A rope of length `L` and mass `M` is being pulled on a rough horizontal floor by a contact horizontal force `F = m g`. The force is acting at one end of the rope in the same direction as the length of the rope. The coefficient of kinetic friction between rope and floor is `1//2`. Then the tension at the midpoint of the rope is

Options

A. `(a)(M g)/(4)`
B. `(b)(2 M g)/(5)`
C. `(c)(M g)/(8)`
D. `(d)(M g)/(2)`

Explanation

To find the tension at the midpoint of the rope being pulled on a rough horizontal floor, we can follow these steps:

### Step 1: Identify the Forces Acting on the Rope
- The rope has a mass \( M \) and is being pulled by a force \( F = mg \) at one end.
- The weight of the rope acting downwards is \( Mg \).
- The normal force acting upwards is also \( Mg \).
- The frictional force opposing the motion is given by \( f = \mu N \), where \( \mu = \frac{1}{2} \) and \( N = Mg \).

### Step 2: Calculate the Frictional Force
- The frictional force \( f \) can be calculated as:
\[ f = \mu N = \frac{1}{2} Mg \]

### Step 3: Determine the Net Force Acting on the Rope
- The net force \( F_{net} \) acting on the rope can be calculated by subtracting the frictional force from the pulling force:
\[ F_{net} = F - f = Mg - \frac{1}{2} Mg = \frac{1}{2} Mg \]

### Step 4: Calculate the Acceleration of the Rope
- The acceleration \( a \) of the rope can be found using Newton's second law \( F = ma \):
\[ a = \frac{F_{net}}{M} = \frac{\frac{1}{2} Mg}{M} = \frac{g}{2} \]

### Step 5: Analyze the Forces at the Midpoint of the Rope
- At the midpoint, the mass of the rope is \( \frac{M}{2} \).
- The normal force at the midpoint is \( N = \frac{Mg}{2} \).
- The frictional force at the midpoint is:
\[ f_{mid} = \mu N = \frac{1}{2} \cdot \frac{Mg}{2} = \frac{Mg}{4} \]

### Step 6: Set Up the Equation for Tension at the Midpoint
- Let \( T \) be the tension at the midpoint. The forces acting on the segment of the rope from the midpoint to the end being pulled are:
\[ T - f_{mid} = \left(\frac{M}{2}\right) a \]
- Substituting the known values:
\[ T - \frac{Mg}{4} = \left(\frac{M}{2}\right) \cdot \frac{g}{2} \]
\[ T - \frac{Mg}{4} = \frac{Mg}{4} \]

### Step 7: Solve for Tension \( T \)
- Rearranging the equation gives:
\[ T = \frac{Mg}{4} + \frac{Mg}{4} = \frac{Mg}{2} \]

### Final Answer
Thus, the tension at the midpoint of the rope is:
\[ T = \frac{Mg}{2} \]