical forces:
\[
N = W_p
\]

5. **Geometric Relationships**:
Let’s denote the length of the ladder as \( L \). We can find the angle \( \theta \) from the relationship with the right triangle formed by the ladder, the wall, and the ground:
- The vertical height \( h \) that the ladder reaches is:
\[
h = L \sin(\theta)
\],
- Hence, \( L \cos(\theta) = d \).

6. **Setting Up and Solving Inequalities**:
The maximum frictional force gives:
\[
N_w \leq 0.45 (173.46 \, \text{N}) = 77.06 \, \text{N}
\]

7. **Moment Calculations**:
\[
173.46 (L \sin(\theta)) = N_w (L \cos(\theta))
\]
Substituting \( N_w \):
\[
173.46 (L \sin(\theta)) = 0.45 (173.46) (L \cos(\theta))
\]
\[
173.46 \sin(\theta) = 0.45 (173.46) \cos(\theta)
\]
Simplifying further:
\[
\sin(\theta) = 0.45 \cos(\theta)
\]
Dividing both sides by \( \cos(\theta) \) (assuming \( \cos(\theta) \neq 0 \)):
\[
\tan(\theta) = 0.45
\]

8. **Finding the Angle**:
\[
\theta = \tan^{-1}(0.45)
\]
\[
\theta \approx 24.23^\circ
\]

Therefore, the minimum angle \( \theta_{\text{min}} \) required for the person to climb up to 1.4 meters without having the ladder slip is approximately \( \boxed{24.23} \) degrees.