Math Problem Statement

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!

Solution

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 m1m_1, m2m_2, and m3m_3 are connected by strings.
  • m1m_1 and m2m_2 are hanging off the edges of the table (each on one side of the table), while m3m_3 is on the surface of the table.
  • There's a coefficient of static friction between m3m_3 and the table, given as μs=0.350\mu_s = 0.350.

We'll follow these steps:

Step 1: Identify forces acting on the system.

  • The weight of each mass is W=mgW = m \cdot g, where g=9.8m/s2g = 9.8 \, \text{m/s}^2.

  • For mass m1m_1 and m2m_2, they will experience gravitational forces pulling them down:
    Fg1=m1gF_{g1} = m_1 \cdot g
    Fg2=m2gF_{g2} = m_2 \cdot g

  • For mass m3m_3, it experiences two forces: tension in the strings and friction from the table surface. The friction force is given by Ff=μsNF_f = \mu_s \cdot N, where N=m3gN = m_3 \cdot g is the normal force.

Step 2: Write the equations of motion.

  1. For mass m1m_1 (moving downward): m1gT1=m1am_1 \cdot g - T_1 = m_1 \cdot a where T1T_1 is the tension in the string attached to m1m_1, and aa is the acceleration of the system.

  2. For mass m2m_2 (moving downward): m2gT2=m2am_2 \cdot g - T_2 = m_2 \cdot a where T2T_2 is the tension in the string attached to m2m_2.

  3. For mass m3m_3 (on the table): T1T2Ff=m3aT_1 - T_2 - F_f = m_3 \cdot a where Ff=μsm3gF_f = \mu_s \cdot m_3 \cdot g.

Step 3: Solve for acceleration aa.

First, let's calculate the friction force on m3m_3: Ff=μsm3g=0.35029.8=6.86NF_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. 49.8T1=4a4 \cdot 9.8 - T_1 = 4 \cdot a T1=39.24aT_1 = 39.2 - 4a

  2. 19.8T2=1a1 \cdot 9.8 - T_2 = 1 \cdot a T2=9.8aT_2 = 9.8 - a

  3. T1T26.86=2aT_1 - T_2 - 6.86 = 2 \cdot a Substitute T1=39.24aT_1 = 39.2 - 4a and T2=9.8aT_2 = 9.8 - a into this equation: (39.24a)(9.8a)6.86=2a(39.2 - 4a) - (9.8 - a) - 6.86 = 2a Simplify: 29.43a6.86=2a29.4 - 3a - 6.86 = 2a 22.54=5a22.54 = 5a a=22.545=4.51m/s2a = \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 m3m_3 is increased?
  3. How would the setup change if m1m_1 and m2m_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 m3m_3 and the table?

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

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Math Problem Analysis

Mathematical Concepts

Newton's Second Law
Forces
Friction
Tension in Strings

Formulas

F = m * a
F_friction = μ * N
N = m * g
Tension equations for m1, m2, and m3

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 10-12