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One-fourth of a rope of length \(l\) is hanging down over the edge of a frictionless table. The rope has a uniform, linear density (mass per unit length) \(\lambda\) (Greek lambda), and the end already on the table is held by a person. How much work does the person do when she pulls on the rope to raise he rest of the rope slowly onto the table? Do the problem in two ways as follows. (a) Find the force that the person must exert to raise the rope and from this the work done. Note that this force is variable because at different times, different amounts of rope are hanging over the edge. (b) Suppose the segment of the rope initially hanging over the edge of the table has all of its mass concentrated at its center of mass. Find the work necessary to raise this to table height. You will probably find this approach simpler than that of part (a). How do the answers compare, and why is this so?

Step by step solution

01

Understanding the Problem

We have a rope of length \( l \) with one-fourth, i.e., \( \frac{l}{4} \), hanging over the edge of a frictionless table. Uniform linear density of the rope is \( \lambda \), and we need to find the work done to pull the rope up onto the table.

02

Calculating Work Using Variable Force

Let \( x \) be the length of the rope still hanging over the edge. The mass of this part of the rope is \( \lambda x \), and the force needed to lift it is \( F = \lambda x g \), where \( g \) is the acceleration due to gravity. The work done to lift an infinitesimal piece of the rope \( dx \) is \( dW = F dx = \lambda x g dx \). Integrate this from \( x = 0 \) to \( x = \frac{l}{4} \) to get the total work: \[ W = \int_{0}^{\frac{l}{4}} \lambda gx dx = \lambda g \left[ \frac{x^2}{2} \right]_0^{\frac{l}{4}} = \lambda g \frac{(l/4)^2}{2} = \frac{\lambda g l^2}{32}. \]

03

Calculating Work Using Center of Mass

The center of mass of the hanging section is situated \( \frac{l}{8} \) below the table's edge. The total mass of the hanging section is \( \frac{\lambda l}{4} \). The work done in raising the center of mass to the level of the table is:\[ W = \text{Force} \times \text{Distance} = \left(\frac{\lambda l}{4} g\right) \times \left(\frac{l}{8}\right) = \frac{\lambda g l^2}{32}. \]

04

Comparing Both Results

Both methods result in the same work value \( \frac{\lambda g l^2}{32} \), as the center of mass approach provides a simplification consistent with integrating variable forces. This equivalence occurs because the average position of mass requiring lifting is captured by the center of mass method.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Variable Force

When lifting a rope that isn't uniformly distributed over a frictionless table, the force needed to pull it up varies. This is because the amount of rope hanging off the edge changes, influencing the mass that needs to be lifted. Here's how it works:

• The bottom part of the rope, still over the edge, has more mass as you start to pull it up.
• The force required to lift it equals the weight of this mass, which is described by the formula: \( F = \lambda x g \).
• Here, \( \lambda \) is the rope's linear density, \( x \) is the length of the rope still hanging, and \( g \) is acceleration due to gravity.

This force is not constant. It reduces as you continue to lift the rope because less of the rope hangs over the table edge. To measure the work done, which is simply the total energy required to lift the rope, the varying force is integrated over the length of the rope that's hanging.
By integrating the force over the distance, we determine the total work done while lifting: \( W = \int_{0}^{\frac{l}{4}} \lambda g x \, dx \). This reveals how much effort is truly needed when forces shift along the distance.

Center of Mass

The center of mass approach greatly simplifies solving mechanics problems like lifting a rope. To visualize this, imagine concentrating all the mass of the hanging section at one point.

• This point, the center of mass, represents the average location of the mass.
• For a rope with the mass concentrated at its center, it is located at \( \frac{l}{8} \) below the table edge.
• The uniform linear density, \( \lambda \), ensures that this distance provides the average lifting work.

The work done to raise the mass solely to table height can then be calculated directly by multiplying the gravitational force on the center of mass by the distance lifted. This method yields the same work value as the integration of variable force: \( W = \left(\frac{\lambda l}{4} g\right) \times \left(\frac{l}{8}\right) \).
This illustrates that the center of mass essentially simplifies the variable nature of force along the rope, delivering consistent results with less complexity.

Integral Calculus

The mathematical tool used to solve for work involving variable forces is integral calculus. This branch of mathematics helps us accumulate small changes to determine a total effect, such as work done on a rope.

• In our rope exercise, the incremental work needed to lift each tiny piece of the rope is calculated using \( dW = \lambda x g \, dx \).
• Integrals sum these small increments over a specific range, giving us the total work: \( W = \int_{0}^{\frac{l}{4}} \lambda g x \, dx \).
• This process accounts for the entire rope length incrementally, reflecting how force gradually transitions along the pull.

By integrating, we're effectively observing how variable elements add up over a continuous stretch. This reveals cumulative work accurately when dealing with changing forces, an essential piece of understanding dynamic systems like pulling ropes and much more in physics.

Mechanics Problem Solving

Solving problems in mechanics often involves different approaches to provide deeper insight or simpler solutions. When dealing with forces and work, it's beneficial to explore various strategies:

• Understanding Variable Forces: Evaluating how force changes with position helps calculate exact work requirements.
• Utilizing Center of Mass: This technique reduces the complexity of calculations by consolidating distributed mass into a single weight point, simplifying work calculations.
• Applying Integral Calculus: This reflects changes in the system over a continuum, highlighting how differential work builds into total work.

Such flexibilities in mechanics enable us to choose the method based on problem conditions, providing clarity and precision. The comparison of different methods, like using variable force versus center of mass, showcases not only equal outcomes but also different facets of understanding physical situations.
Exploring these tools deepens comprehension and enhances practical problem-solving skills in physics, enabling more adept handling of real-world mechanics scenarios.