Problem 64 - College Physics

Question

A block with mass \(m_{1}\) is placed on an inclined plane with slope angle \(\alpha\) and is connected to a second hanging block with mass \(m_{2}\) by a cord passing over a small, frictionless pulley (Figure 5.73). The coefficient of static friction is \(\mu_{\mathrm{s}}\) and the coefficient of kinetic fric- tion is \(\mu_{\mathrm{k}}\) (a) Find the mass \(m_{2}\) for which block \(m_{1}\) moves up the plane at constant speed once it is set in motion. (b) Find the mass \(m_{2}\) for which block \(m_{1}\) moves down the plane at constant speed once it is set in motion. (c) For what range of values of \(m_{2}\) will the blocks remain at rest if they are released from rest?

Step-by-Step Solution

Answer

(a) \(m_2 = m_1(\sin \alpha + \mu_k \cos \alpha)\), (b) \(m_2 = m_1(\sin \alpha - \mu_k \cos \alpha)\), (c) \(m_1(\sin \alpha - \mu_s \cos \alpha) \leq m_2 \leq m_1(\sin \alpha + \mu_s \cos \alpha)\).

Step 1: Analyze Forces on Inclined Plane

Identify the forces acting on the block on the incline (\(m_1\)): gravitational force, normal force, frictional force, and tension from the rope. Apply Newton's second law. The gravitational force component pulling down the incline is \(m_1 g \sin \alpha\), and the normal force is \(m_1 g \cos \alpha\). The frictional force, opposing motion when moving at constant speed, is \(\mu_k m_1 g \cos \alpha\).

Step 2: Analyze Forces on Hanging Block

The hanging block \(m_2\) is affected by its weight \(m_2 g\) pulling it down and the tension \(T\) in the rope pulling it up. Since \(m_1\) is moving at constant speed, the net force on \(m_2\) must also be zero: \(m_2 g = T\).

Step 3: Write Equations for Constant Speed Solutions

For \(m_1\) moving up the incline at constant speed, apply \(T + \mu_k m_1 g \cos \alpha = m_1 g \sin \alpha\). Solving \(m_2 g = m_1 g \sin \alpha + \mu_k m_1 g \cos \alpha\), find \(m_2 = m_1(\sin \alpha + \mu_k \cos \alpha)\).

Step 4: Write Equation for Moving Down Incline

For \(m_1\) moving down at constant speed, the equation is adjusted since friction opposes gravity: \(m_1 g \sin \alpha = T + \mu_k m_1 g \cos \alpha\). Solving, \(m_2 g = m_1 g \sin \alpha - \mu_k m_1 g \cos \alpha\) gives \(m_2 = m_1(\sin \alpha - \mu_k \cos \alpha)\).

Step 5: Determine Static Equilibrium Range

For static equilibrium, both blocks must be motionless, which means friction balances gravity perfectly. Static friction satisfies \(-\mu_s m_1 g \cos \alpha \leq T - m_1 g \sin \alpha \leq \mu_s m_1 g \cos \alpha\). Solving, \(m_1(\sin \alpha - \mu_s \cos \alpha) \leq m_2 \leq m_1(\sin \alpha + \mu_s \cos \alpha)\).

Key Concepts

Newton's Second Law

Newton's Second Law states that the acceleration of an object depends on the net force acting upon it and its mass. In equation form, it is expressed as \( F = ma \), where \( F \) is the net force, \( m \) is the mass, and \( a \) is the acceleration. This fundamental law helps us understand the motion of objects by connecting force, mass, and acceleration together.

For our inclined plane scenario, Newton's Second Law is used to analyze forces on each block. For the block on the incline, the net force is balanced so that it moves at constant speed. This means the sum of forces in both the direction of motion and perpendicular to the motion is zero. Such forces include gravity, friction, and tension. Understanding how these forces interact underlines the physics of objects on inclined planes.

Static Friction

Static friction is the force that keeps an object at rest when a force tries to move it. Think of it as the grip of the surface holding the object tightly in place until enough force is applied. The coefficient of static friction \( \mu_s \) determines how strong this grip is, depending on the surfaces in contact.

• Static friction is what you need to overcome to start moving an object.
• It acts opposite to the direction of the applied force.
• It's maximal just before the object starts moving.

For our blocks, static friction keeps block \( m_1 \) from sliding down the incline when at rest. The static frictional force can be calculated using the formula \( f_s = \mu_s N \), where \( N \) is the normal force. In this scenario, \( N = m_1 g \cos \alpha \). The range of masses for block \( m_2 \) which keeps the system stationary requires the forces of static friction to balance the gravitational pull on \( m_1 \).

Kinetic Friction

When an object is moving, kinetic friction steps in. This force acts against the direction of motion and is generally less than static friction. The coefficient of kinetic friction \( \mu_k \) describes how much frictional force resists the motion of sliding objects.

• Kinetic friction keeps slowing down moving objects.
• It is constant and does not vary with speed.
• The force of kinetic friction is calculated as \( f_k = \mu_k N \).

In our scenario, as block \( m_1 \) moves up or down the incline at constant speed, kinetic friction plays a crucial role. It counters the movement, helping determine the necessary tension in the rope to maintain that consistent speed. The equations from the original solution show how kinetic friction balances the gravitational forces on the plane and influences the tension needed for steady motion.

Tension in Ropes

Tension is the force conducted along the rope or cable when it pulls objects from each side. It ensures the connected blocks work as a single system. Tension adjusts based on forces acting on both ends, like gravity or friction.

• Tension is uniform throughout if the rope is massless and frictionless.
• It acts opposite to the weight force on the hanging block.
• On the incline, it complements friction to pull objects upward.

In the inclined plane problem, tension in the connecting rope between the blocks links the equations of motion for both blocks. For moving blocks, tension must overcome kinetic friction on the incline, calculated considering weights and angles involved. Consistent tension allows for constant speed movement as determined in the solution steps, whether the block moves up or down.