Problem 94 - Physics

Question

An \(85.0\) -N backpack is hung from the middle of an aluminum wire, as the drawing shows. The temperature of the wire then drops by \(20.0 \mathrm{C}^{\circ} .\) Find the tension in the wire at the lower temperature. Assume that the distance between the supports does not change, and ignore any thermal stress.

Step-by-Step Solution

Answer

The tension increases from thermal contraction. Calculate thermal stress and add to initial tension for final tension.

Step 1: Understand the Problem

Initially, the tension in the wire is due to the weight of the backpack. When the temperature drops, the wire contracts, adding tension to support the same weight. We need to calculate this additional tension due to thermal contraction.

Step 2: Calculate Initial Tension

The initial tension in the wire is only due to the weight of the backpack. So,\[ T_{ ext{initial}} = 85.0 \, \text{N} \]

Step 3: Determine Linear Expansion Coefficient

For aluminum, the linear expansion coefficient \( \alpha \) is approximately \( 23 \times 10^{-6} \, \text{C}^{-1} \).

Step 4: Calculate Change in Length

The change in length \( \Delta L \) of the wire can be calculated using the formula: \[ \Delta L = L \alpha \Delta T \] where \( L \) is the original length of the wire, \( \Delta T = -20.0 \degree C \). For this problem, exact \( L \) is not needed since tension depends on stress, not absolute length.

Step 5: Calculate Thermal Stress

Thermal stress \( \sigma \) in the wire can be calculated using Hooke's Law \( \sigma = E \varepsilon \) where \( E \) is Young's modulus for aluminum (approximately \( 70 \times 10^9 \, \text{Pa} \)) and \( \varepsilon = \alpha \Delta T \) is the strain. Insert given values to find the additional stress.

Step 6: Determine Final Tension

The total tension is a combination of the initial tension due to weight and the additional tension from thermal stress. Additional tension due to thermal stress is \( T_{ ext{thermal}} = \sigma A \), where \( A \) is the cross-sectional area of the wire. Final tension is:\[ T = 85.0 \, \text{N} + T_{\text{thermal}} \]

Key Concepts

Tension Calculation

Calculating tension involves understanding the forces acting on the wire. Initially, the tension is simply the force needed to support the weight of the backpack. This force is expressed as the weight of the backpack, which is \(85.0 \, \text{N}\).
When the temperature drops, things get a bit more interesting. The wire contracts due to thermal contraction, and this contraction adds more tension because the length between the supports doesn’t change.
To find the new total tension, we calculate both the initial tension and the additional tension from thermal contraction. Finally, these are summed to get the final tension in the wire.

Linear Expansion Coefficient

The Linear Expansion Coefficient is crucial when dealing with temperature changes in materials. It's denoted by \(\alpha\) and represents how much a material expands per degree change in temperature, typically measured in \(\text{C}^{-1}\).
For aluminum, the linear expansion coefficient is approximately \(23 \times 10^{-6} \, \text{C}^{-1}\). With a drop in temperature, the material contracts instead of expanding.
To compute the change in length, you can use the formula: \[\Delta L = L \alpha \Delta T\] Here \(L\) is the original length, and \(\Delta T\) is the change in temperature. In the given exercise, this drop is \(-20.0 \, ^{\circ}\text{C}\). Note that
length \(L\) is not required in exact numbers as it cancels out when calculating stress and strain.

Thermal Stress

Thermal stress is the stress developed in a constrained material when it undergoes a temperature change. It can be calculated using Hooke’s Law, relating stress to strain as follows: \[ \sigma = E \varepsilon \] where \(\sigma\) is thermal stress, \(E\) is Young’s Modulus, and \(\varepsilon\) is the strain.Strain in this scenario is calculated using the linear expansion coefficient and temperature change: \(\varepsilon = \alpha \Delta T\). With Young's modulus for aluminum being \(70 \times 10^9 \, \text{Pa}\), we can thus find how much additional stress the thermal contraction contributes.This additional stress is
then used to calculate the added tension in the wire
using the cross-sectional area as a multiplication factor.

Young's Modulus

Young’s Modulus, symbolized by \(E\), is an indicator of the stiffness of a material. For aluminum, Young’s Modulus is approximately \(70 \times 10^9 \, \text{Pa}\). This value is a constant that relates stress and strain within the elastic limit of the material.In our exercise, Young’s Modulus helps us calculate the thermal stress that develops due to the temperature drop.
It forms part of Hooke’s Law equation, where stress \(\sigma\) equals \(E\) multiplied by strain \(\varepsilon\). Understanding Young's Modulus is essential
because it determines how much the material will deform under stress.In essence, a high Young’s Modulus means the material is stiff and doesn't deform easily, while a lower modulus implies the opposite. For
our wire, this modulus ensures that aluminum reacts predictably
to the contraction caused by the cooling.