Animated Solution for Physics - Laws of Motion: A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force $F$ is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of $45^{\circ }$ with the vertical. Then, $F$ equals (Take, $g=10{\text{ ms}}^{-2}$ and the rope to be massless)

## The Tug of War: Mastering Equilibrium in a Suspended Rope

Imagine a $10\text{ kg}$ mass hanging peacefully from a ceiling. It's a perfectly static, boring system. But then, you grab the exact midpoint of the rope and pull it horizontally with a force $F$. You pull until the top half of the rope tilts, making a $45^{\circ }$ angle with the vertical.

The question is: exactly how much force $F$ are you applying?

This classic physics problem is a beautiful demonstration of static equilibrium and the power of resolving vectors. Let's break it down step-by-step.

Visualizing the Forces

To find this unknown force $F$, we need to analyze the forces acting exactly at the midpoint of the rope. Why the midpoint? Because that is the central junction where all the action is happening.

Since the system is completely stationary, this point is in perfect equilibrium. Let's draw its free body diagram. We have three primary forces acting on this junction: 1. The Tension ($T$) pulling up and to the left along the top half of the rope. 2. The Weight ($Mg$) of the mass pulling straight down via the bottom half of the rope. 3. Our applied Horizontal Force ($F$) pulling to the right.

Resolving the Tension

Since the tension $T$ is acting at an angle of $45^{\circ }$ to the vertical, it's a bit tricky to work with directly. In physics, whenever a force acts at an angle to our primary axes, our first instinct should be to break it down into its vertical and horizontal components.

Using basic trigonometry, we can resolve the tension $T$: - The vertical component pulling upwards is $T\cos 45^{\circ }$. - The horizontal component pulling to the left is $T\sin 45^{\circ }$.

Balancing the Scales

Now, because the midpoint is perfectly still, the forces in every direction must perfectly balance out. It's like a multi-directional tug-of-war where nobody is winning.

Let's look at the vertical direction first. The upward force must exactly balance the downward pull of the $10\text{ kg}$ mass:

Similarly, in the horizontal direction, the leftward pull from the rope's tension must be perfectly equal to our applied rightward force $F$:

Solving for F

We now have two beautiful equations. We want to find $F$, but we have an annoying unknown variable: the tension $T$.

A clever mathematical trick here is to divide the horizontal equation by the vertical equation. Watch what happens:

Notice how the $T$ simply cancels out! Furthermore, we know that $\frac{\sin \theta }{\cos \theta }=\tan \theta$. This leaves us with a highly elegant relation:

We are almost there! The tangent of $45^{\circ }$ is simply $1$. Rearranging our equation, we find that $F$ is exactly equal to the weight of the mass:

Plugging in our mass of $10\text{ kg}$ and the acceleration due to gravity $g=10{\text{ m/s}}^2$:

And there is our answer! You are pulling with exactly $100\text{ Newtons}$ of force.

Can We Make It Horizontal?

Think about this fascinating consequence of our derived formula: $F=Mg\tan \theta$.

What if we wanted to pull the rope so hard that the top half became completely horizontal? The angle $\theta$ with the vertical would approach $90^{\circ }$.

But mathematically, as $\theta \to 90^{\circ }$, $\tan \theta \to \infty$. This means that no matter how hard you pull—even if you had the strength of a million elephants—you would need an infinite amount of force to make the rope perfectly straight! The vertical component of tension must always exist to balance the downward weight. Physics dictates that a perfectly horizontal loaded rope is an absolute impossibility.