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To solve this problem, we need to calculate the minimum coefficient of static friction \(\mu_s\) required to prevent the log from sliding off the truck during acceleration. Here's a breakdown of how to approach this:

### Given:
- Mass of the log: \(m = 789 \, \text{kg}\)
- Mass of the truck: \(M = 9750 \, \text{kg}\)
- Truck's acceleration: From \(0 \, \text{mph}\) to \(55 \, \text{mph}\) in \(24.0 \, \text{s}\)
  - Convert \(55 \, \text{mph}\) to \(\text{m/s}\): 
    \[
    55 \, \text{mph} = 24.5872 \, \text{m/s}
    \]
  - The acceleration \(a\) of the truck is:
    \[
    a = \frac{\Delta v}{\Delta t} = \frac{24.5872 \, \text{m/s}}{24.0 \, \text{s}} = 1.024 \, \text{m/s}^2
    \]
  
### Forces acting on the log:
- The force required to keep the log in place is the force of static friction \(f_s\).
- The force causing the log to slide is the inertial force due to acceleration, which is given by:
  \[
  f = m \cdot a = 789 \, \text{kg} \times 1.024 \, \text{m/s}^2 = 808.94 \, \text{N}
  \]

### Static friction equation:
The static friction \(f_s\) must be greater than or equal to the force trying to move the log. The frictional force is:
\[
f_s = \mu_s \cdot N
\]
where \(N\) is the normal force. Since the log is on a flat surface, \(N\) is simply the weight of the log:
\[
N = m \cdot g = 789 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 7732.29 \, \text{N}
\]

Setting the frictional force equal to the force required to prevent the log from sliding:
\[
\mu_s \cdot N = f
\]
\[
\mu_s = \frac{f}{N} = \frac{808.94 \, \text{N}}{7732.29 \, \text{N}} = 0.1046
\]

### Final Answer:
The minimum coefficient of static friction \(\mu_s\) needed is approximately \(0.105\).

### Comparison to the literature:
The problem states that the typical coefficient of static friction between the log and the trailer bed is around \(0.840\). Since \(0.105\) is significantly less than \(0.840\), this suggests that the log would not slide under the given conditions.

Would you like a deeper explanation or have any specific questions?

Here are five related questions:
1. How would the solution change if the truck's acceleration were doubled?
2. What would happen if the mass of the log increased?
3. How does kinetic friction differ from static friction in this context?
4. Why is it important to convert units properly when calculating acceleration?
5. How does the incline of a truck affect the normal force and friction?

**Tip:** Always check if the coefficient of friction given in problems is realistic by comparing it to typical values for different surfaces!

Calculate the minimum coefficient of static friction needed to keep the log in the back of the truck. The log has a mass of 789 kg, and the truck has a mass of 9750 kg. The truck accelerates from 0 to 55 mph in 24.0 s.

In this problem, we calculate the minimum coefficient of static friction (μ_s) required to keep a 789 kg log in the back of a truck during acceleration. The truck, with a mass of 9750 kg, accelerates from 0 to 55 mph in 24 seconds. Using Newton's second law and the static friction equation, we determine that μ_s is approximately 0.105.

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