A uniform chain of length L and mass M is lying on a smooth table and one-third of its length is han

Q: Wait, why don't we consider the part of the chain already on the table?

The part of the chain on the table is moving horizontally on a smooth surface. Gravity acts downwards, and the normal force balances it. Since the displacement is perpendicular to gravity and there is no friction, no work is done on the horizontal part.

Q: Why is the height $\frac{L}{6}$ and not $\frac{L}{3}$?

$\frac{L}{3}$ is the total length of the hanging part. The lowest link has to be lifted by $\frac{L}{3}$, but the link just below the edge barely needs to be lifted at all. On average, the mass is lifted by half that distance. This is exactly what the Center of Mass represents: the average position of the mass, which is at $\frac{L}{6}$.

Q: Can I solve this using integration instead of Center of Mass?

Absolutely! You can take a small element $dx$ at a depth $x$. Its mass is $dm = (\frac{M}{L})dx$. The work to lift it is $dW = (dm)gx$. Integrating this from $x=0$ to $x=\frac{L}{3}$ will give you the exact same result, $\frac{MgL}{18}$. The Center of Mass method is just a beautiful shortcut!

Q: What if the table had friction?

If the table had friction, you would have to do work against both gravity and the frictional force. As you pull the chain, the length of the chain on the table increases, so the normal force and the frictional force would also increase. You would definitely need integration to solve that!

Q: Why is the mass of the hanging part exactly $\frac{M}{3}$?

The problem states that the chain is "uniform". This means its mass is distributed evenly along its entire length. Therefore, the mass of any segment is directly proportional to its length. Since the hanging length is $\frac{1}{3}$ of the total length, its mass must be $\frac{1}{3}$ of the total mass.

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 to the table is

Options:

(A) $MgL$

(B) $\frac{MgL}{3}$

(C) $\frac{MgL}{9}$

(D) $\frac{MgL}{18}$

Sigma Percentile · Accepted Answer

### The Deceptive Simplicity of the Chain Problem

Imagine a smooth table with a heavy chain resting on it. One-third of this chain is dangling over the edge, just waiting to slide off. Our job is to pull this hanging part back onto the table. It looks like a simple mechanics problem, but it hides a beautiful application of the Center of Mass concept.

### The Physics of Pulling

Why does it take work to pull the chain? The table is smooth, so there is no friction. The only force we are fighting is gravity. As we pull the chain, we are lifting the hanging links against the gravitational pull.

According to the Work-Energy Theorem, the work done by our external pulling force will be exactly equal to the change in the gravitational potential energy of the chain.

### The Integration Route (The Hard Way)

One might be tempted to solve this using calculus. You could take a small element $dx$ at a distance $x$ from the table. Its mass would be $dm = \frac{M}{L} dx$, and the work to lift it would be $dW = (dm)gx$. Integrating this from $0$ to $L/3$ gives:

$$ W = \int_0^{L/3} \left(\frac{M}{L} dxight) g x = \frac{Mg}{L} \left[ \frac{x^2}{2} ight]_0^{L/3} = \frac{MgL}{18} $$

This is rigorous but time-consuming. In competitive exams like JEE, time is of the essence.

### The Center of Mass Shortcut (The Elegant Way)

Here is where the magic happens. Since the chain is uniform, we can treat the entire hanging portion as a single point mass located exactly at its center of mass.

First, what is the mass of the hanging part? Since exactly $1/3$ of the length is hanging, its mass is simply:

$$ m = \frac{M}{3} $$

Next, where is its center of mass? For a uniform rod or chain of length $l$, the center of mass is exactly in the middle, at $l/2$. Here, the hanging length is $L/3$, so the center of mass is at a depth of:

$$ h = \frac{1}{2} 	imes \frac{L}{3} = \frac{L}{6} $$

### The Final Execution

Now, the problem is reduced to lifting a single point mass $m$ by a height $h$. The work done is simply:

$$ W = mgh $$

Substituting our values into this master equation:

$$ W = \left(\frac{M}{3}ight) g \left(\frac{L}{6}ight) $$

Multiplying the denominators, we arrive at our final, elegant result:

$$ W = \frac{MgL}{18} $$

**This is the power of the Center of Mass concept.** It takes a problem that seems to require calculus and turns it into a two-step algebra calculation. Always look for these symmetries and shortcuts!

Select Answer:

Visualized Solution

The Setup

Work-Energy Theorem

Mass of Hanging Part

Center of Mass

Position of COM

Work Done Equation

Substitution

Final Calculation

The Sigma Insight: Work Done by Forces

The Deceptive Simplicity of the Chain Problem

The Physics of Pulling

The Integration Route (The Hard Way)

The Center of Mass Shortcut (The Elegant Way)

The Final Execution

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