Solving for the Weight of a Uniform Meter Ruler Using Moments and Torque
Introduction
Understanding the principles of moments and torque is crucial in engineering and physics. This article explores how to determine the weight of a uniform meter ruler using the principles of equilibrium. We will apply these principles to a specific problem involving a ruler with a hanging weight.
Problem Statement
A uniform meter ruler has a 3 N weight hanging from one end. The ruler balances when suspended from a point 0.2 m from that end. We need to find the weight of the ruler.
Principle of Moments and Torque
The principle of moments states that for an object to be in equilibrium about a pivot point, the sum of the clockwise moments must equal the sum of the counterclockwise moments. Moments are calculated as the product of the force and the perpendicular distance from the pivot point.
Given Values
Weight of the hanging mass, ( W_h 3 , text{N} ) Distance from the hanging mass to the pivot, ( d_h 0.2 , text{m} ) Length of the ruler, ( L 1 , text{m} )Assumptions
The weight of the ruler acts at its center of mass, located 0.5 m from either end. The distance from the center of the ruler to the pivot point is ( 0.5 , text{m} - 0.2 , text{m} 0.3 , text{m} ).Calculation and Equilibrium
To solve this problem, we can use the principle of moments about the pivot point where the ruler is suspended.
Uneven Torque
The clockwise moment from the hanging weight is given by: ( text{Moment}_h W_h times d_h 3 , text{N} times 0.2 , text{m} 0.6 , text{Nm} ) The counterclockwise moment from the ruler's weight is given by: ( text{Moment}_r W_r times d_r W_r times 0.3 , text{m} )For the ruler to be in equilibrium, the clockwise moment must equal the counterclockwise moment:
$$ W_h times d_h W_r times d_r $$
Substitute the given values:
$$ 3 , text{N} times 0.2 , text{m} W_r times 0.3 , text{m} $$
Simplifying:
$$ 0.6 , text{Nm} W_r times 0.3 , text{m} $$
Solving for ( W_r ):
$$ W_r frac{0.6 , text{Nm}}{0.3 , text{m}} 2 , text{N} $$
Conclusion
The weight of the ruler is ( 2 , text{N} ).
Advanced Understanding
To further explore the concepts, let's analyze the gravitational torque on the ruler.
The gravitational torque by a section of length ( x ) at a pivot point at one of its ends is equivalent to the torque from a point mass ( x cdot m / L ) at the center of mass of this section. Since the ruler is homogeneous, the center of mass of a section is exactly halfway the section at ( x/2 ). Therefore, the torque from such a section is:
$$ tau_g F_g cdot d frac{x cdot m cdot g}{L} cdot frac{x}{2} frac{x^2 mg}{2L} $$
Given the position of the pivot point, the torques by both sides must be equal. On the side where the force is applied, the torque by gravity is:
$$ tau_1 3 , text{N} times 0.2 , text{m} frac{0.2^2 mg}{2} 0.6 , text{Nm} 0.02 mg $$
The torque of the opposing side is:
$$ tau_2 frac{0.8^2 mg}{2} 0.32 mg $$
Setting the torques equal for equilibrium:
$$ 0.6 , text{Nm} 0.02 mg 0.32 mg $$
Solving for ( m ):
$$ 0.6 0.3 mg $$
$$ m frac{0.6}{0.3g} approx 0.2 , text{kg} $$
Conclusion
Understanding the principles of moments and torque allows us to solve problems involving uniform objects with hanging weights. Drawing a picture and visualizing the torques helps in grasping these concepts more effectively.