Animated Solution for Physics - Properties of Solids and Liquids: Two blocks of masses $3\text{ kg}$ and $5\text{ kg}$ are connected by a metal wire going over a smooth pulley. The breaking stress of the metal is $\frac{24}{\pi }\times 10^2{\text{ Nm}}^{-2}$. What is the minimum radius of the wire? (Take, $g=10{\text{ ms}}^{-2}$)

The Pulley and the Breaking Wire

Imagine you are an engineer tasked with designing a simple elevator system. You have a pulley, a wire, and two masses. Your primary concern isn't just how fast the masses will move, but whether the wire can actually survive the forces pulling on it. This problem beautifully bridges the gap between classical mechanics (Newton's Laws) and the material properties of solids (Elasticity and Stress).

Let's break down the thought process step-by-step.

We are given a classic Atwood machine setup: a smooth, frictionless pulley with two blocks hanging from it. On the left, we have a $3\text{ kg}$ block, and on the right, a heavier $5\text{ kg}$ block.

Because the $5\text{ kg}$ block is heavier, gravity will pull it down harder than the $3\text{ kg}$ block. This creates an imbalance, causing the entire system to accelerate. The $5\text{ kg}$ block will accelerate downwards, and the $3\text{ kg}$ block will accelerate upwards with the exact same magnitude of acceleration, $a$.

Connecting them is a metal wire. This wire is under tension, $T$, which is the internal force trying to keep the wire from stretching and snapping.

To find the tension in the wire, we must first determine the acceleration of the system. We do this by drawing Free Body Diagrams (FBDs) for each block.

For the heavier $5\text{ kg}$ block, gravity pulls down with a force of $5g$, and tension $T$ pulls up. Since it accelerates downwards, Newton's Second Law gives us:

For the lighter $3\text{ kg}$ block, tension $T$ pulls up, and gravity pulls down with $3g$. Since it accelerates upwards, the equation is:

By adding these two equations together, the tension $T$ elegantly cancels out, allowing us to solve for the acceleration $a$:

Now that we know the acceleration, we can substitute it back into our first equation to find the tension $T$:

This $37.5\text{ N}$ is the crucial force that is actively trying to rip our metal wire apart.

Now we transition from mechanics to material science. The problem states that the wire has a breaking stress of $\frac{24}{\pi }\times 10^2{\text{ N/m}}^2$.

Stress is defined as the internal restoring force per unit area. In this case, the restoring force is the tension $T$, and the area is the cross-sectional area of the wire, $A=\pi r^2$. Therefore, the stress experienced by the wire is:

To find the minimum radius $r$ that prevents the wire from breaking, we must equate the stress experienced by the wire to its maximum allowable breaking stress:

Notice how the $\pi$ on both sides of the denominator cancels out perfectly. This is a classic hallmark of a well-designed physics problem! Rearranging the equation to solve for $r^2$, we get:

Simplifying this fraction might look intimidating, but if we multiply the numerator and denominator by $10$, we get $\frac{375}{24000}$, which reduces beautifully to $\frac{1}{64}$.

Taking the square root of both sides gives us the radius in meters:

Finally, since our multiple-choice options are in centimeters, we multiply by $100$:

The minimum radius of the wire must be $12.5\text{ cm}$ to safely support the dynamic forces of the accelerating blocks. Always remember, physics isn't just about finding numbers; it's about ensuring the structures we build can withstand the realities of the physical world!