Two men `A` and `B` are carrying a uniform bar of length `L` on their shoulders. The bar is held horizontally such that A gets one-fourth load. If `A` is at one end of the bar, the distance of `B` from that end is

Options

A. `L/3`
B. `L/4`
C. `(2L)/3`
D. `(3L)/4`

Answer

To solve the problem, we need to determine the distance of man B from the end of the bar where man A is located. Here’s a step-by-step solution:

### Step 1: Understand the Load Distribution
Given that man A carries one-fourth of the total load, we can denote the total force exerted by the bar as \( F \). Therefore, the force on man A (denoted as \( F_A \)) is:
\[ F_A = \frac{1}{4} F \]
And the force on man B (denoted as \( F_B \)) is:
\[ F_B = F - F_A = F - \frac{1}{4} F = \frac{3}{4} F \]

### Step 2: Set Up the Torque Equation
Since the bar is uniform and horizontal, we can analyze the torques about the center of mass of the bar. The center of mass of a uniform bar of length \( L \) is located at \( \frac{L}{2} \).

The torque exerted by A about the center of mass is given by:
\[ \tau_A = F_A \cdot d_A \]
where \( d_A \) is the distance from A to the center of mass, which is \( \frac{L}{2} \).

The torque exerted by B about the center of mass is given by:
\[ \tau_B = F_B \cdot d_B \]
where \( d_B \) is the distance from B to the center of mass.

### Step 3: Equate the Torques
Since the bar is in equilibrium, the torques must be equal:
\[ \tau_A = \tau_B \]
Substituting the expressions for torque:
\[ F_A \cdot \frac{L}{2} = F_B \cdot d_B \]
Substituting \( F_A \) and \( F_B \):
\[ \left(\frac{1}{4} F\right) \cdot \frac{L}{2} = \left(\frac{3}{4} F\right) \cdot d_B \]

### Step 4: Simplify the Equation
We can cancel \( F \) from both sides (assuming \( F \neq 0 \)):
\[ \frac{1}{4} \cdot \frac{L}{2} = \frac{3}{4} \cdot d_B \]
This simplifies to:
\[ \frac{L}{8} = \frac{3}{4} \cdot d_B \]

### Step 5: Solve for \( d_B \)
To find \( d_B \), multiply both sides by \( \frac{4}{3} \):
\[ d_B = \frac{L}{8} \cdot \frac{4}{3} = \frac{L}{6} \]

### Step 6: Calculate the Distance of B from the End
The distance of B from the end of the bar (where A is located) is the distance from A to the center of mass plus the distance from the center of mass to B:
\[ \text{Distance from A to B} = \frac{L}{2} + d_B = \frac{L}{2} + \frac{L}{6} \]
To add these fractions, find a common denominator (which is 6):
\[ \frac{L}{2} = \frac{3L}{6} \]
Thus,
\[ \text{Distance from A to B} = \frac{3L}{6} + \frac{L}{6} = \frac{4L}{6} = \frac{2L}{3} \]

### Final Answer
The distance of B from the end where A is located is:
\[ \boxed{\frac{2L}{3}} \]