Question

Part 1: A ladder rests against a vertical wall. There is no friction between the wall and the ladder. The coefficient of static friction between the ladder and the ground is µ = 0.439 . Consider the following expressions:

A1: f = Fw
A2: f = Fw sin θ
B1: N = W/2
B2: N = W
C1: ℓ Fw sin θ = 2 Fw cos θ
C2: ℓ Fw sin θ = ℓ W cos θ
C3: ℓ Fw sin θ = 1/2 ℓ W cos θ ,
where f: force of friction between the ladder and the ground, Fw: normal force on the ladder due to the
wall, θ: angle between the ladder and the ground, N: normal force on the ladder due to the ground, W: weight of the ladder, and ℓ: length of the ladder. Identify the set of equations which is correct.

1. A2, B1, C2
2. A1, B2, C1
3. A2, B1, C3
4. A1, B1, C3
5. A2, B1, C1
6. A1, B1, C1
7. A1, B2, C2
8. A2, B2, C1
9. A1, B1, C2
10. A1, B2, C3

Part 2: Determine the smallest angle θ for which the ladder remains stationary. Answer in units of degrees.

Answer 1

Let's analyze each expression to identify which set of equations is correct.

### Part 1: Analyzing the Statements
1. **A1: \( f = F_w \)**
This implies that the frictional force equals the normal force from the wall. This does not hold true because the frictional force at the base counteracts the component of the wall force, so **A1 is incorrect**.

2. **A2: \( f = F_w \sin \theta \)**
The frictional force is related to the wall's normal force and the angle at which the ladder is inclined. This seems incorrect as friction at the base is determined by \( \mu N \).

3. **B1: \( N = \frac{W}{2} \)**
Assuming a uniform ladder where forces balance horizontally and vertically, this can only be correct if the weight acts at the midpoint and the loads are symmetrically distributed. If there's no other lateral force affecting \( N \), it will not simply be half of \( W\). Thus, **B1 may not hold true** without increased context.

4. **B2: \( N = W \)**
This indicates that the normal force on the ground equals the weight of the ladder. This is generally valid assuming the ladder is in static equilibrium, so **B2 is correct**.

5. **C1: \( \ell F_w \sin \theta = 2 F_w \cos \theta \)**
This equation does not balance properly considering torques/moments and the ladder system.

6. **C2: \( \ell F_w \sin \theta = \ell W \cos \theta \)**
This condition does not express the proper balance of forces; for static equilibrium, moments about the base must balance.

7. **C3: \( \ell F_w \sin \theta = \frac{1}{2} \ell W \cos \theta \)**
This introduces an incorrect factor of \( \frac{1}{2} \) when considering torque about the base.

After analysis, we can summarize:
- **Correct expressions**: B2.
- **Currently unclear or more specifics needed**: A1 (incorrect), A2 (needs relationship), B1 (requires conditions), C1 (incorrect), C2 & C3 (examined but not conditions).

Considering these points of failure, **the equations that could be inferred** as potentially valid could lead us to only accepting **B2**.

Without a perfect match, we could have **option 7**, which is the best candidate if we consider \( N = W \) true and choose \( A1 \) and \( A2 \) requiring further verification in different setups.

### Part 2: Determining the Smallest Angle θ
To find the smallest angle θ for which the ladder remains stationary:

1. **For equilibrium**,
- \( F_w = \mu N \)
- Normal force: \( N = W \) leads \( \mu W = F_w \)

2. **Using the moments:**
- Sum of torques about the base:
\( F_w \cdot \ell \cdot \sin \theta = W \cdot \frac{\ell}{2} \cdot \cos \theta \)

Substituting \( F_w = \mu W \) into it gives:
\( \mu W \cdot \ell \cdot \sin \theta = W \cdot \frac{\ell}{2} \cdot \cos \theta \)

Simplifying cancels \( W \) and \( \ell \):

\[
\mu \sin \theta = \frac{1}{2} \cos \theta
\]

3. **Rearranging the equation yields**:

\[
\tan \theta = \frac{1}{2\mu}
\]
Plugging in \( \mu = 0.439\):

\[
\tan \theta = \frac{1}{2 \times 0.439} = \frac{1}{0.878} \approx 1.138
\]

4. **Calculating θ**:

Finding \( θ \) gives:

\[
θ = \tan^{-1}(1.138) \approx 48.37°.
\]

### Final Answers
- **Part 1**: Best set of equations could not isolate forms; closest would be **option 7** based on common ground.
- **Part 2**: Smallest angle \( θ \) for which the ladder remains stationary is approximately **48.37 degrees**.