A steel rope has length L area of cross-section A young's modulus Y [density =d] (i) it is pulled on a horizotnal fritionless floor with a constant horizontal force `F=|dALg|//2` applied at one end find the strain at the midpoint.
(ii). If the steel rope is vertical and moving with the force acting vertically up at the upper end find the strain at distance L/3 from lowerr end.

Answer

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (i): Finding the strain at the midpoint of the steel rope on a horizontal frictionless floor.

1. **Understanding the Setup**:
- We have a steel rope of length \( L \) and cross-sectional area \( A \).
- A force \( F = \frac{d A L g}{2} \) is applied at one end of the rope on a frictionless surface.

2. **Finding the Acceleration**:
- The total mass \( m \) of the rope can be expressed as:
\[ m = d \cdot A \cdot L \]
- The net force acting on the entire rope is equal to the applied force \( F \):
\[ F = m \cdot a \]
- Substituting the mass:
\[ \frac{d A L g}{2} = (d A L) \cdot a \]
- Simplifying gives us the acceleration \( a \):
\[ a = \frac{g}{2} \]

3. **Finding the Inertial Force at the Midpoint**:
- The mass of the half rope (from the midpoint to the end) is:
\[ m_{\text{half}} = \frac{m}{2} = \frac{d A L}{2} \]
- The inertial force \( F_{\text{inertial}} \) acting on this half due to acceleration is:
\[ F_{\text{inertial}} = m_{\text{half}} \cdot a = \frac{d A L}{2} \cdot \frac{g}{2} = \frac{d A L g}{4} \]

4. **Calculating the Strain**:
- Strain \( \epsilon \) is defined as:
\[ \epsilon = \frac{\text{Stress}}{Y} \]
- Stress is given by:
\[ \text{Stress} = \frac{F_{\text{inertial}}}{A} = \frac{\frac{d A L g}{4}}{A} = \frac{d L g}{4} \]
- Therefore, the strain at the midpoint is:
\[ \epsilon = \frac{\frac{d L g}{4}}{Y} = \frac{d L g}{4 Y} \]

### Part (ii): Finding the strain at a distance \( \frac{L}{3} \) from the lower end when the rope is vertical.

1. **Understanding the Setup**:
- The steel rope is now vertical, and the force \( F \) is acting vertically upward at the upper end.
- The weight of the rope acts downward.

2. **Finding the Weight and Net Force**:
- The weight \( W \) of the entire rope is:
\[ W = d A L g \]
- The force acting upward is:
\[ F = \frac{d A L g}{2} \]
- The net force acting downward is:
\[ F_{\text{net}} = W - F = d A L g - \frac{d A L g}{2} = \frac{d A L g}{2} \]

3. **Finding the Acceleration**:
- The acceleration \( a \) of the entire rope is given by:
\[ a = \frac{F_{\text{net}}}{m} = \frac{\frac{d A L g}{2}}{d A L} = \frac{g}{2} \]

4. **Finding the Inertial Force at \( \frac{L}{3} \)**:
- The mass of the section of the rope from the lower end to \( \frac{L}{3} \) is:
\[ m_{\frac{L}{3}} = \frac{1}{3} \cdot d A L \]
- The inertial force \( F_{\text{inertial}} \) acting on this section is:
\[ F_{\text{inertial}} = m_{\frac{L}{3}} \cdot a = \frac{1}{3} \cdot d A L \cdot \frac{g}{2} = \frac{d A L g}{6} \]

5. **Calculating the Strain**:
- The stress in this section is:
\[ \text{Stress} = \frac{F_{\text{inertial}}}{A} = \frac{\frac{d A L g}{6}}{A} = \frac{d L g}{6} \]
- Therefore, the strain at \( \frac{L}{3} \) from the lower end is:
\[ \epsilon = \frac{\frac{d L g}{6}}{Y} = \frac{d L g}{6 Y} \]

### Final Answers:
- (i) The strain at the midpoint is \( \frac{d L g}{4 Y} \).
- (ii) The strain at a distance \( \frac{L}{3} \) from the lower end is \( \frac{d L g}{6 Y} \).