To solve the problem of finding the Young's modulus \( E_2 \) for wire Y, we need to analyze the forces acting on both wires and how they relate to the Young's modulus of each wire. Given that the weight of 30 N is placed at point O on the rod XY, where the ratio of distances from O to X and O to Y is 1:2, we can use this information to set up our equations.
Understanding the Setup
We have two vertical wires, X and Y, connected by a horizontal rod. The weight of 30 N is acting downwards at point O, which divides the rod into two segments: XO and OY. Since the ratio of these segments is 1:2, we can denote the lengths as follows:
- Length of XO = \( l_1 = \frac{l}{3} \)
- Length of OY = \( l_2 = \frac{2l}{3} \)
Force Distribution
When the weight is applied, it creates tension in both wires. The force in wire X, denoted as \( F_X \), and the force in wire Y, denoted as \( F_Y \), must satisfy the equilibrium condition:
Since the total downward force is 30 N, we have:
\( F_X + F_Y = 30 \, \text{N} \)
Relating Forces to Young's Modulus
The Young's modulus \( E \) is defined as the ratio of stress to strain. The stress in each wire can be expressed as:
\( \text{Stress} = \frac{F}{A} \)
And the strain is given by:
\( \text{Strain} = \frac{\Delta l}{l} \)
For wire X, we can express the Young's modulus \( E_1 \) as:
\( E_1 = \frac{F_X / A}{\Delta l_X / l} \)
Rearranging gives us:
\( F_X = E_1 \cdot \frac{\Delta l_X}{l} \cdot A \)
Calculating the Extension
Since both wires are stretched and the rod remains horizontal, the extensions \( \Delta l_X \) and \( \Delta l_Y \) must be proportional to their lengths:
\( \Delta l_X = \frac{l}{3} \cdot \frac{F_X}{E_1 A} \)
\( \Delta l_Y = \frac{2l}{3} \cdot \frac{F_Y}{E_2 A} \)
Equilibrium of Extensions
For the rod to remain horizontal, the extensions must be equal:
\( \Delta l_X = \Delta l_Y \)
Substituting the expressions for \( \Delta l_X \) and \( \Delta l_Y \) gives:
\( \frac{l}{3} \cdot \frac{F_X}{E_1 A} = \frac{2l}{3} \cdot \frac{F_Y}{E_2 A} \)
We can cancel \( A \) and \( l \) from both sides:
\( \frac{F_X}{E_1} = \frac{2F_Y}{E_2} \)
Substituting Forces
From the equilibrium condition \( F_X + F_Y = 30 \, \text{N} \), we can express \( F_Y \) in terms of \( F_X \):
\( F_Y = 30 - F_X \)
Substituting this into our earlier equation gives:
\( \frac{F_X}{E_1} = \frac{2(30 - F_X)}{E_2} \)
Solving for Young's Modulus \( E_2 \)
Cross-multiplying leads to:
\( E_2 \cdot F_X = 2E_1(30 - F_X) \)
Expanding this gives:
\( E_2 \cdot F_X = 60E_1 - 2E_1F_X \)
Rearranging terms results in:
\( E_2 \cdot F_X + 2E_1F_X = 60E_1 \)
Factoring out \( F_X \) yields:
\( F_X(E_2 + 2E_1) = 60E_1 \)
Thus, we can express \( F_X \) as:
\( F_X = \frac{60E_1}{E_2 + 2E_1} \)
Finding \( F_Y \)
Now substituting \( F_X \) back into the equation for \( F_Y \):
\( F_Y = 30 - \frac{60E_1}{E_2 + 2E_1} \)
Substituting \( F_Y \) into the equation for \( E_2 \) gives us a system of equations that can be solved simultaneously. However, we can simplify the calculations by substituting known values.
Final Calculation
Given \( E_1 = 1.0 \times 10^{11} \, \text{N/m}^2 \), we can substitute this value into our equations to find \( E_2 \). After performing the calculations, we find:
\( E_2 = \frac{60E_1}{F_Y} - 2E_1 \)
By substituting the values and solving, we can determine the Young's modulus \( E_2 \) for wire Y.
In summary, the Young's modulus \( E_2 \) can be calculated using the relationships established between the forces and the Young's moduli of the two wires, leading to a clear understanding of how the system behaves under the applied load.
