A uniform rope of length L, resting on a frictionless horizontal table is pulled at one end by a force F. What is the tension in the rope at a distance x form the end where the force is applied?

Options

A. F(L -x)
B. Fx/L
C. (L-x) / F
D. F(1- x/L)

Answer

To find the tension in a uniform rope of length \( L \) that is resting on a frictionless horizontal table and is pulled at one end by a force \( F \), we can follow these steps:

### Step 1: Understand the System
We have a uniform rope of length \( L \) on a frictionless table. A force \( F \) is applied at one end of the rope. We need to find the tension \( T \) at a distance \( x \) from the end where the force is applied.

### Step 2: Define the Mass of the Rope
Let the total mass of the rope be \( m \). The mass per unit length of the rope is given by:
\[ \text{mass per unit length} = \frac{m}{L} \]

### Step 3: Determine the Mass of the Rope Segments
- The mass of the segment of the rope from the end where the force is applied to the point \( x \) is:
\[ m_a = \frac{m}{L} \cdot x \]
- The mass of the remaining segment of the rope from point \( x \) to the other end is:
\[ m_b = \frac{m}{L} \cdot (L - x) \]

### Step 4: Apply Newton's Second Law
Assume the entire rope moves with an acceleration \( a \) when the force \( F \) is applied.

**For the segment from the end to point \( x \):**
Using Newton's second law for this segment:
\[ F - T = m_a \cdot a \]
Substituting \( m_a \):
\[ F - T = \left(\frac{m}{L} \cdot x\right) a \quad \text{(Equation 1)} \]

**For the segment from point \( x \) to the end of the rope:**
Using Newton's second law for this segment:
\[ T = m_b \cdot a \]
Substituting \( m_b \):
\[ T = \left(\frac{m}{L} \cdot (L - x)\right) a \quad \text{(Equation 2)} \]

### Step 5: Solve for Acceleration \( a \)
Adding Equation 1 and Equation 2 gives:
\[ F = m_a \cdot a + m_b \cdot a \]
Substituting the expressions for \( m_a \) and \( m_b \):
\[ F = \left(\frac{m}{L} \cdot x + \frac{m}{L} \cdot (L - x)\right) a \]
This simplifies to:
\[ F = \frac{m}{L} \cdot L \cdot a = m \cdot a \]
Thus, we can express acceleration as:
\[ a = \frac{F}{m} \]

### Step 6: Substitute \( a \) Back to Find Tension \( T \)
Now substitute \( a \) back into Equation 2:
\[ T = \left(\frac{m}{L} \cdot (L - x)\right) \cdot \frac{F}{m} \]
The mass \( m \) cancels out:
\[ T = \frac{F}{L} \cdot (L - x) \]
This can be rewritten as:
\[ T = F \cdot \left(1 - \frac{x}{L}\right) \]

### Final Answer
The tension in the rope at a distance \( x \) from the end where the force is applied is:
\[ T = F \cdot \left(1 - \frac{x}{L}\right) \]