Problem 15 - University Physics with Modern Physics

Question

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. E5.15. 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 tension in the rope while the load is moving? How does the tension compare to the weight of the load of bricks? To the weight of the counterweight?

Step-by-Step Solution

Answer

The bricks accelerate at 3.15 m/s²; tension is 100.95 N, less than both weights.

Step 1: Understand the Problem

We have two masses connected by a rope over a frictionless pulley. One is the load of bricks weighing 15.0 kg, and the other is a counterweight of 28.0 kg. We are asked to determine the free-body diagrams for each mass, the magnitude of the acceleration of the bricks, and the tension in the rope as the system moves.

Step 2: Draw Free-Body Diagrams

For the load of bricks (15.0 kg), the forces acting are the gravitational force downward (weight) and the tension in the rope upward. For the counterweight (28.0 kg), the forces are the tension in the rope upward and the gravitational force downward.

Step 3: Formulate Equations of Motion

Use Newton's second law for both masses. For the load of bricks:\[ m_1 g - T = m_1 a \] where \( m_1 = 15.0 \text{ kg} \), \( g = 9.8 \text{ m/s}^2 \), and \( T \) is the tension.For the counterweight:\[ T - m_2 g = -m_2 a \] where \( m_2 = 28.0 \text{ kg} \).

Step 4: Solve for Acceleration

Add the equations from step 3:\[ m_1 g - m_2 g = m_1 a + m_2 a \] Simplify and solve:\[ (m_2 - m_1)g = (m_1 + m_2)a \] \[ a = \frac{(m_2 - m_1)g}{(m_1 + m_2)} \] Plug in the values:\[ a = \frac{(28.0 - 15.0) \, \times \, 9.8}{28.0 + 15.0} \approx 3.15 \, \text{m/s}^2 \]

Step 5: Solve for Tension in the Rope

Substitute the value of \( a \) back into one of the equations from Step 3. We'll use the equation for the load of bricks:\[ T = m_1 g - m_1 a \]\[ T = 15.0 \, \times \, 9.8 - 15.0 \, \times \, 3.15 \approx 100.95 \, \text{N} \]

Step 6: Compare Tension to Weights

The tension \( T \) is 100.95 N:- The weight of the bricks \( W_1 = m_1 g = 15.0 \times 9.8 = 147 \, \text{N} \), so \( T < W_1 \).- The weight of the counterweight \( W_2 = m_2 g = 28.0 \times 9.8 = 274.4 \, \text{N} \), so \( T < W_2 \).

Key Concepts

Understanding Free-Body Diagrams

A free-body diagram is a simple yet essential tool in physics to visualize the forces acting on an object. Let's consider our two objects: the 15.0 kg load of bricks and the 28.0 kg counterweight.
For the load of bricks, the free-body diagram includes:

• The gravitational force (weight) acting downward, calculated as the mass of the bricks multiplied by the gravitational acceleration (\( m_1g \)).
• The tension in the rope acting upward, which is the force transmitted through the rope.

On the other hand, for the counterweight, the free-body diagram shows:

• The gravitational force (weight) acting downward (\( m_2g \)).
• The tension in the rope acting upward.

By creating these diagrams, we can clearly see all forces involved and set the stage for applying Newton's Second Law to solve for unknowns like tension and acceleration.

Explaining Tension in the Rope

Tension in a rope is a force transmitted through a string, rope, cable, or similar when it is pulled tight by forces acting from opposite ends.
In this system involving a pulley, the tension remains constant throughout the rope because the pulley is frictionless and the rope's mass is negligible.
To find the tension while the brick load is moving, we applied Newton's Second Law. We used the equations:

• For the load of bricks: \( T = m_1g - m_1a \)
• For the counterweight: \( T = m_2g + m_2a \)

By solving these equations, we determined that the tension in the rope is approximately 100.95 N when the system moves.
It is crucial to compare this tension with the weights of both masses to understand its influence on the system dynamics.
The tension is less than both the weight of the bricks (\(147 ext{ N}\)) and the counterweight (\(274.4 ext{ N}\)), serving as a balancing force but not a complete one.

Calculating Acceleration of Objects

The acceleration of an object measures how quickly it changes velocity. When calculating the acceleration of objects connected by a rope over a pulley, we need to consider the difference in their weights.
Applying Newton’s Second Law to our scenario, we derived that:

• The net force acting on the entire system is the difference in gravitational forces of each mass (\((m_2g - m_1g)\)).
• Thus, the net effective force is used to calculate acceleration using the combined mass of both objects.

The formula is given as: \[ a = \frac{(m_2 - m_1)g}{(m_1 + m_2)} \]
Plugging in the values, we find the acceleration of the bricks is approximately 3.15 m/s².
This is a critical insight because it indicates how the difference in mass affects the motion, accelerating the smaller mass (bricks) upward as the heavier mass (counterweight) descends.