Animated Solution for Physics - Laws of Motion: The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle $\theta$ should be (2001)

Imagine you are an architect designing a delicate suspension bridge. Every cable, every weight, and every angle must be perfectly calculated to ensure the structure stands still against the relentless pull of gravity. This problem presents us with a miniature version of such a challenge—a beautiful, symmetric pulley system in perfect equilibrium.

The beauty of physics lies in its demand for balance. When a system is in static equilibrium, it means that the net force acting on every single component is exactly zero. No acceleration, no movement—just a perfect stalemate of forces.

Let's break down the forces at play. We have a single, continuous string passing over two smooth pulleys. Because the string is massless and the pulleys are frictionless, the tension throughout the entire string is uniform. Let's call this tension $T$.

Our first step is to isolate one of the side masses, $m$. If we draw a Free Body Diagram for this mass, we see a very simple tug-of-war. Gravity pulls it downwards with a force equal to its weight, $mg$. Simultaneously, the string pulls it upwards with the tension $T$.

Since this mass is perfectly at rest, these two forces must be equal in magnitude. This gives us our first crucial piece of the puzzle:

Now, let's shift our focus to the star of the show: the central mass, $\sqrt{2}m$. This block is heavier, and it is supported not by one, but by two segments of the string.

Gravity pulls this central block downwards with a weight of $\sqrt{2}mg$. To counteract this, the two string segments pull upwards. However, they don't pull straight up; they pull at an angle $\theta$ with respect to the vertical.

Because these tension forces are angled, we must resolve them into their horizontal and vertical components. The horizontal components are $T\sin \theta$ pulling to the left and $T\sin \theta$ pulling to the right. Because the system is symmetric, these horizontal forces perfectly cancel each other out.

The vertical components, however, work together as a team. Each string segment contributes an upward force of $T\cos \theta$. Therefore, the total upward force supporting the central mass is $2T\cos \theta$.

For the central mass to remain in equilibrium, this total upward force must exactly balance its downward weight. This leads us to our master equation:

We now have a system of equations. We know that $T=mg$, and we know that $2T\cos \theta =\sqrt{2}mg$. The path forward is clear: we must substitute the value of $T$ into our master equation.

Substituting $T=mg$ gives us:

Notice how elegantly the physics simplifies. The term $mg$ appears on both sides of the equation. Since neither the mass nor gravity is zero, we can divide both sides by $mg$, effectively canceling it out:

Now, we simply isolate $\cos \theta$ by dividing both sides by 2:

To make this recognizable, we can simplify the fraction. Knowing that $2=\sqrt{2}\times \sqrt{2}$, we can rewrite the expression as:

This is a classic trigonometric value. We must ask ourselves: for what angle is the cosine equal to $\frac{1}{\sqrt{2}}$? The answer, drawn from our standard trigonometric tables, is exactly $45^{\circ }$.

Therefore, for this delicate system to maintain its perfect balance, the angle $\theta$ must be $45^{\circ }$. The forces align, the math simplifies, and the structure holds firm.