Question

Consider the situation shown in figure. The force F is equal to the m2g/2. If the area of cross section of the string is A and its Young modulus Y, find the strain developed in it. The string is light and there is no friction anywhere.
Figure

Answer 1

Strain developed is $\frac {m_2g(2m_1 + m_2)}{2AY(m_1 + m_2)}$

Explanation:

Given, Force $(F) = \frac {m_2g}{2}$

• Area of the cross-section of string = A

• Young's Modulus = Y

Let us suppose 'a' as acceleration produced in block $m_2$ in downward direction and T be tension in the string.

• From the Free Body Diagram-  

$m_2g - T = m_2a$ '''''''(1)

$T - F = m_1a$ """""(2)

* Refer figure for reference as attachment

• From equations (1) & (2) we get -

$a = \frac {m_2g - F} {m_1 + m_2}$

Applying $F = \frac {m_2g}{2}$

$a = \frac {m_2g}{2(m_1 + m_2)}$

Again $T = F + m_1a$

• On Applying the values of F and a, we will get -  

$T = \frac {m_2g}{2} + m_1\times \frac {m_2g}{2(m_1 + m_2)}$

We know that,  

$Y = \frac {FL}{A\Delta L}$

Strain = $\frac {\Delta L}{L} = \frac {F}{AY}$

Strain = $\frac {(m_2^2 + 2m_1m_2)g}{2(m_1 + m_2)AY}$

          = $\frac {m_2g(2m_1 + m_2)}{2AY(m_1 + m_2)}$

So, Required Strain Developed in String =  $\frac {m_2g(2m_1 + m_2)}{2AY(m_1 + m_2)}$

Answer 2

Answer:

Please refer to the given attachment.

Hope it helps you.

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