Each of three blocks \( P \), \( Q \), and \( R \) shown in the figure has a mass of 3 kg. Each of the wires \( A \) and \( B \) has a cross-sectional area of 0.005 cm\(^2\) and Young's modulus \( 2 \times 10^{11} \, \text{N} \, \text{m}^{-2} \). Neglecting friction, the longitudinal strain on wire \( B \) is \( \ldots \times 10^{-4} \). (Take \( g = 10 \, \text{m/s}^2 \))
Correct Answer: 2
Approach Solution - 1
The problem asks for the longitudinal strain on wire B in the given system of blocks and wires. We are given the masses of the blocks, the cross-sectional area and Young's modulus of the wires, and the acceleration due to gravity.
Concept Used:
The solution involves a combination of concepts from mechanics and properties of matter:
1. Newton's Second Law of Motion: To find the tension in the wires, we must first determine the acceleration of the system of blocks. The acceleration (\(a\)) is given by \(a = \frac{F_{net}}{M_{total}}\), where \(F_{net}\) is the net external force and \(M_{total}\) is the total mass being accelerated.
2. Young's Modulus (Y): Young's modulus relates the stress (\(\sigma\)) applied to a material to the resulting longitudinal strain (\(\epsilon\)). The formula is:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\epsilon} \]where \(F\) is the tension in the wire and \(A\) is its cross-sectional area. From this, the strain can be calculated as:
\[ \epsilon = \frac{\text{Stress}}{Y} = \frac{F}{A \cdot Y} \]Step-by-Step Solution:
Step 1: Calculate the acceleration of the system.
The system consists of three blocks, P, Q, and R, each with a mass of 3 kg. The driving force is the weight of the hanging block R. The total mass being accelerated is the sum of the masses of all three blocks.
Net driving force, \(F_{net} = m_R \cdot g = 3 \, \text{kg} \times 10 \, \text{m/s}^2 = 30 \, \text{N}\).
Total mass, \(M_{total} = m_P + m_Q + m_R = 3 + 3 + 3 = 9 \, \text{kg}\).
The acceleration of the system is:
\[ a = \frac{F_{net}}{M_{total}} = \frac{30 \, \text{N}}{9 \, \text{kg}} = \frac{10}{3} \, \text{m/s}^2 \]Step 2: Calculate the tension in wire B (\(T_B\)).
Wire B is connected to block R. We can find the tension by considering the free-body diagram of block R. The forces acting on R are its weight (\(m_R g\)) downwards and the tension (\(T_B\)) upwards. The net force results in the downward acceleration \(a\).
\[ m_R g - T_B = m_R a \] \[ 30 \, \text{N} - T_B = 3 \, \text{kg} \times \frac{10}{3} \, \text{m/s}^2 \] \[ 30 - T_B = 10 \] \[ T_B = 20 \, \text{N} \]This tension \(T_B\) is the force acting on wire B.
Step 3: Calculate the longitudinal strain on wire B.
First, convert the cross-sectional area to SI units (m²).
\[ A = 0.005 \, \text{cm}^2 = 0.005 \times (10^{-2} \, \text{m})^2 = 0.005 \times 10^{-4} \, \text{m}^2 = 5 \times 10^{-7} \, \text{m}^2 \]The given Young's modulus is \(Y = 2 \times 10^{11} \, \text{N/m}^2\).
Now, use the formula for strain:
\[ \text{Strain on B} = \frac{T_B}{A \cdot Y} \] \[ \text{Strain on B} = \frac{20 \, \text{N}}{(5 \times 10^{-7} \, \text{m}^2) \times (2 \times 10^{11} \, \text{N/m}^2)} \]Final Computation & Result:
Simplify the expression for the strain.
\[ \text{Strain on B} = \frac{20}{5 \times 2 \times 10^{-7} \times 10^{11}} \] \[ \text{Strain on B} = \frac{20}{10 \times 10^{4}} = \frac{2}{10^{4}} = 2 \times 10^{-4} \]The problem asks for the value of x, where the strain is \(x \times 10^{-4}\).
By comparison, \(x = 2\).
The longitudinal strain on wire B is 2 \( \times 10^{-4} \).
Approach Solution -2
Step 1. Calculate the Total Force Acting on Block R: The total force on block R due to its weight is:
\( F = m \times g = 3 \, \text{kg} \times 10 \, \text{m/s}^2 = 30 \, \text{N} \)
Step 2. Determine the Tension T1 in Wire B: Assuming the system is in equilibrium, the net force acting on P, Q, and R needs to balance out, with wire B supporting the tension:
\( T_1 = F - T_2 = 20 \, \text{N} \)
Step 3. Calculate Longitudinal Strain: Strain = \( \frac{\text{stress}}{Y} \) where stress = \( \frac{T_1}{A} \) and \( A = 0.005 \, \text{cm}^2 = 0.5 \times 10^{-6} \, \text{m}^2 \):
\( \text{strain} = \frac{T_1}{A \times Y} = \frac{20}{0.5 \times 10^{-6} \times 2 \times 10^{11}} = 2 \times 10^{-4} \)
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