Question

An acrobat of mass \(55 \mathrm{kg}\) is going to hang by her teeth from a steel wire and she does not want the wire to stretch beyond its elastic limit. The elastic limit for the wire is \(2.5 \times 10^{8} \mathrm{Pa} .\) What is the minimum diameter the wire should have to support her?

Short answer

The minimum diameter of the wire should be 5.2 mm.

Step-by-step solution

Understanding the Problem

We need to calculate the minimum diameter of a steel wire that can support an acrobat of mass 55 kg without exceeding the elastic limit of the wire, which is given as \(2.5 \times 10^{8} \, \text{Pa}\).

Gather Necessary Formulas

We will use the formula for stress: \( \text{Stress} = \frac{\text{Force}}{\text{Area}} \). The force due to the acrobat's weight is \( F = mg \), where \( m = 55 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). The area of the wire is the cross-sectional area of a circle, calculated as \( A = \pi \left( \frac{d}{2} \right)^2 \), where \(d\) is the diameter of the wire.

Calculate the Force

Calculate the force exerted by the acrobat due to gravity: \[ F = mg = 55 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 539 \, \text{N} \]

Relate Stress to Diameter

Set up the equation for stress: \[ \text{Stress} = \frac{F}{A} = \frac{F}{\pi \left( \frac{d}{2} \right)^2} \] We know the maximum allowable stress (elastic limit) is \(2.5 \times 10^{8} \, \text{Pa}\), so: \[ 2.5 \times 10^{8} = \frac{539}{\pi \left( \frac{d}{2} \right)^2} \]

Solve for Diameter \(d\)

Rearrange the equation to solve for \(d\): \[ \pi \left( \frac{d}{2} \right)^2 = \frac{539}{2.5 \times 10^{8}} \] \[ \left( \frac{d}{2} \right)^2 = \frac{539}{2.5 \times 10^{8} \pi} \]\[ \frac{d}{2} = \sqrt{\frac{539}{2.5 \times 10^{8} \pi}} \] \[ d = 2 \times \sqrt{\frac{539}{2.5 \times 10^{8} \pi}} \] Calculate \(d\): \[ d \approx 0.0052 \, \text{m} = 5.2 \, \text{mm} \]

Conclusion

The minimum diameter the wire should have to support the acrobat without exceeding the elastic limit is approximately 5.2 mm.

Key concepts

Stress Calculation

In the world of physics, stress is an essential concept, especially when dealing with materials like wires. Stress is defined as the force applied over an area, and it helps us determine how much pressure a material can withstand before it faces deformation. This comes into play significantly when calculating the safe limit for materials under force.

We use the formula for stress:

• \( \text{Stress} = \frac{\text{Force}}{\text{Area}} \)

The stress exerted on a material is expressed in Pascals (Pa). If stress exceeds a material's elastic limit, it stretches or deforms irreversibly. In our exercise with the acrobat, it's crucial the steel wire doesn’t stretch beyond its elastic limit of \(2.5 \times 10^8 \, \text{Pa}\). To prevent this, we calculate a safe diameter for the wire, ensuring it withstands the force without permanent deformation.

Cross-sectional Area

The cross-sectional area is a geometric property that describes the area of an object as viewed along an axis. In the case of a steel wire, it's the area of the circle formed when you look at its end, which significantly impacts the wire's ability to handle stress.

For a circular wire, the cross-sectional area \(A\) is calculated using the formula:

• \( A = \pi \left( \frac{d}{2} \right)^2 \)

where \(d\) is the diameter of the wire. This mathematical expression tells us how the increase in the diameter affects the area and consequently, the stress the wire can bear. A larger diameter results in a larger area, which reduces the stress for the same amount of force. This concept is vital as it determines the minimum diameter necessary to keep the wire within its elastic limits, ensuring safety and structural integrity.

Force Due to Gravity

In physics, the force due to gravity is the force with which Earth attracts a body towards its center. This force is directly proportional to the mass of the object and is calculated using the gravitational constant.

• \( F = mg \)

where \(m\) is the mass of the object (in kilograms), and \(g\) is the acceleration due to gravity, approximately \(9.8 \, \text{m/s}^2\).

For the acrobat, with a mass of 55 kg, the force due to gravity can be calculated as:

• \( F = 55 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 539 \, \text{N} \)

This force represents the weight of the acrobat and is pivotal in our calculations since it influences the stress on the wire. Understanding this force helps us ensure the wire's diameter is ample enough to handle the force safely, without surpassing the wire's elastic limit.