Problem 72 - University Physics with Modern Physics

Question

CP 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 the spring is attached to a wall (Fig. Pl4.72). A second block with mass \(m\) rests on top of the first block. The coefficient of static friction between the blocks is \(\mu_{\mathrm{s}}\) . Find the maximum amplitude of oscillation such that the top block will not slip on the bot tom block.

Step-by-Step Solution

Answer

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

Step 1: Understand the System

We have two blocks. Block with mass \(M\) is on a frictionless surface connected to a spring with force constant \(k\), and block with mass \(m\) is on top of it. Only static friction prevents block \(m\) from slipping off block \(M\). As the spring oscillates, the system undergoes simple harmonic motion.

Step 2: Identify Forces Involved

The force of static friction \(F_f\) that keeps block \(m\) from slipping is given by \(F_f = \mu_s N\), where \(N = m \cdot g\) is the normal force on block \(m\). The force that tends to make block \(m\) slip is the maximum inertial force exerted by the oscillation of the spring, \(F_{max} = m \cdot a_{max}\), where \(a_{max}\) is the maximum acceleration.

Step 3: Determine Maximum Acceleration

In simple harmonic motion, the maximum acceleration \(a_{max}\) of the block is given by \(a_{max} = \omega^2 A\), where \(\omega = \sqrt{\frac{k}{M}}\) is the angular frequency of the system and \(A\) is the amplitude of oscillation. Thus, we have \(a_{max} = \frac{k}{M} A\).

Step 4: Equate Forces for Non-Slip Condition

For the block \(m\) not to slip, the static friction force must be equal to or greater than the inertial force (\(F_f \geq F_{max}\)). Substituting for these forces gives: \(\mu_s m g \geq m \cdot \frac{k}{M} A\). Simplifying, we get \(\mu_s g \geq \frac{k}{M} A\).

Step 5: Solve for Maximum Amplitude

Rearrange the inequality to solve for amplitude \(A\). This results in: \(A \leq \frac{\mu_s g M}{k}\). This expression gives us the maximum amplitude of oscillation such that block \(m\) does not slip on block \(M\).

Key Concepts

Static Friction

Static friction is a quintessential concept when you have two surfaces in contact and one surface tends to slide over the other. It is this hidden hero that prevents motion when an object is at rest. Imagine placing a book on a table. Even if you tilt the table slightly, the book stays put thanks to static friction.

Static friction depends on two main factors:

• The nature of the surfaces in contact (smooth, rough, etc.).
• The normal force, which is generally the weight of the object on a horizontal surface.

In our problem, static friction comes into play between the two blocks. The top block will remain stationary relative to the bottom block as long as the static friction force is greater than or equal to the force that attempts to move it. The formula we use for static friction here is:\[F_f = \mu_s N\]where \(\mu_s\) is the static friction coefficient and \(N\) is the normal force, calculated as \(m \cdot g\), with \(g\) representing the acceleration due to gravity. The challenge is to ensure that this frictional force can counterbalance the movement caused by the oscillating spring.

Amplitude of Oscillation

Oscillation involves repetitive back-and-forth motion, such as the swinging of a pendulum or even the vibrations of a guitar string. In simple harmonic motion, amplitude represents the furthest point that an object moves from its starting position — essentially, its max reach.

For our problem, the block-spring system oscillates horizontally, and amplitude \(A\) signifies this maximum displacement. When calculating amplitude in the context of the non-slipping condition, we must balance it against the static friction capability. The amplitude must be small enough that the movement does not exceed the maximum static friction force keeping the top block in place.

The formula derived from our step-by-step solution is:\[A \leq \frac{\mu_s g M}{k}\]This tells us the highest possible amplitude, where \(\mu_s\) is the static friction coefficient, \(g\) is gravity, \(M\) is the mass of the bottom block, and \(k\) is the spring constant. A larger amplitude risks surpassing the static friction, causing the top block to slip.

Angular Frequency

The angular frequency \(\omega\) is a measure of how quickly an object moves back and forth in simple harmonic motion, serving a similar purpose to frequency in rotational motion but without requiring circular paths.

In simple harmonic motion, the angular frequency can be calculated using the spring constant \(k\) and the mass it's moving, in this case, the mass of block \(M\). The formula for angular frequency is:\[\omega = \sqrt{\frac{k}{M}}\]This equation is critical because it is directly related to the acceleration of our system, specifically in the maximum acceleration formula \(a_{max} = \omega^2 A\).

A higher angular frequency implies a system that oscillates faster. In our problem, we use angular frequency to find the maximum acceleration that still allows the top block to stay without slipping. Balancing the amplitude and angular frequency with static friction ensures the system's stability.