A uniform chain of 6 m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is _______________ m.

Explanation

To solve the problem of finding the maximum length of a uniform chain that can hang over the edge of a table without slipping, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a uniform chain of total length \( L = 6 \, \text{m} \). A part of this chain, denoted as \( x \), is hanging over the edge of the table. The coefficient of static friction between the chain and the table is \( \mu = 0.5 \).

2. **Weight of the Hanging Chain**:
The weight of the hanging part of the chain (length \( x \)) can be expressed as:
\[ W_{\text{hanging}} = \frac{m}{L} \cdot x \cdot g \]
where \( m \) is the total mass of the chain, \( g \) is the acceleration due to gravity, and \( \frac{m}{L} \) is the mass per unit length.

3. **Normal Force on the Table**:
The normal force \( N \) acting on the part of the chain that is on the table (length \( L - x \)) is given by:
\[ N = \frac{m}{L} \cdot (L - x) \cdot g \]

4. **Frictional Force**:
The maximum static frictional force \( F_{\text{friction}} \) that can act on the chain is:
\[ F_{\text{friction}} = \mu \cdot N = \mu \cdot \left(\frac{m}{L} \cdot (L - x) \cdot g\right) \]

5. **Setting Up the Equation**:
For the chain to remain at rest, the frictional force must balance the weight of the hanging part:
\[ F_{\text{friction}} = W_{\text{hanging}} \]
Substituting the expressions we derived:
\[ \mu \cdot \left(\frac{m}{L} \cdot (L - x) \cdot g\right) = \frac{m}{L} \cdot x \cdot g \]

6. **Canceling Common Terms**:
We can cancel \( \frac{m}{L} \) and \( g \) from both sides (assuming \( m \) and \( g \) are non-zero):
\[ \mu \cdot (L - x) = x \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ \mu L - \mu x = x \]
\[ \mu L = x + \mu x \]
\[ \mu L = x(1 + \mu) \]

8. **Solving for \( x \)**:
We can now solve for \( x \):
\[ x = \frac{\mu L}{1 + \mu} \]

9. **Substituting Known Values**:
Substitute \( \mu = 0.5 \) and \( L = 6 \, \text{m} \):
\[ x = \frac{0.5 \cdot 6}{1 + 0.5} = \frac{3}{1.5} = 2 \, \text{m} \]

### Final Answer:
The maximum length of the chain hanging from the table is \( \boxed{2 \, \text{m}} \).