Three objects are connected on a table. The coefficient of static friction between m3 and the table is 0.350. The objects have a masses m1=4 kg, m2=1 kg, and m3=2 kg. Determine the acceleration of the objects. Round to the nearest tenths, units!



Let's break down the problem step by step. We have three masses connected on a table. The goal is to find the acceleration of the system. We'll assume the setup is something like this:

- Masses \(m_1\), \(m_2\), and \(m_3\) are connected by strings.
- \(m_1\) and \(m_2\) are hanging off the edges of the table (each on one side of the table), while \(m_3\) is on the surface of the table. 
- There's a coefficient of static friction between \(m_3\) and the table, given as \(\mu_s = 0.350\).

We'll follow these steps:

### Step 1: Identify forces acting on the system.
- The weight of each mass is \( W = m \cdot g \), where \( g = 9.8 \, \text{m/s}^2 \).
- For mass \(m_1\) and \(m_2\), they will experience gravitational forces pulling them down:  
  \( F_{g1} = m_1 \cdot g \)  
  \( F_{g2} = m_2 \cdot g \)
  
- For mass \(m_3\), it experiences two forces: tension in the strings and friction from the table surface. The friction force is given by \( F_f = \mu_s \cdot N \), where \( N = m_3 \cdot g \) is the normal force.

### Step 2: Write the equations of motion.
1. For mass \(m_1\) (moving downward):
   \[
   m_1 \cdot g - T_1 = m_1 \cdot a
   \]
   where \( T_1 \) is the tension in the string attached to \(m_1\), and \(a\) is the acceleration of the system.

2. For mass \(m_2\) (moving downward):
   \[
   m_2 \cdot g - T_2 = m_2 \cdot a
   \]
   where \( T_2 \) is the tension in the string attached to \(m_2\).

3. For mass \(m_3\) (on the table):
   \[
   T_1 - T_2 - F_f = m_3 \cdot a
   \]
   where \( F_f = \mu_s \cdot m_3 \cdot g \).

### Step 3: Solve for acceleration \(a\).

First, let's calculate the friction force on \(m_3\):
\[
F_f = \mu_s \cdot m_3 \cdot g = 0.350 \cdot 2 \cdot 9.8 = 6.86 \, \text{N}
\]

Now we can set up the system of equations:
1. \( 4 \cdot 9.8 - T_1 = 4 \cdot a \)
   \[
   T_1 = 39.2 - 4a
   \]

2. \( 1 \cdot 9.8 - T_2 = 1 \cdot a \)
   \[
   T_2 = 9.8 - a
   \]

3. \( T_1 - T_2 - 6.86 = 2 \cdot a \)
   Substitute \( T_1 = 39.2 - 4a \) and \( T_2 = 9.8 - a \) into this equation:
   \[
   (39.2 - 4a) - (9.8 - a) - 6.86 = 2a
   \]
   Simplify:
   \[
   29.4 - 3a - 6.86 = 2a
   \]
   \[
   22.54 = 5a
   \]
   \[
   a = \frac{22.54}{5} = 4.51 \, \text{m/s}^2
   \]

### Final Answer:
The acceleration of the system is approximately **4.5 m/s²**.

Would you like more details on the steps or equations? Here are five follow-up questions to consider:

1. How does the coefficient of friction affect the overall acceleration of the system?
2. What happens to the acceleration if the mass \(m_3\) is increased?
3. How would the setup change if \(m_1\) and \(m_2\) were on different surfaces with different friction coefficients?
4. Can you calculate the tensions in the strings after finding the acceleration?
5. How would the results differ if there was no friction between \(m_3\) and the table?

**Tip**: Always check the direction of friction. It opposes the motion of the object on the surface!

Three objects are connected on a table. The coefficient of static friction between m3 and the table is 0.350. The objects have masses m1=4 kg, m2=1 kg, and m3=2 kg. Determine the acceleration of the objects. Round to the nearest tenths, units!

Find the acceleration of three connected objects with masses m1=4 kg, m2=1 kg, and m3=2 kg, including frictional forces between m3 and the table. This detailed solution explains the application of Newton's Second Law, tension forces, and friction in the system. Ideal for students in Grades 10-12.

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