2.8 Spring Forces

Spring forces are a fundamental concept in physics, describing how objects respond to stretching or compression. These forces follow Hooke's law, which states that the force exerted by a spring is proportional to its displacement from equilibrium.

Understanding spring forces is crucial for analyzing oscillatory motion and energy storage in mechanical systems. The direction of the force always opposes the displacement, acting as a restoring force that tries to bring the system back to its equilibrium position.

Force of Ideal Spring

An ideal spring has specific properties that make it useful for modeling physical systems and understanding elastic behavior.

Characteristics of Ideal Spring

An ideal spring is a theoretical concept that simplifies our analysis of spring behavior. In reality, no spring is perfectly ideal, but many springs approximate these characteristics closely enough for practical purposes.

The mass of an ideal spring is considered negligible compared to other objects in the system.

• This means we can ignore the spring's weight in our calculations
• We can treat the spring as if all its mass is concentrated at the ends
• This simplification allows us to focus on the interaction between the spring and attached objects

An ideal spring follows a linear force-displacement relationship throughout its range of motion.

• The force exerted by the spring is directly proportional to how much it's stretched or compressed
• If you stretch a spring twice as far, it pulls back with twice the force
• This relationship remains consistent regardless of how many times you use the spring
• The spring's relaxed length is its natural length when no external forces are applied
• The equilibrium position is where the net force on the system equals zero

Hooke's Law

Hooke's law mathematically describes how springs behave when stretched or compressed from their natural length.

The force exerted by an ideal spring is given by:

$F = -k\Delta x$

Where:

• $F$ is the force exerted by the spring (measured in Newtons)
• $k$ is the spring constant, which measures the spring's stiffness (in N/m)
• $\Delta x$ is the displacement from the spring's equilibrium position (in meters)
• The negative sign indicates the force acts in the opposite direction of the displacement

A spring with a large spring constant (k) is stiffer and requires more force to stretch or compress than one with a small spring constant. For example, a spring with k = 100 N/m will exert twice the force of a spring with k = 50 N/m when stretched by the same amount.

Direction of Spring Force

The spring force always acts as a restoring force, pulling or pushing the system back toward its equilibrium position.

When a spring is stretched:

• The displacement ($\Delta x$) is positive
• The spring force is negative, pulling back toward equilibrium
• The force tries to return the spring to its natural length

When a spring is compressed:

• The displacement ($\Delta x$) is negative
• The spring force is positive, pushing away from the compressed position
• The force works to extend the spring back to its relaxed state

This restoring nature of spring forces is what enables springs to store potential energy and is responsible for oscillatory motion in spring systems. When an object attached to a spring is displaced and released, the spring force continually pulls it back toward equilibrium, causing it to oscillate back and forth.

Practice Problem 1: Spring Force Calculation

Solution

To solve this problem, we need to apply Hooke's law: $F = -k\Delta x$

Given:

• Spring constant $k = 25$ N/m
• Displacement $\Delta x = 0.15$ m (positive because the spring is stretched)

Substituting these values into Hooke's law: $F = -k\Delta x = -(25 \text{ N/m})(0.15 \text{ m}) = -3.75 \text{ N}$

The negative sign indicates that the force is directed opposite to the displacement. Since the spring is stretched in the positive direction, the force acts in the negative direction, trying to pull the spring back to its equilibrium position.

Therefore, the spring exerts a force of 3.75 N directed toward the equilibrium position.

Practice Problem 2: Finding Spring Constant

Solution

In this problem, we need to find the spring constant $k$ using Hooke's law and the fact that at equilibrium, the spring force balances the weight of the mass.

At equilibrium:

• The weight of the mass is $F_g = mg = (2 \text{ kg})(9.8 \text{ m/s}^2) = 19.6 \text{ N}$
• This weight is balanced by the spring force: $F_{\text{spring}} = k\Delta x$
• Since the system is at rest, $F_g = F_{\text{spring}}$

Therefore: $19.6 \text{ N} = k(0.08 \text{ m})$

Solving for $k$: $k = \frac{19.6 \text{ N}}{0.08 \text{ m}} = 245 \text{ N/m}$

The spring constant is 245 N/m.