Question

Two crates connected by a rope lie on a horizontal surface (\(\textbf{Fig. E5.37}\)). Crate \(A\) has mass \(m_A\), and crate \(B\) has mass \(m_B\). The coefficient of kinetic friction between each crate and the surface is \(\mu_k\). The crates are pulled to the right at constant velocity by a horizontal force \(F\). Draw one or more free-body diagrams to calculate the following in terms of \(m_A\), \(m_B\), and \(\mu_k\): (a) the magnitude of \(F\) and (b) the tension in the rope connecting the blocks.

Short answer

(a) \( F = \mu_k (m_A + m_B) g \), (b) \( T = \mu_k m_A g \).

Step-by-step solution

Draw Free-Body Diagrams

First, draw the free-body diagrams for both crates, Crate A and Crate B. Identify all the forces acting on each crate: gravitational force (weight), normal force, tension (T), kinetic frictional force (f_k), and applied force (F, only on Crate B). Since the velocity is constant, the net force on each crate is zero.

Calculate the Frictional Force (f_k)

The kinetic frictional force acting on each crate is calculated by the formula: \( f_k = \mu_k \cdot N \), where \(N\) is the normal force. Since the crates lie on a horizontal surface, the normal force on each crate equals its weight: \( N_A = m_A \cdot g \) and \( N_B = m_B \cdot g \). Thus, \( f_k = \mu_k \cdot m_A \cdot g \) for Crate A and \( f_k = \mu_k \cdot m_B \cdot g \) for Crate B.

Analyze Crate A

Equate the tension and the frictional force for Crate A as it moves at constant velocity, meaning net force is zero: \( T = \mu_k \cdot m_A \cdot g \). This means the tension in the rope is equal to the kinetic frictional force acting on Crate A.

Analyze Crate B

For Crate B, apply Newton’s second law in the horizontal direction: \( F - T - f_{kB} = 0 \). Since we've already calculated \( T \), substitute \( T = \mu_k \cdot m_A \cdot g \) and \( f_{kB} = \mu_k \cdot m_B \cdot g \) into this equation to solve for \( F \): \( F = \mu_k \cdot m_A \cdot g + \mu_k \cdot m_B \cdot g \). Thus, \( F = \mu_k \cdot (m_A + m_B) \cdot g \).

Summarize the Results

To find the force and tension: (a) the magnitude of the force \( F = \mu_k (m_A + m_B) g \); (b) the tension in the rope connecting the blocks \( T = \mu_k \cdot m_A \cdot g \).

Key concepts

Free-Body Diagram

In physics, a free-body diagram is a simple representation of an object and all the forces acting on it. It helps to visualize the problem and makes the process of solving it much easier.
When working with free-body diagrams, we represent the object as a dot or a box, and draw arrows originating from it to depict forces. These arrows indicate the direction and relative magnitude of the forces.

• In our exercise, the free-body diagrams for Crates A and B include multiple forces: gravitational force (weight), normal force, tension in the rope, kinetic frictional force, and an applied force on Crate B.

Drawing precise free-body diagrams is crucial because it sets the stage for accurate calculations required to solve the problem effectively. They allow us to see which forces need to be balanced or calculated to understand the system's movement.

Kinetic Friction

Kinetic friction is a force that acts between moving surfaces. It contrasts with static friction, which acts when surfaces are at rest. Kinetic friction always opposes the direction of movement.
The formula for calculating kinetic frictional force is:\[ f_k = \mu_k \cdot N \]where \( \mu_k \) is the coefficient of kinetic friction, and \( N \) is the normal force, which usually equals the weight of the object if it lies on a horizontal surface.

• For Crate A and B, their weights become \( m_A \cdot g \) and \( m_B \cdot g \) respectively, because gravity pulls them downward, resulting in respective frictional forces of \( \mu_k \cdot m_A \cdot g \) and \( \mu_k \cdot m_B \cdot g \).

Kinetic friction plays a vital role in determining how much force is needed to keep an object moving at a constant velocity.

Tension in Ropes

Tension is the pulling force transmitted through a string, rope, or cable when it is pulled tight by forces acting at each end. In the context of our exercise, tension (\( T \)) is present in the rope connecting the crates.
At constant velocity, the tension in the rope equals the force needed to overcome the kinetic friction acting on Crate A, thus we have:\[ T = \mu_k \cdot m_A \cdot g \]

• This equation tells us that tension is directly influenced by the frictional resistance of Crate A. Higher friction or mass would demand greater tension.

Tension is crucial in everyday scenarios, too, determining how objects like elevators or swings are able to move or stay suspended.

Force Analysis

Force analysis involves assessing all the forces acting on objects to understand their effects. This technique is grounded in Newton's Laws of Motion, specifically focusing on ensuring the net force is balanced when objects move at constant velocity.
For Crate A, force analysis showed that tension equaled kinetic friction, highlighting no net force. But for Crate B, we considered the applied force \( F \), the kinetic frictional force on Crate B, and the tension from Crate A:\[ F - T - f_{kB} = 0 \]

• This equation helps us solve for \( F \), by substituting previously calculated terms to get the force needed to pull both crates.

Force analysis is used commonly in engineering and physics to set up and simplify problem-solving by understanding how forces interact to influence motion.

Constant Velocity

Constant velocity means an object moves at a steady speed in a straight line. According to Newton’s first law, this implies no net force is acting on the object.
Our crates being pulled at constant velocity implies the applied force is exactly balancing out the frictional forces acting against the direction of movement.

• This situation is framed in the problem by setting net forces to zero. For both crates, this equilibrium explains why no acceleration occurs, maintaining the constant velocity.

Grasping the concept of constant velocity helps predict and calculate the type and magnitude of forces necessary for equilibrium in real-world applications like transport or conveyor systems.