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Chapter 11: Problem 37

In a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of 90.8 \(\mathrm{N}\) is applied perpendicular to each end. If the diameter of the wire is \(1.84 \mathrm{mm},\) what is the breaking stress of the alloy?

Short Answer

Expert verified
The breaking stress of the alloy is approximately \(3.41 \times 10^7 \text{ N/m}^2\).

Step by step solution

01

Understand the Problem

We need to calculate the breaking stress of a metal wire when a specific force is applied before it breaks. We are given the force and the diameter of the wire.
02

Recall the Formula for Stress

Stress is defined as the force applied per unit area. The formula for stress (\( \sigma \)) is:\[ \sigma = \frac{F}{A} \]where \( F \) is the force applied, and \( A \) is the cross-sectional area of the wire.
03

Calculate the Cross-sectional Area of the Wire

The wire has a circular cross-section. The area (\( A \)) of a circle is calculated using the formula:\[ A = \pi \left(\frac{d}{2}\right)^2 \]where \( d \) is the diameter of the circle. Here, \( d = 1.84 \text{ mm} = 1.84 \times 10^{-3} \text{ m} \).
04

Substitute Diameter in the Area Formula

Substitute \( d = 1.84 \times 10^{-3} \text{ m} \) into the formula for area:\[ A = \pi \left(\frac{1.84 \times 10^{-3} \text{ m}}{2}\right)^2 \]Simplifying, we get the area in square meters.
05

Calculate the Cross-sectional Area

Calculate:\[ A = \pi \left(0.92 \times 10^{-3} \right)^2 = \pi (0.8464 \times 10^{-6}) = 2.6605 \times 10^{-6} \text{ m}^2 \]
06

Substitute into the Stress Formula

Substitute \( F = 90.8 \text{ N} \) and \( A = 2.6605 \times 10^{-6} \text{ m}^2 \) into the stress formula:\[ \sigma = \frac{90.8 \text{ N}}{2.6605 \times 10^{-6} \text{ m}^2} \]
07

Calculate the Breaking Stress

Calculate:\[ \sigma = 3.41 \times 10^7 \text{ N/m}^2 \]This value is the stress at which the alloy wire breaks.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tensile Force
Understanding tensile force is crucial when studying the mechanical properties of materials. Tensile force is a type of force that attempts to stretch a material.
It acts along the length of the material and tries to pull it apart. In the context of the given problem, the tensile force applied to the wire is 90.8 N. This means a force of 90.8 newtons is pulling each end of the wire in opposite directions, trying to stretch it till it breaks.
Materials have different capacities to withstand tensile forces without breaking, characterized by a property known as tensile strength. Larger tensile forces exert more stress, which could lead to breaking under high force. Therefore, understanding how different materials respond to tensile force helps in selecting the appropriate material for specific applications where strength is critical.
Cross-sectional Area
The cross-sectional area of a wire is the surface area of a slice of the wire as if it were cut at any random point perpendicular to its length. For circular wires, such as the one in the exercise, the cross-sectional area determines the amount of surface area over which a force acts.
To calculate the cross-sectional area, one uses the formula for the area of a circle: \[A = \pi \left( \frac{d}{2} \right)^2\]where \(d\) is the diameter of the wire. In the problem, the wire's diameter is 1.84 mm. We convert this to meters before using it in the formula because standard units for area in stress calculations are square meters.
Calculating the area involves converting 1.84 mm to meters \[1.84 \text{ mm} = 1.84 \times 10^{-3} \text{ m},\]then applying it in the area formula, yielding a cross-sectional area of \[2.6605 \times 10^{-6} \text{ m}^2.\] Understanding cross-sectional area is crucial for stress calculations. Larger cross-sectional areas can distribute tensile forces over a bigger area, potentially reducing the stress experienced by the material. Smaller areas can increase stress on the material, influencing its ability to withstand force.
Stress Calculation
Stress is a measure of how much force is applied to a specific area of a material. It is calculated by dividing the force (\(F\)) applied by the cross-sectional area (\(A\)) over which the force acts:\[\sigma = \frac{F}{A}\]This formula represents the internal resistance offered by the material to withstand the force. It is crucial for determining the load a material can bear without failing.
In the exercise, we calculated the stress as the tensile force (90.8 N) applied on the small cross-sectional area we already found (\(2.6605 \times 10^{-6} \text{ m}^2\)).
Substituting these values into our stress formula results in:\[\sigma = \frac{90.8 \text{ N}}{2.6605 \times 10^{-6} \text{ m}^2}\]The calculation yields a breaking stress of \[3.41 \times 10^7 \text{ N/m}^2.\]Stress measurements help engineers decide whether a material is suitable for construction or a specific application. High stress can lead to material failure, so ensuring that the material's breaking stress surpasses the expected working stresses is essential for safety and reliability.

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Most popular questions from this chapter

A petite young woman distributes her 500 \(\mathrm{N}\) weight equally over the heels of her high-heeled shoes. Each heel has an area of 0.750 \(\mathrm{cm}^{2}\) (a) What pressure is exerted on the floor by each heel? (b) With the same pressure, how much weight could be supported by two flat- bottomed sandals, each of area 200 \(\mathrm{cm}^{2} ?\)

A bar with cross-sectional area \(A\) is subjected to equal and opposite tensile forces \(\vec{F}\) at its ends. Consider a plane through the bar making an angle through the bar making an angle \(\theta\) with a plane at right angles to the bar (Fig. 11.64\()\) . (a) What is the tensile (normal) stress at this plane in terms of \(F, A,\) and \(\theta ?(b)\) What is the shear (tangential) stress at the plane in terms of \(F, A,\) and \(\theta ?\) (c) For what value of \(\theta\) is the tensile stress a maximum? (d) For what value of \(\theta\) is the shear stress a maximum?

A \(15,000-N\) crane pivots around a friction-free axle at its base and is supported by a cable making a \(25^{\circ}\) angle with the crane (Fig. 11.29 ). The crane is 16 \(\mathrm{m}\) long and is not uniform, its center of gravity being 7.0 \(\mathrm{m}\) from the axle as measured along the crane. The cable is attached 3.0 \(\mathrm{m}\) from the upper end of the crane. When the crane is raised to \(55^{\circ}\) above the horizontal holding an \(11,000-N\) pallet of bricks by a 2.2 -in very light cord, find (a) the tension in the cable, and \((b)\) the horizontal and vertical components of the force that the axle exerts on the crane. Start with a free-body diagram of the crane.

A door 1.00 \(\mathrm{m}\) wide and 2.00 \(\mathrm{m}\) high weighs 280 \(\mathrm{N}\) and is supported by two hinges, one 0.50 \(\mathrm{m}\) from the top and the other 0.50 \(\mathrm{m}\) m from the bottom.Each hinge supports half the total weight of the door. Assuming that the door's center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

Forearm. In the human arm, the forearm and hand pivot about the elbow joint. Consider a simplified model in which the biceps muscle is attached to the forearm 3.80 \(\mathrm{cm}\) from the elbow joint. Assume that the person's hand and forearm together weigh 15.0 \(\mathrm{N}\) and that their center of gravity is 15.0 \(\mathrm{cm}\) from the elbow (not quite halfway to the hand). The forearm is held horizontally at a a right angle to the upper arm, with the biceps muscle exerting its force perpendicular to the foreann. (a) Draw a free-body diagram for the forearm, and find the force exerted by the biceps when the hand is empty. (b) Now the person holds a \(80.0-\mathrm{N}\) weight in his hand, with the forearm still horizontal. Assume that the center of gravity of this weight is 33.0 \(\mathrm{cm}\) from the elbow. Construct a freebody diagram for the forearm, and find the force now exerted by the biceps. Explain why the biceps muscle needs to be very strong. (c) Under the conditions of part ( \((b),\) find the magnitude and direction of the force that the elbow joint exerts on the forearm. (d) While holding the 80.0 -N weight, the person raises his forearm. until it is at an angle of \(53.0^{\circ}\) above the horizontal. If the biceps muscle continues to exert its force perpendicular to the forearm, what is this force when the forearmis in this position? Has the force increased or decreased fromits value in part (b)? Explain why this is so, and test your answer by actually doing this with your own arm.
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