https://file.aiquickdraw.com/mathbot/file/1727733276501qcic4tki.jpg

Let's walk through the problem based on the image you've uploaded, which involves the free body diagram (FBD) and forces acting on a crate.

### Part (a): Free Body Diagram (FBD)
We need to consider the following forces acting on the crate:

1. **\( F_g \)**: The gravitational force acting downward (weight of the crate).
2. **\( F_n \)**: The normal force exerted by the wall, acting perpendicular to the surface of contact (in this case, horizontally to the left).
3. **\( F_s \)**: The static friction force acting vertically upward or downward depending on the situation (since the crate isn't sliding, this holds it in place vertically).
4. **\( F_{\text{app}} \)**: The applied force pushing the crate horizontally to the right.
5. **\( F_k \)**: Kinetic friction force is not needed here, since the crate is not sliding.

You can label the axes as follows:
- \( x \)-axis is horizontal (to the right), positive direction where \( F_{\text{app}} \) acts.
- \( y \)-axis is vertical (upward).

### Part (b): x-component of Net Force (\( F_{\text{net},x} \))
In the horizontal direction, we have:

- Applied force \( F_{\text{app}} \) acting to the right.
- Normal force \( F_n \) acting to the left.

Since the crate isn't moving horizontally (no sliding), the net force in the x-direction is zero:

\[
F_{\text{net},x} = F_{\text{app}} - F_n = 0
\]

Thus:

\[
F_{\text{app}} = F_n
\]

### Part (c): y-component of Net Force (\( F_{\text{net},y} \))
In the vertical direction, we have:

- Gravitational force \( F_g \) acting downward.
- Static friction \( F_s \) acting upward.

Since the crate isn’t moving vertically, the net force in the y-direction is zero:

\[
F_{\text{net},y} = F_s - F_g = 0
\]

Thus:

\[
F_s = F_g
\]

### Part (d): Minimum Magnitude of Applied Force
To prevent the crate from sliding, the static friction must be strong enough to balance the gravitational force. The maximum static friction is given by:

\[
F_s = \mu_s F_n
\]

From Part (b), we know \( F_n = F_{\text{app}} \), so:

\[
F_s = \mu_s F_{\text{app}}
\]

To prevent sliding, the static friction must equal the weight of the crate:

\[
\mu_s F_{\text{app}} = F_g
\]

Thus, the minimum applied force \( F_{\text{app}} \) is:

\[
F_{\text{app}} = \frac{F_g}{\mu_s}
\]

---

Do you have any questions or want to explore any of these steps in more detail?

### 5 Related Questions:
1. How do different angles of applied force affect the friction required to keep the crate stationary?
2. What happens to the forces if the coefficient of static friction changes?
3. How would kinetic friction come into play if the crate started to slide?
4. Can the applied force be resolved into components if it's applied at an angle?
5. How would the problem change if the surface wasn't perfectly vertical?

### Tip:
Always start by drawing a clear free body diagram when solving problems involving forces, as it helps visualize all interactions.

A student of Physics I who is moving a heavy wooden crate is becoming well-versed concerning friction. This student momentarily shifts the burden by pressing the crate against a wooden wall with a horizontally applied force, Fapp as shown in the figure. Assume that the crate does not slide. The coefficients of static and kinetic friction are μs and μk, respectively. Coordinate axes have been indicated.

This detailed solution explains how to analyze the forces acting on a wooden crate pressed against a wall using applied force, including static friction and equilibrium conditions. Suitable for introductory physics students.

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