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We need to find the maximum acceleration \(a_{\text{max}}\) that the truck can have before the crate starts slipping backward. This involves analyzing the forces acting on the crate and the role of friction. The key concepts are:

1. The force due to static friction, \( f_s \), which prevents the crate from slipping.
2. The gravitational force components acting on the crate, specifically parallel (\( mg \sin \theta \)) and perpendicular (\( mg \cos \theta \)) to the incline.
3. The maximum static friction is given by \( f_s^{\text{max}} = \mu_s F_N \), where \( \mu_s \) is the coefficient of static friction and \( F_N \) is the normal force.

### Step-by-Step Analysis:

1. **Forces in the \(y\)-direction (perpendicular to the incline):**
   The normal force \( F_N \) balances the perpendicular component of the gravitational force:
   \[
   F_N = mg \cos \theta
   \]
   where:
   - \( m \) is the mass of the crate,
   - \( g \) is the acceleration due to gravity,
   - \( \theta = 7^\circ \).

2. **Maximum static friction:**
   The maximum static friction force is:
   \[
   f_s^{\text{max}} = \mu_s F_N = \mu_s mg \cos \theta
   \]
   where \( \mu_s = 0.310 \).

3. **Forces in the \(x\)-direction (parallel to the incline):**
   The static friction force opposes the relative motion (slipping), while the gravitational force component \( mg \sin \theta \) pulls the crate down the incline. To find the maximum acceleration \( a_{\text{max}} \), we set up the following equation of motion along the incline:
   \[
   ma = f_s^{\text{max}} - mg \sin \theta
   \]
   Substituting \( f_s^{\text{max}} \) into this equation:
   \[
   ma = \mu_s mg \cos \theta - mg \sin \theta
   \]
   Simplifying, the mass \( m \) cancels out:
   \[
   a = g (\mu_s \cos \theta - \sin \theta)
   \]

### Substituting the values:
   - \( g = 9.81 \, \text{m/s}^2 \),
   - \( \mu_s = 0.310 \),
   - \( \theta = 7^\circ \).

I'll now calculate the maximum acceleration \( a_{\text{max}} \).The maximum acceleration \( a_{\text{max}} \) that the truck can attain before the crate begins to slip backward is approximately \( 1.82 \, \text{m/s}^2 \).

Would you like further clarification or details on any part of the process?

Here are some related questions you might find helpful:
1. How does the angle of inclination affect the maximum acceleration?
2. What would happen if the coefficient of friction were higher?
3. How does the mass of the crate impact the maximum acceleration?
4. How is the frictional force related to the normal force?
5. What if the truck were moving downhill instead of uphill?

**Tip:** When solving problems involving inclined planes, always break forces into components parallel and perpendicular to the incline for simpler calculations.

A flatbed truck is carrying a crate up a hill of angle of inclination θ = 7.00°, as the figure illustrates. The coefficient of static friction between the truck bed and the crate is μs = 0.310. Find the maximum acceleration that the truck can attain before the crate begins to slip backward relative to the truck.

Learn how to find the maximum acceleration of a truck carrying a crate up a 7° incline using static friction. The problem covers the application of Newton's second law and friction formulas. Suitable for high school students in physics classes.

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