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To solve the given problem, we need to determine:

### A. **The friction force**  
### B. **The minimum required coefficient of friction (μ) so that the box does not slip**

#### Problem Breakdown:
- The box weighs **1.7 slugs**, and a force **P = 24 lb** is applied at an incline of **51°**.
- The friction force and normal force act to prevent the box from moving.

Let’s use the following physics principles to solve the problem:

---

### 1. **Weight of the box**:
The mass of the box is given as **1.7 slugs**. To calculate the weight (force due to gravity):

\[
W = m \cdot g = 1.7 \, \text{slugs} \times 32.2 \, \frac{\text{ft}}{\text{s}^2} = 54.74 \, \text{lb}
\]

### 2. **Forces acting on the box**:
The forces acting on the box are:
- **P = 24 lb** (force applied parallel to the incline)
- **Weight, W = 54.74 lb** (acts vertically downward)
- **Normal force (N)** (acts perpendicular to the incline)
- **Friction force (f)** (acts parallel to the incline, opposing motion)

---

### 3. **Force components**:
Break the weight of the box into two components:
- Parallel to the incline (down the slope):
  \[
  W_{\parallel} = W \cdot \sin(51^\circ) = 54.74 \cdot \sin(51^\circ)
  \]
- Perpendicular to the incline:
  \[
  W_{\perp} = W \cdot \cos(51^\circ) = 54.74 \cdot \cos(51^\circ)
  \]

---

### 4. **Equilibrium conditions**:
For the box to not slip, the forces along both the parallel and perpendicular axes must balance.

- Along the **parallel direction (x-axis)**:
  \[
  P + f = W_{\parallel}
  \]
  Solving for **f**:
  \[
  f = W_{\parallel} - P
  \]

- Along the **perpendicular direction (y-axis)**:
  \[
  N = W_{\perp}
  \]

The friction force is related to the normal force by:
\[
f = \mu N
\]
Where **μ** is the coefficient of friction.

---

Now let's compute these values. I'll calculate them for you.### Solution:

#### A. The friction force:
\[
f = 18.54 \, \text{lb}
\]

#### B. The minimum required coefficient of friction (μ):
\[
\mu_{\text{min}} = \frac{f}{N} = \frac{18.54}{34.45} \approx 0.54
\]

So, the minimum coefficient of friction required to prevent the box from slipping is approximately **0.54**.

Would you like more details on any step? Here are some related questions to expand on this topic:

1. How would the friction force change if the incline angle increased?
2. How is the friction coefficient affected by the mass of the box?
3. What if the force \( P \) was directed at a different angle?
4. How do we handle dynamic friction if the box were sliding instead of static?
5. How does the gravitational acceleration affect the problem if done on a different planet?

**Tip**: When solving problems involving forces on an incline, always break the weight into parallel and perpendicular components for easier calculations.

A 24 lb force, P, is being applied to keep a 1.7-slug box from moving on a 51° incline. Determine the friction force and the minimum required coefficient of friction so the box does not slip.

Solve a statics problem involving a 1.7-slug box on a 51° incline, held by a 24 lb force. Determine the friction force and the minimum required coefficient of friction to prevent slipping, using key concepts of inclined planes and equilibrium.

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