Problem 60 - College Physics

Question

\(\bullet\) \(\bullet\) A block with mass \(M\) rests on a frictionless surface and is connected to a horizontal spring of force constant \(k,\) the other end of which is attached to a wall (Figure 11.41 ). A second block with mass \(m\) rests on top of the first block. The coefficient of static friction between the blocks is \(\mu_{s}\) . Find the maximum amplitude of oscillation such that the top block will not slip on the bottom block.

Step-by-Step Solution

Answer

The maximum amplitude of oscillation is \( A = \frac{M\mu_s g}{k} \).

Step 1: Identify forces and system dynamics

We begin by analyzing the forces acting on the system. The spring pulls the mass block $M$ with force $F_s = -kx$, where $x$ is the displacement from the equilibrium position. The top block $m$ will experience the force of static friction as it rests on block $M$ while the system oscillates.

Step 2: Understand static friction criteria

For block \(m\) to not slip on block \(M\), the static friction force \(f_s\) needs to be sufficient to counteract the maximum acceleration of block \(m\). The maximum static friction is \(f_s = \mu_s N = \mu_s mg\), where \(N\) is the normal force, which equals \(mg\) because the blocks are on a horizontal surface.

Step 3: Use Newton's Second Law

The maximum acceleration \(a_{max}\) that block \(m\) can have without slipping is given by \(a_{max} = \frac{f_s}{m} = \mu_s g\). During the maximum displacement, \(a_{max}\) is also the maximum acceleration of the spring-block system, so \(a_{max} = \frac{kA}{M}\), where \(A\) is the amplitude of the oscillation.

Step 4: Solve for maximum amplitude

Equating the expressions for maximum acceleration from both the spring and static friction criteria, we get: \(a_{max} = \mu_s g = \frac{kA}{M}\). Solve this for the maximum amplitude \(A\): \(A = \frac{M\mu_s g}{k}\). This is the maximum amplitude such that block \(m\) does not slip on block \(M\).

Key Concepts

Static Friction

Static friction is the force that prevents two surfaces from sliding past each other. It acts when objects are at rest relative to each other. In the context of the spring-block system, static friction is crucial because it keeps the top block from slipping off the bottom block. To understand static friction, we need to focus on the concept of the maximum static friction force, which is the greatest force that static friction can exert. This is given by the formula:

• \( f_s = \mu_s N \)

where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. Since the problem involves a horizontal surface, the normal force \( N \) is simply the weight of the top block, or \( mg \), where \( m \) is the mass of the top block and \( g \) is the acceleration due to gravity. Static friction ensures the top block moves together with the bottom block without slipping, which is necessary for analyzing the oscillating system.

Spring-Block System

A spring-block system consists of blocks attached to a spring, which can oscillate horizontally or vertically. In our exercise, a spring-block system oscillates on a frictionless surface. This allows us to explore the dynamics of harmonic oscillation where only the spring force and static friction are at play. The essential feature of a spring is its force constant, represented by \( k \), which indicates how stiff the spring is. The reactive force the spring exerts is described by Hooke's law:

• \( F_s = -kx \)

where \( x \) is the displacement from the equilibrium position. This force aims to restore the spring to its natural length, resulting in oscillatory motion. The spring-block system's task is to maintain balance such that the top block, with the influence of static friction, doesn't slip.

Maximum Amplitude

The maximum amplitude of the spring-block system is the largest distance from the equilibrium point that the system can achieve during oscillation without causing the top block to slip. The maximum amplitude is crucial because it marks the system's limit for non-slipping motion. To find this amplitude, we must equate the maximum static friction force to the maximum force required to keep the mass from slipping. The formula derived is:

• \( A = \frac{M \mu_s g}{k} \)

where \( A \) is the maximum amplitude, \( M \) is the mass of the block, \( \mu_s \) is the coefficient of static friction, \( g \) is gravitational acceleration, and \( k \) is the spring constant. This equation tells us how far the spring can stretch or compress at its maximum oscillation while ensuring the integrity of the system and preventing slipping.

Newton's Second Law

Newton's Second Law is pivotal in solving problems involving dynamics and force. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This is mathematically represented as:

• \( F = ma \)

In the spring-block system, this principle helps relate the forces due to the spring and static friction to the motion of the blocks. Specifically, it's used to determine the maximum acceleration that the top block can sustain without slipping. When static friction is balanced with the spring's maximum restoring force, Newton's Second Law enables us to find the relationship:\( a_{max} = \mu_s g = \frac{kA}{M} \). This relation allows us to calculate the maximum amplitude \( A \) by analyzing how far the system can oscillate within the limits of static friction.