A uniform elastic plank moves due to a constant force `F_(0)` applied at one end whose area is `S`. The Young's modulus of the plank is `Y`. The strain produced in the direction of force is

Options

A. `(F_(0)Y)/(S)`
B. `(F_(0))/(SY)`
C. `(F_(0))/(2SY)`
D. `(F_(0)Y)/(2S)`

Answer

To find the strain produced in the direction of the force on a uniform elastic plank subjected to a constant force \( F_0 \), we can follow these steps:

### Step 1: Understand the Given Information
We are given:
- A constant force \( F_0 \) applied at one end of a uniform elastic plank.
- The cross-sectional area of the plank is \( S \).
- The Young's modulus of the plank is \( Y \).
- The total length of the plank is \( L \).

### Step 2: Consider a Small Element of the Plank
Let's consider a small element of length \( dx \) at a distance \( x \) from the free end of the plank. The internal restoring force acting on this element due to the applied force will be denoted as \( T \).

### Step 3: Relate the Internal Force to the Applied Force
The internal force \( T \) at a distance \( x \) can be expressed in terms of the applied force \( F_0 \). Since the free end has zero tension, we can express \( T \) as:
\[ T = F_0 \cdot \frac{L}{L} \cdot x = F_0 \cdot x \]

### Step 4: Apply Newton's Second Law
For the small element \( dx \), the mass \( dm \) of the element can be expressed as:
\[ dm = \frac{m}{L} \cdot dx \]
where \( m \) is the total mass of the plank. The acceleration \( a \) of the element can be expressed as:
\[ a = \frac{F_0}{m} \]
Thus, we can write:
\[ T \cdot dx = dm \cdot a \]
Substituting for \( dm \) gives:
\[ F_0 \cdot x \cdot dx = \left(\frac{m}{L} \cdot dx\right) \cdot \frac{F_0}{m} \]

### Step 5: Calculate the Elongation of the Element
The elongation \( d\delta L \) of the small element \( dx \) can be expressed using Young's modulus:
\[ d\delta L = \frac{T \cdot dx}{A \cdot Y} \]
Substituting \( T = F_0 \cdot x \):
\[ d\delta L = \frac{F_0 \cdot x \cdot dx}{S \cdot Y} \]

### Step 6: Integrate to Find Total Elongation
To find the total elongation \( \delta L \) of the entire plank, we integrate \( d\delta L \) from 0 to \( L \):
\[ \delta L = \int_0^L \frac{F_0 \cdot x \cdot dx}{S \cdot Y} = \frac{F_0}{S \cdot Y} \cdot \int_0^L x \, dx = \frac{F_0}{S \cdot Y} \cdot \left[\frac{x^2}{2}\right]_0^L = \frac{F_0 \cdot L^2}{2 \cdot S \cdot Y} \]

### Step 7: Calculate the Strain
Strain \( \epsilon \) is defined as the ratio of the change in length to the original length:
\[ \epsilon = \frac{\delta L}{L} = \frac{F_0 \cdot L}{2 \cdot S \cdot Y} \]

### Final Answer
Thus, the strain produced in the direction of the force is:
\[ \epsilon = \frac{F_0}{2 \cdot S \cdot Y} \]