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 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.
Answers
GPT-4o mini
To determine the minimum angle \(\theta_{min}\) at which the ladder must be positioned to prevent slipping while a person climbs it, we need to consider forces acting on the ladder and the torques about the base of the ladder.
1. **Identify Forces**:
- The weight of the person, \(W_p = m_p \cdot g = 17.7 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 173.46 \, \text{N}\).
- The normal force at the base of the ladder, \(N\), acting vertically upwards.
- The frictional force \(F_f\) at the base of the ladder, which prevents slipping. This frictional force can be maximized as follows:
\[
F_f = \mu N,
\]
where \(\mu = 0.45\) is the coefficient of friction.
2. **Weight Distribution**:
The weight of the person will act downward vertically at a distance of 1.4 m along the ladder. The height above the ground can be calculated by:
\[
h = 1.4 \sin \theta.
\]
Where \(d\) is the distance horizontally covered by the person:
\[
d = 1.4 \cos \theta.
\]
3. **Torque Calculations**:
For the ladder to be in equilibrium about the base where it rests on the floor, we set up the torque balance. The torque due to the weight of the person about the base of the ladder is:
\[
\tau_{person} = W_p \cdot (1.4 \cos \theta),
\]
The normal force will create no torque as it acts at the pivot point (the base). The torque due to the frictional force will also be considered in this equilibrium.
The force exerted by the wall will create a counteracting torque. Therefore, we set up the equation (assuming the length of the ladder is \(L\)):
\[
W_p \cdot (1.4 \cos \theta) = F_f \cdot L \cdot \sin \theta.
\]
4. **Finding Normal and Frictional Forces**:
The total vertical forces must balance:
\[
N = W_p.
\]
Therefore,
\[
F_f = \mu W_p = 0.45 \cdot 173.46 \approx 78.06 \, \text{N}.
\]
5. **Setting up Equation**:
We can substitute \(F_f\) into the torque equation:
\[
173.46 \cdot (1.4 \cos \theta) = (0.45 \cdot 173.46) \cdot L \cdot \sin \theta.
\]
Since the ladder's length \(L\) is the hypotenuse, \(L = \frac{1.4}{\sin \theta}\). Substituting gives:
\[
173.46 \cdot (1.4 \cos \theta) = (0.45 \cdot 173.46) \cdot \left(\frac{1.4}{\sin \theta}\right) \cdot \sin \theta.
\]
6. **Simplify**:
The equation simplifies to:
\[
173.46 \cdot 1.4 \cos \theta = 0.45 \cdot 173.46 \cdot 1.4.
\]
Dividing both sides by \(173.46 \cdot 1.4\):
\[
\cos \theta = 0.45.
\]
7. **Solve for \(\theta\)**:
Thus,
\[
\theta = \cos^{-1}(0.45).
\]
Calculating this yields:
\[
\theta \approx 63.6^\circ.
\]
Thus, the minimum angle \(\theta_{min}\) required for the person to climb 1.4 m up the ladder without slipping is approximately \( \boxed{63.6} \) degrees.
1. **Identify Forces**:
- The weight of the person, \(W_p = m_p \cdot g = 17.7 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 173.46 \, \text{N}\).
- The normal force at the base of the ladder, \(N\), acting vertically upwards.
- The frictional force \(F_f\) at the base of the ladder, which prevents slipping. This frictional force can be maximized as follows:
\[
F_f = \mu N,
\]
where \(\mu = 0.45\) is the coefficient of friction.
2. **Weight Distribution**:
The weight of the person will act downward vertically at a distance of 1.4 m along the ladder. The height above the ground can be calculated by:
\[
h = 1.4 \sin \theta.
\]
Where \(d\) is the distance horizontally covered by the person:
\[
d = 1.4 \cos \theta.
\]
3. **Torque Calculations**:
For the ladder to be in equilibrium about the base where it rests on the floor, we set up the torque balance. The torque due to the weight of the person about the base of the ladder is:
\[
\tau_{person} = W_p \cdot (1.4 \cos \theta),
\]
The normal force will create no torque as it acts at the pivot point (the base). The torque due to the frictional force will also be considered in this equilibrium.
The force exerted by the wall will create a counteracting torque. Therefore, we set up the equation (assuming the length of the ladder is \(L\)):
\[
W_p \cdot (1.4 \cos \theta) = F_f \cdot L \cdot \sin \theta.
\]
4. **Finding Normal and Frictional Forces**:
The total vertical forces must balance:
\[
N = W_p.
\]
Therefore,
\[
F_f = \mu W_p = 0.45 \cdot 173.46 \approx 78.06 \, \text{N}.
\]
5. **Setting up Equation**:
We can substitute \(F_f\) into the torque equation:
\[
173.46 \cdot (1.4 \cos \theta) = (0.45 \cdot 173.46) \cdot L \cdot \sin \theta.
\]
Since the ladder's length \(L\) is the hypotenuse, \(L = \frac{1.4}{\sin \theta}\). Substituting gives:
\[
173.46 \cdot (1.4 \cos \theta) = (0.45 \cdot 173.46) \cdot \left(\frac{1.4}{\sin \theta}\right) \cdot \sin \theta.
\]
6. **Simplify**:
The equation simplifies to:
\[
173.46 \cdot 1.4 \cos \theta = 0.45 \cdot 173.46 \cdot 1.4.
\]
Dividing both sides by \(173.46 \cdot 1.4\):
\[
\cos \theta = 0.45.
\]
7. **Solve for \(\theta\)**:
Thus,
\[
\theta = \cos^{-1}(0.45).
\]
Calculating this yields:
\[
\theta \approx 63.6^\circ.
\]
Thus, the minimum angle \(\theta_{min}\) required for the person to climb 1.4 m up the ladder without slipping is approximately \( \boxed{63.6} \) degrees.