Two particles of mass m each are tied at the ends of a light string of length 2a. The whole system i

Q: Wait, why is the net force on point $P$ exactly zero even though it's moving upwards?

This is a fantastic question! Point $P$ is just a specific location on a light (massless) string. According to Newton's second law, $F_{\text{net}} = m a$. Since the mass $m = 0$ for point $P$, the net force on it must mathematically be zero, regardless of its acceleration. Thus, the applied upward force $F$ must be perfectly and instantaneously balanced by the downward tension components.

Q: Why do we take the angle $\theta$ with the vertical instead of the horizontal surface?

You can absolutely take the angle with the horizontal if you prefer! If you call the horizontal angle $\alpha$, then $\theta = 90^\circ - \alpha$. The vertical force balance would then be $F = 2T \sin\alpha$, and the horizontal driving force would be $T \cos\alpha$. The final result will be exactly the same. Taking $\theta$ with the vertical just makes the vertical balance $F = 2T \cos\theta$ slightly more direct to write.

Q: How do we know the tension $T$ is the same on both sides of the string?

We know this because the setup is perfectly symmetric. The masses are identical, they are at the exact same distance $x$ from the center, and the force $F$ is applied exactly at the midpoint. Due to this perfect left-right symmetry, the tension must distribute itself equally on both sides of the string.

Q: Does the normal force from the table affect the horizontal motion of the masses?

No, it doesn't. The normal force acts purely in the vertical direction to balance the gravitational force ($mg$) pulling down on the masses. Since the problem states the table is frictionless, there is no horizontal friction force generated by this normal force. The only horizontal force acting on the masses is the component of the string's tension.

Q: What physically happens to the acceleration as the masses get very close to each other?

As the masses get closer, $x$ approaches $0$. If you look at our final formula, the numerator approaches $0$, meaning the acceleration $a_0$ approaches $0$. Physically, as the masses meet in the middle, the string becomes completely vertical. The tension pulls straight up, meaning its horizontal component vanishes. With no horizontal force, there is no horizontal acceleration.

Question

Two particles of mass $m$ each are tied at the ends of a light string of length $2a$. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance $a$ from the centre $P$ (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force $F$. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes $2x$, is (2007)

Options:

(A) $\frac{F}{2m} \frac{a}{\sqrt{a^2 - x^2}}$

(B) $\frac{F}{2m} \frac{x}{\sqrt{a^2 - x^2}}$

(C) $\frac{F}{2m} \frac{x}{a}$

(D) $\frac{F}{2m} \frac{\sqrt{a^2 - x^2}}{x}$

Sigma Percentile · Accepted Answer

### The Pull of the String: Unraveling the Kinematics of Constrained Motion

Imagine a classic tug-of-war, but instead of pulling against each other from the ends of a rope, you are standing in the middle, pulling the center of the rope vertically upwards. This problem is a beautiful and classic example of **constrained motion**, where the rigid geometry of a system strictly dictates the relationship between the applied forces and the resulting accelerations.

At first glance, the setup might seem slightly abstract. We have two identical masses, each of mass $m$, resting on a perfectly frictionless horizontal table. They are connected by a light, inextensible string of total length $2a$. The magic happens when we pinch the exact midpoint of this string, let's call it point $P$, and pull it vertically upwards with a constant force $F$.

As point $P$ rises, the string forms a tent-like shape, and the two masses are inevitably dragged towards each other along the frictionless surface. Our goal is to find the instantaneous acceleration of these masses when the separation between them is exactly $2x$.

### The 2D Cross-Section and the Geometry of the Tent

To solve this, we don't need to visualize the entire 3D space. We can simply look at the vertical cross-sectional plane that contains the string and the two masses. In this 2D view, the horizontal table forms our x-axis, and the vertical direction of the pull forms our y-axis.

Let's analyze the geometry at the specific instant when the separation between the masses is $2x$. Because we are pulling the exact midpoint, the system is perfectly symmetric. Each mass is at a horizontal distance $x$ from the center line. The string is divided into two equal segments, each of length $a$.

This setup forms two identical right-angled triangles. Let $	heta$ be the angle between the vertical y-axis (the direction of force $F$) and the string. Looking at one of these triangles, the hypotenuse is the string segment $a$, and the horizontal base is $x$.

From basic trigonometry, we can immediately write:
$$ \sin	heta = \frac{x}{a} $$

Using the Pythagorean theorem, the vertical height of point $P$ above the table is $\sqrt{a^2 - x^2}$. This allows us to write the cosine and tangent of the angle:
$$ \cos	heta = \frac{\sqrt{a^2 - x^2}}{a} $$
$$ 	an	heta = \frac{\sin	heta}{\cos	heta} = \frac{x}{\sqrt{a^2 - x^2}} $$

Keep this geometric relationship safe; it is the key that will unlock our final answer.

### The Magic of the Massless Point

Now, let's dive into the dynamics. We start by analyzing point $P$. What exactly is point $P$? It is just a geometric location on a light string. It has absolutely no mass ($m = 0$).

According to Newton's Second Law, the net force on any object is equal to its mass times its acceleration ($F_{	ext{net}} = ma$). If the mass is zero, then the net force *must* be exactly zero, regardless of how violently the point might be accelerating!

This is a profound and highly useful concept in physics. It tells us that the upward force $F$ applied at point $P$ must be perfectly and instantaneously balanced by the downward forces acting on it. What are these downward forces? They are the vertical components of the tension $T$ from the two halves of the string.

The vertical component of tension from one side is $T \cos	heta$. Since there are two sides pulling down symmetrically, the total downward force is $2T \cos	heta$. Equating the upward and downward forces gives us our master equilibrium equation for point $P$:
$$ F = 2T \cos	heta $$

From this, we can easily isolate the tension $T$ in the string:
$$ T = \frac{F}{2 \cos	heta} $$

### The Horizontal Drive

With the tension known, we can now shift our focus to the masses sliding on the table. Let's look at the right mass $m$. It is constrained to move only along the horizontal surface.

Since the table is frictionless, there are no horizontal resistive forces. The only force driving the mass towards the center is the horizontal component of the string's tension. Looking at our angle $	heta$, this horizontal pulling force is $T \sin	heta$.

Applying Newton's Second Law to the mass $m$ in the horizontal direction, we get:
$$ T \sin	heta = m a_0 $$

Where $a_0$ is the horizontal acceleration we are looking for. Rearranging for $a_0$:
$$ a_0 = \frac{T \sin	heta}{m} $$

### Synthesizing Geometry and Dynamics

We are now in the endgame. We have an expression for acceleration in terms of tension, and an expression for tension in terms of the applied force $F$. Let's merge them. Substitute the expression for $T$ into the acceleration equation:

$$ a_0 = \frac{1}{m} \left( \frac{F}{2 \cos	heta} ight) \sin	heta $$

Notice how the sine and cosine terms beautifully combine into a tangent:
$$ a_0 = \frac{F}{2m} 	an	heta $$

This is a remarkably elegant intermediate result. It tells us that the acceleration is directly proportional to the tangent of the angle the string makes with the vertical.

Finally, we bring back our geometric key. Substitute the expression for $	an	heta$ that we derived at the very beginning:

$$ a_0 = \frac{F}{2m} \frac{x}{\sqrt{a^2 - x^2}} $$

And there we have it! This is the exact magnitude of the acceleration of the particles when their separation is $2x$.

### Physical Intuition: The Limiting Cases

It is always a great practice to test our final mathematical expression against physical intuition by looking at extreme cases.

**Case 1: What happens as the masses get very close to each other ($x 	o 0$)?**
If we look at our formula, as $x$ approaches zero, the numerator goes to zero, and the acceleration $a_0$ becomes zero. Does this make sense? Yes! As the masses meet in the center, the string becomes perfectly vertical. The tension is pulling straight up, with zero horizontal component. Without a horizontal force, the horizontal acceleration must be zero.

**Case 2: What happens when the string is almost flat ($x 	o a$)?**
If the masses are far apart, $x$ is very close to $a$. The denominator $\sqrt{a^2 - x^2}$ approaches zero, which means the acceleration $a_0$ shoots up towards infinity! Why? Because when a string is almost perfectly horizontal, pulling it vertically requires an immense amount of tension to balance even a small vertical force $F$. This massive tension translates into a massive horizontal force on the masses, causing extreme acceleration.

By blending geometric constraints with Newton's laws, we've not only solved the problem but also gained a deep understanding of how the system behaves under different conditions.

Select Answer:

Visualized Solution

Visualizing the System

Geometry of the Setup

Free Body Diagram of Point $P$

Tension in the String

Free Body Diagram of Mass $m$

Acceleration of Mass $m$

Substituting Tension

Final Expression

The Sigma Insight: Newton's Laws of Motion

The Pull of the String

The 2D Cross-Section and the Geometry of the Tent

The Magic of the Massless Point

The Horizontal Drive

Synthesizing Geometry and Dynamics

Physical Intuition

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