Problem 15 - Applied Physics

Question

A bar weighing \(26.0 \mathrm{~N}\) is supported horizontally on each end by two hanging springs, each \(15.0 \mathrm{~cm}\) long, with spring constants \(0.970 \mathrm{~N} / \mathrm{cm}\) and \(1.45 \mathrm{~N} / \mathrm{cm}\), respectively. The bar is \(6.00 \mathrm{~m}\) long and has a center of mass \(2.00 \mathrm{~m}\) from the spring with constant \(0.970 \mathrm{~N} / \mathrm{cm} .\) How far does each spring stretch?

Step-by-Step Solution

Answer

Each spring stretches by 13.4 cm and 8.97 cm respectively.

Step 1: Identify Forces and Torques

The bar is in equilibrium, therefore the sum of vertical forces and torques around any point should equal zero. The forces acting are the weight of the bar and the spring forces. The sum of torques around the point where the spring with constant 0.970 N/cm is located will be used to find the force in the other spring.

Step 2: Calculate the Force on Each Spring

The weight of the bar (26.0 N) is supported by two springs, so the sum of the forces from the springs equals the weight of the bar: \[ F_1 + F_2 = 26.0 \text{ N} \] where \( F_1 \) and \( F_2 \) are the forces from the springs with spring constants 0.970 N/cm and 1.45 N/cm respectively.

Step 3: Use Torque Equation to Solve for Forces

Taking torques about the point where the 0.970 N/cm spring is attached:\[ 2.0 F_2 = 2.0 \times 26.0 \text{ N} \] because the center of mass of the bar is 2.0 m from this spring, and the total length of the bar is 6.00 m. This equation gives the force on the spring with constant 1.45 N/cm as \( F_2 = 13.0 \text{ N} \). Substituting into the earlier equation gives \( F_1 = 13.0 \text{ N} \).

Step 4: Calculate Stretch of Each Spring

Using Hooke's Law, which states \( F = kx \), where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the stretch:- For the first spring with \( k = 0.970 \text{ N/cm} \): \[ x_1 = \frac{F_1}{k_1} = \frac{13.0 \text{ N}}{0.970 \text{ N/cm}} = 13.4 \text{ cm} \]- For the second spring with \( k = 1.45 \text{ N/cm} \): \[ x_2 = \frac{F_2}{k_2} = \frac{13.0 \text{ N}}{1.45 \text{ N/cm}} = 8.97 \text{ cm} \]

Key Concepts

Torque and Rotational Motion

Torque and rotational motion are fundamental concepts when analyzing objects in equilibrium. Torque, also known as the moment of force, refers to the rotational effect of a force applied to a pivot point or fulcrum. When a force is applied at a distance from this pivot, it creates a turning effect, which we define as torque.
A key principle involving torque is that for an object to be in rotational equilibrium, the net torque acting on it must be zero. This is essential when solving problems involving beams or bars supported by multiple forces, including springs.

• In our problem, the bar is 6.00 m long, and its weight acts at the center of mass, 2.00 m from the spring with constant 0.970 N/cm.
• By applying the concept of torque, the equilibrium condition is such that the clockwise torque about the pivot must equal the counterclockwise torque.
• We use the equation: \( \tau = r \times F \), where \( \tau \) is the torque, \( r \) is the perpendicular distance from the pivot, and \( F \) is the force applied.

Understanding how to calculate torque and apply it to determine the forces exerted by springs is crucial to solving equilibrium problems effectively.

Hooke's Law

Hooke's Law is a fundamental principle used to describe how springs and other elastic materials respond to applied forces. The law is mathematically formulated as \( F = kx \), where \( F \) is the force applied to the spring, \( x \) is the displacement or stretch of the spring, and \( k \) is the spring constant. This constant is a measure of the spring's stiffness.
In equilibrium scenarios involving springs, understanding Hooke's Law allows us to determine how much a spring stretches or compresses when a certain force is applied.

• Springs in our exercise have constants of 0.970 N/cm and 1.45 N/cm, indicating their relative stiffness.
• The use of Hooke's Law helps calculate the exact stretch of each spring when supporting the weight of the bar.
• By rearranging the equation, \( x = \frac{F}{k} \), we can find the stretch, where equilibrium conditions ensure forces are balanced.

Remember, the direct proportionality between force and stretch allows for easy computation of changes in spring length under various forces.

Spring Force Calculation

Spring force calculation is vital for assessing how springs react when subjected to different forces, especially in problems requiring equilibrium analysis. In the context of our problem, we calculate how much each spring stretches under the weight of the bar.
By knowing the spring constants and ensuring the sum of forces equals the weight of the bar, we can deduce individual spring forces.

• First, equilibrium dictates that the total force exerted by two springs equals the weight of the 26.0 N bar.
• Torque considerations help us assign parts of this total force to each spring. For instance, the spring with 1.45 N/cm is found to exert a force of 13.0 N.
• Once we know the forces, Hooke's Law unveils the springs' elongations: the first spring stretches 13.4 cm, and the second spring stretches 8.97 cm.

Through understanding the dynamics of spring force calculations, the interplay of multiple forces on a mechanical system becomes clear. This assists in comprehending not just theoretical problems, but real-world applications involving spring mechanisms.