Question

Three blocks A, B and C each of mass 4 kg are attached as shown in figure. Both the wires has equal cross sectional area 5 x 10-7 m². The surface is smooth. Find the longitudinal strain in each wire if Young modulus of both the wires is 2 x 10¹¹ N/m² (Take g = 10 m/s²)

Answer 1

The longitudinal strain in wire 1 is $\frac{4}{3} \times 10^{-4}$ and in wire 2 is $\frac{8}{3} \times 10^{-4}$.

• Strain in wire 1: $1.333 \times 10^{-4}$
• Strain in wire 2: $2.667 \times 10^{-4}$

Answer 2

Explanation of the Solution:

1. Analyze the System and Forces:

   • Let the mass of each block be $m = 4 \text{ kg}$.
   • The surface is smooth, so there is no friction.
   • The system (blocks A, B, and C) will accelerate. Let the acceleration be $a$.
   • Let $T_1$ be the tension in wire 1 (between A and B).
   • Let $T_2$ be the tension in wire 2 (between B and C).

2. Apply Newton's Second Law to Each Block:

   • For Block A: The only horizontal force is $T_1$. $T_1 = m a$ (Equation 1)
   • For Block B: The forces are $T_2$ to the right and $T_1$ to the left. $T_2 - T_1 = m a$ (Equation 2)
   • For Block C: The forces are its weight $mg$ downwards and tension $T_2$ upwards. $mg - T_2 = m a$ (Equation 3)

3. Solve for Acceleration ($a$) and Tensions ($T_1, T_2$):

   • Substitute (1) into (2): $T_2 - ma = ma \implies T_2 = 2ma$.
   • Substitute $T_2 = 2ma$ into (3): $mg - 2ma = ma \implies mg = 3ma$.
   • Since $m \neq 0$, we get $a = \frac{g}{3}$.
   • Given $g = 10 \text{ m/s}^2$, $a = \frac{10}{3} \text{ m/s}^2$.
   • Now find tensions:
     • $T_1 = ma = 4 \text{ kg} \times \frac{10}{3} \text{ m/s}^2 = \frac{40}{3} \text{ N}$.
     • $T_2 = 2ma = 2 \times 4 \text{ kg} \times \frac{10}{3} \text{ m/s}^2 = \frac{80}{3} \text{ N}$.

4. Calculate Longitudinal Strain for Each Wire:

   • Longitudinal strain ($\epsilon$) is given by $\epsilon = \frac{\text{Stress}}{\text{Young's Modulus}} = \frac{Tension}{Area \times Young's Modulus}$.

   • Given cross-sectional area $A = 5 \times 10^{-7} \text{ m}^2$.

   • Given Young's modulus $Y = 2 \times 10^{11} \text{ N/m}^2$.

   • The product $AY = (5 \times 10^{-7} \text{ m}^2) \times (2 \times 10^{11} \text{ N/m}^2) = 10 \times 10^4 \text{ N} = 10^5 \text{ N}$.

   • Strain in Wire 1 ($\epsilon_1$): $\epsilon_1 = \frac{T_1}{AY} = \frac{40/3 \text{ N}}{10^5 \text{ N}} = \frac{40}{3 \times 10^5} = \frac{4}{3} \times 10^{-4}$. $\epsilon_1 \approx 1.333 \times 10^{-4}$.

   • Strain in Wire 2 ($\epsilon_2$): $\epsilon_2 = \frac{T_2}{AY} = \frac{80/3 \text{ N}}{10^5 \text{ N}} = \frac{80}{3 \times 10^5} = \frac{8}{3} \times 10^{-4}$. $\epsilon_2 \approx 2.667 \times 10^{-4}$.