Animated Solution for Physics - Laws of Motion: A uniform rope of length $L$ and mass $M$ lying on a smooth table is pulled by a constant force $F$. What is the tension in the rope at a distance $l$ from the end where the force is applied?

The problem of finding tension inside a continuous moving body is a classic test of how well you understand Newton's Second Law. It forces us to stop looking at objects as single, unbreakable dots and start seeing them as a collection of connected parts.

Imagine a thick, heavy rope resting on a perfectly smooth, frictionless table. We grab one end of this rope and start pulling it with a constant force, $F$. Because there is no friction to hold it back, the entire rope is going to accelerate together as a single, rigid body.

Since every single particle of the rope is moving together, we can treat the whole rope as one single mass, $M$. To find out how fast it is speeding up, we pull out our trusty tool: Newton's Second Law of Motion. The net force equals mass times acceleration.

The only horizontal force acting on the entire rope is our applied force, $F$. So, we simply divide this force by the total mass $M$ to get the acceleration.

This $a$ is the acceleration of every single point on the rope. Keep this value safe; we will need it very soon.

Now comes the main point. We do not want the force at the ends; we want the tension inside the rope, at a specific point $C$. This point is at a distance $l$ from the end where we are pulling. Imagine slicing the rope mentally at this exact point. This divides our rope into two distinct parts: the front part $CB$, and the back part $AC$.

Let's look closely at the back segment, $AC$. If the total length of the rope is $L$, and the front piece has a length of $l$, then simple geometry tells us that the remaining back piece must have a length of $L-l$. This back piece is what we will focus on.

Because the rope is uniform, its mass is distributed evenly. The mass per unit length is simply total mass $M$ divided by total length $L$. To find the mass of just the back segment $AC$, we multiply this density by its length, $L-l$.

Now, let's draw a free body diagram just for the back segment, $AC$. What is physically pulling this piece forward? It is not the external force $F$—that is applied way at the other end! The only thing pulling segment $AC$ is the tension $T$ from the front segment $CB$ at point $C$. This tension is the internal force we are looking for.

Let's apply Newton's Second Law specifically to this back segment. The net force on it is just the tension $T$. And this must equal its mass, $m_{AC}$, multiplied by its acceleration. Remember, the whole rope shares the exact same acceleration $a$ that we found earlier.

It is time to bring everything together. Let's substitute the expressions we found for the mass of the segment and the acceleration of the system into our tension equation.

Do you see the magic of physics here? The total mass $M$ is in the numerator from the mass term, and in the denominator from the acceleration term. They perfectly cancel each other out! This means the tension does not even depend on the total mass of the rope, just the force and the lengths.

Finally, we can rearrange the remaining terms. Dividing $L-l$ by $L$ gives us $1-\frac{l}{L}$. So, the tension $T$ at a distance $l$ from the pulled end is $F$ times the quantity $1-\frac{l}{L}$.

Notice that if $l$ is zero, tension is $F$, and if $l$ is $L$, tension is zero. It makes perfect physical sense!