Animated Solution for Physics - Properties of Solids and Liquids: A wire elongates by $l$ mm when a load $w$ is hanged from it. If the wire goes over a pulley and two weights $w$ each are hung at the two ends, the elongation of the wire will be (in mm)

This is a classic problem that tests your conceptual understanding of tension and Hooke's Law. It is very easy to fall into the trap of thinking that two weights mean double the tension, but let's break down exactly why that isn't the case.

Imagine a wire of length $L$, cross-sectional area $A$, and Young's modulus $Y$. In the first scenario, it is hanging from a rigid ceiling with a weight $w$ attached to its bottom end.

The ceiling is simply providing a reaction force to keep the wire from falling. The tension throughout the wire is exactly $w$.

According to Hooke's Law, the elongation $l$ is given by the formula:

This is our master equation. We will use this to compare the second scenario.

Now, we take the exact same wire and drape it over a smooth, frictionless pulley. We hang a weight $w$ on both ends.

Here is where the conceptual trap lies. Many students intuitively think, "There are two weights $w$, so the tension must be $2w$." But let's pause and think about equilibrium.

If you look at just one of the weights, it is being pulled down by gravity with a force $w$. To keep it from accelerating downwards, the wire must pull it up with a tension $T=w$. Because the pulley is smooth and the system is static, this tension $T=w$ is uniform throughout the entire wire.

The second weight is essentially doing the exact same job that the rigid ceiling was doing in the first scenario—it is just providing the necessary reaction force to maintain the tension $w$.

Even though the tension is the same, the geometry has changed. The pulley divides the wire into two halves, each of length $L/2$.

Let's look at just the left half of the wire. It has a length of $L/2$ and is subjected to a tension $w$. Its individual elongation $l^{\prime }$ will be:

Since we know that $\frac{wL}{AY}=l$, we can substitute this in:

So, the left half of the wire stretches by $l/2$. By symmetry, the right half of the wire also stretches by $l/2$.

To find the total elongation of the entire wire, we simply add the elongations of the two halves together:

Fascinating, isn't it? The total elongation remains exactly the same as in the first case.

A great way to internalize this is to remember: pulling a wire from both ends with a force $F$ produces the exact same stretch as tying one end to a wall and pulling the other end with a force $F$. The relative separation of the ends is what matters!