Problem 24
Question
A uniform chain of length \(L\) and mass \(M\) is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If \(g\) is acceleration due to gravity, the work required to pull the hanging part on the the table is (a) \(M g L\) (b) \(\mathrm{Mgl} / 3\) (c) \(\mathrm{Mg} \mathrm{L} / 9\) (d) \(\mathrm{Mgl} / 18\)
Step-by-Step Solution
Verified Answer
The work required is \( \frac{MgL}{18} \), option (d).
1Step 1: Determine the Length of the Hanging Chain
The problem states that one-third of the chain's length is hanging over the edge. So if the total length of the chain is \( L \), the hanging part of the chain has a length of \( \frac{L}{3} \).
2Step 2: Calculate the Mass of the Hanging Part
Since the chain is uniform, its mass per unit length is \( \frac{M}{L} \). Therefore, the mass of the hanging part is given by \( \frac{M}{L} \times \frac{L}{3} = \frac{M}{3} \).
3Step 3: Determine the Gravitational Force on the Hanging Part
The gravitational force acting on the hanging part of the chain is given by \( \text{mass} \times g = \frac{M}{3} \times g = \frac{Mg}{3} \).
4Step 4: Calculate the Work Done to Pull the Hanging Part onto the Table
To pull the chain onto the table, we need to do work against gravity. The center of mass of the hanging part is at a height \( \frac{1}{2} \times \frac{L}{3} = \frac{L}{6} \) below the table. The work done is \( ext{force} \times ext{distance} = \frac{Mg}{3} \times \frac{L}{6} = \frac{MgL}{18} \).
Key Concepts
Uniform Chain ProblemGravitational ForceCenter of MassKinematicsAccelaration due to Gravity
Uniform Chain Problem
The uniform chain problem is a popular exercise in mechanics that helps illustrate concepts of work and energy. In this problem, we often deal with a chain that is draped over an edge, such as a table. The chain is uniform, meaning its mass is evenly distributed along its length. This assumption simplifies calculations:
- The mass per unit length of the chain is constant.
- Sections of the chain can be considered separately while maintaining proportional relationships of mass and length.
- Determine the length of the hanging part.
- Calculate the corresponding mass of the hanging section.
- Use these to find other physical quantities like forces or work done.
Gravitational Force
Gravitational force is a fundamental concept in physics, describing the attraction between two bodies with mass. In the uniform chain problem, this force is what pulls the hanging part of the chain downward.For any mass \( m \) close to the Earth's surface, gravitational force can be calculated by:\[ F = m \times g \]Where:
- \( F \) is the gravitational force.
- \( m \) is the mass.
- \( g \) is the acceleration due to gravity, approximately equal to \( 9.8 \text{ m/s}^2 \).
Center of Mass
Center of mass is an essential concept when discussing the work done on objects. It is the point where the mass of an object is considered to be concentrated, affecting how we compute forces and resultant movements. For the hanging part of the uniform chain:
- The center of mass will lie halfway along its length.
- If one-third of the chain is hanging, its center of mass is at \( \frac{L}{6} \) below the table's edge.
- Balance and stability of objects.
- The effort needed to lift or move sections of an object.
Kinematics
Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces causing the motion. In the context of pulling the chain back onto the table, we are interested in how the chain's position changes over time.
While solving this problem, we implicitly use kinematics by examining how the height (or position) of the center of mass changes as the chain is lifted. For this motion:
- Initial and final positions of the chain are considered.
- Movement occurs vertically against gravity.
- Grasping concepts of displacement, velocity, and acceleration.
- Analyzing how an object's position changes over time.
Accelaration due to Gravity
Acceleration due to gravity is a constant that greatly simplifies solving problems involving mass and force near the Earth's surface. It's the rate at which an object's velocity changes due to Earth's gravitational pull, denoted by \( g \).For problems like the uniform chain, this constant allows us to relate gravitational force to mass:
- Typically valued at \( 9.8 \text{ m/s}^2 \).
- Affects how we calculate other forces and work done due to gravity.
- It’s independent of the object's mass and only depends on its interaction with the Earth.
- Simplifies equations when estimating motion under gravity.
Other exercises in this chapter
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