A uniform rope of length L , resting on frictionless horizontal table is pulled at one end by a force F . What is the tension in the rope at a distance x from the end where the force is applied ?
[ Hint : Consider the motion of the entire rope and the motion of x length of rope using P = ma formula and third law of motion ]

Answer

To solve the problem of finding the tension in a uniform rope of length \( L \) being pulled by a force \( F \) at one end, we can follow these steps:

### Step 1: Analyze the entire rope
When the force \( F \) is applied to the end of the rope, the entire rope will accelerate. We can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (\( F = ma \)).

Let the mass per unit length of the rope be \( \lambda \). Therefore, the total mass \( m \) of the rope is given by:
\[ m = \lambda L \]
The acceleration \( a \) of the entire rope can be expressed as:
\[ F = ma \implies a = \frac{F}{m} = \frac{F}{\lambda L} \]

### Step 2: Consider a segment of the rope
Now, let’s consider a segment of the rope that is \( x \) long from the end where the force is applied. The mass of this segment \( m_x \) is:
\[ m_x = \lambda x \]

### Step 3: Apply Newton's second law to the segment
The segment of the rope will also experience the same acceleration \( a \) as the entire rope. Therefore, applying Newton's second law to this segment, we have:
\[ T = m_x a \]
where \( T \) is the tension in the rope at a distance \( x \) from the end where the force is applied.

Substituting \( m_x \) and \( a \) into the equation gives:
\[ T = \lambda x \cdot \frac{F}{\lambda L} \]

### Step 4: Simplify the expression for tension
Now, we can simplify the expression for tension \( T \):
\[ T = \frac{\lambda x F}{\lambda L} = \frac{x F}{L} \]

### Final Result
Thus, the tension in the rope at a distance \( x \) from the end where the force is applied is given by:
\[ T = \frac{x F}{L} \]