Problem 66 - College Physics

Question

\(\bullet$$\bullet\) Atwood's Machine. A 15.0 -kg load of bricks hangs from one end of a rope that passes over a small, frictionless pulley. A 28.0 -kg counterweight is suspended from the other end of the rope, as shown in Fig. 5.60 . The system is released from rest. (a) Draw two free-body diagrams, one for the load of bricks and one for the counterweight. (b) What is the magnitude of the upward acceleration of the load of bricks? (c) What is the ten- sion in the rope while the load is moving? How does the ten- sion compare to the weight of the load of bricks? To the weight of the counterweight?

Step-by-Step Solution

Answer

The acceleration of the load of bricks is approximately 3.16 m/s² upwards. The tension in the rope is about 194.4 N, which is more than the weight of the bricks but less than the weight of the counterweight.

Step 1: Understand the Problem

This is an Atwood's Machine problem involving two masses hanging over a pulley. The goal is to determine the system's dynamics by analyzing forces on both masses.

Step 2: Draw Free-Body Diagrams

For the 15 kg load of bricks (mass 1), two forces act: gravitational force downward, \( F_1 = m_1 g = 15 \times 9.8 \) N, and tension \( T \) upward. For the 28 kg counterweight (mass 2), gravitational force down is \( F_2 = m_2 g = 28 \times 9.8 \) N, and tension \( T \) upward.

Step 3: Set Up Newton’s Second Law for Each Mass

Apply Newton's second law: for mass 1 (bricks), \( m_1 a = T - m_1 g \); and for mass 2 (counterweight), \( m_2 a = m_2 g - T \).

Step 4: Solve for Acceleration

Add the two equations from step 3:\[ m_1 a + m_2 a = m_2 g - m_1 g \]\[ a(m_1 + m_2) = (m_2 - m_1)g \]\[ a = \frac{(m_2 - m_1)g}{m_1 + m_2} \]Plug in values: \( a = \frac{(28 - 15) \times 9.8}{15 + 28} \approx 3.16 \text{ m/s}^2 \).

Step 5: Solve for Tension

Using equation for mass 1, \( T = m_1 a + m_1 g \). Plug in the previously found acceleration:\[ T = 15 \times 3.16 + 15 \times 9.8 \approx 194.4 \text{ N} \].

Step 6: Compare Tension to Weights

For bricks: weight \( 147 \text{ N} \) and for counterweight: weight \( 274.4 \text{ N} \). Tension (\( 194.4 \text{ N} \)) is more than the bricks' weight but less than the counterweight's weight.

Key Concepts

Free-Body Diagrams

To understand the dynamics of Atwood's Machine, the first step is to draw free-body diagrams for the involved masses. A free-body diagram is a visual tool used to illustrate all external forces acting on an object. For each mass in Atwood's Machine, two primary forces must be considered:

• Gravitational Force: This force pulls the mass downward and is equal to the mass times the gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \, \) . For the 15 kg load of bricks, this is \( 15 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 147 \, \text{N} \). Similarly, for the 28 kg counterweight, it is \( 28 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 274.4 \, \text{N} \).
• Tension in the Rope: The tension \( T \) acts upward on both masses, countering the force of gravity.

By clearly depicting these forces, we can better analyze the system's mechanics. In Atwood's Machine, these diagrams help set the stage for applying Newton's second law.

Newton's Second Law

Newton's Second Law is crucial for calculating the motion in Atwood's Machine. This law states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass:

• Mathematically, it is expressed as \( F = ma \), where \( F \) is the net force, \( m \) is mass, and \( a \) is acceleration.
• To apply this to our system, we write separate equations for each mass.

For the 15 kg load of bricks, the equation becomes \( m_1 a = T - m_1 g \). This implies that the upward force due to tension minus the gravitational force gives the net force, determining the acceleration upward.
Similarly, for the 28 kg counterweight, the equation is \( m_2 a = m_2 g - T \), where the force of gravity is reduced by tension.
These equations can then be combined to solve for the unknowns – acceleration and tension – of the system.

Acceleration

Acceleration in Atwood's Machine determines how fast the load of bricks and the counterweight will move. Using the formula derived from combining the equations from Newton's Second Law, we find:

• The acceleration \( a \) is calculated as \[ a = \frac{(m_2 - m_1)g}{m_1 + m_2}\]where \( m_1 = 15 \, \text{kg} \) and \( m_2 = 28 \, \text{kg} \).
• Substituting the given values, the acceleration is approximately \( a \approx 3.16 \, \text{m/s}^2 \).

This result indicates that the entire system accelerates due to the gravitational pull on the heavier mass.
The positive sign of the acceleration confirms that the net motion will be in the direction of the larger mass (the counterweight). By understanding acceleration, we gain insight into the motion's timing and dynamics within Atwood's Machine.

Tension

Tension in the rope of Atwood's Machine is a force that continuously adjusts to ensure both masses move synchronously over the pulley. Calculating the tension involves using the acceleration found earlier:

• For the 15 kg load of bricks, the tension is calculated as \( T = m_1 a + m_1 g \).
  Plugging in the values, we find \[T = 15 \times 3.16 + 15 \times 9.8 \approx 194.4 \, \text{N}\]
• This shows that the tension force not only supports the weight of the load of bricks but also contributes to its upward acceleration.

Comparing this tension to the weights:

• The tension force is more than the weight of the bricks (147 N), because it has to lift the bricks and provide acceleration.
• However, it is less than the counterweight's gravitational force (274.4 N), which is why the counterweight moves downwards, pulling the bricks upward.

Understanding tension is vital for ensuring the safe and effective design of systems like Atwood's Machine, highlighting how forces interact dynamically.