A brass wire is to withstand a tensile force of 350 N without breaking. What minimum diameter must the wire have?

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First, identify the material properties of brass, specifically its tensile strength. Tensile strength is the maximum stress that a material can withstand while being stretched or pulled before breaking.

Next, use the formula for stress, which is defined as force divided by the area over which the force is applied: \( \text{Stress} = \frac{F}{A} \). Here, \( F \) is the force applied, and \( A \) is the cross-sectional area of the wire.

Since the wire is cylindrical, the cross-sectional area \( A \) can be expressed in terms of its diameter \( d \) using the formula for the area of a circle: \( A = \frac{\pi d^2}{4} \).

Rearrange the stress formula to solve for the diameter \( d \): \( d = \sqrt{\frac{4F}{\pi \times \text{Tensile Strength}}} \).

Substitute the given force (350 N) and the tensile strength of brass into the equation to find the minimum diameter required for the wire to withstand the force without breaking.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Tensile Strength

Tensile strength is the maximum amount of tensile stress that a material can withstand before failure. It is a critical property for materials subjected to stretching forces, such as wires. Understanding tensile strength helps determine the material's ability to resist breaking under tension.

Stress and Strain

Stress is the force applied per unit area within materials, while strain is the deformation or displacement it causes. In this context, stress is calculated by dividing the tensile force by the cross-sectional area of the wire. The relationship between stress and strain is crucial for determining the wire's ability to handle the applied force.

Cross-Sectional Area

The cross-sectional area of a wire is the area of its cut surface perpendicular to its length. For a circular wire, it is calculated using the formula A = π(d/2)^2, where d is the diameter. This area is essential for calculating stress, as a larger area can distribute the force more effectively, reducing stress and preventing breakage.

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