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Ch. 12 - Static Equilibrium; Elasticity and Fracture
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 12 - Static Equilibrium; Elasticity and FractureProblem 75b
Chapter 12, Problem 75b

A steel rod of radius R = 15 cm and length ℓ₀ stands upright on a firm surface. A 78-kg man climbs atop the rod. When a metal is compressed, each atom throughout its bulk moves closer to its neighboring atom by exactly the same fractional amount. If iron atoms in steel are normally 2.0 x 10⁻¹⁰ m apart, by what distance did this interatomic spacing have to change in order to produce the normal force required to support the man? [Note: Neighboring atoms repel each other, and this repulsion accounts for the observed normal force.]

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Step 1: Understand the problem. The man exerts a force equal to his weight on the steel rod, which compresses the rod slightly. This compression causes the interatomic spacing of the steel atoms to decrease. The goal is to calculate the change in interatomic spacing required to support the man's weight.
Step 2: Calculate the force exerted by the man on the rod. The force is given by the man's weight, which is the product of his mass (m = 78 kg) and the acceleration due to gravity (g ≈ 9.8 m/s²). Use the formula: F = m × g.
Step 3: Relate the force to the stress in the rod. Stress is defined as force per unit area. The cross-sectional area of the rod is circular, so use the formula for the area of a circle: A = π × R², where R = 15 cm = 0.15 m. Then calculate the stress: σ = F / A.
Step 4: Use the relationship between stress and strain. Strain is the fractional change in length of the material and is given by: strain = stress / Young's modulus (Y). For steel, the Young's modulus is approximately Y ≈ 2.0 × 10¹¹ N/m². Calculate the strain: strain = σ / Y.
Step 5: Relate the strain to the change in interatomic spacing. Since strain is the fractional change in length, it is also the fractional change in interatomic spacing. If the original interatomic spacing is d₀ = 2.0 × 10⁻¹⁰ m, the change in interatomic spacing Δd is given by: Δd = strain × d₀. Substitute the values to find Δd.

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Key Concepts

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

Normal Force

Normal force is the support force exerted by a surface perpendicular to an object resting on it. In this scenario, the normal force counteracts the weight of the man on the steel rod, ensuring he does not fall through. It is crucial for understanding how the rod supports the man's weight and how this force affects the interatomic spacing in the material.
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Interatomic Spacing

Interatomic spacing refers to the distance between neighboring atoms in a material. In steel, this spacing is typically around 2.0 x 10⁻¹⁰ m. When a force is applied, such as the weight of the man, this spacing changes slightly, which is essential for understanding how materials deform under load and how they maintain structural integrity.
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Elastic Deformation

Elastic deformation is the reversible change in shape or size of a material when a force is applied. In the context of the steel rod, when the man climbs on it, the rod experiences compression, causing the interatomic distances to decrease. This concept is vital for analyzing how the rod can support weight without permanent deformation, as it returns to its original shape once the load is removed.
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Related Practice
Textbook Question

A uniform 95-kg flagpole of length 8.4 m is being erected by pulling on a rope attached 2/3 of the way to the top (Fig. 12–94). When the pole is inclined at 35° and the rope makes an angle with the ground of 18°, what is the tension in the rope?

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Textbook Question

A uniform beam of mass M and length ℓ is mounted on a hinge at a wall as shown in Fig. 12–101. It is held in a horizontal position by a wire making an angle θ as shown. A mass m is placed on the beam a distance x from the wall, and this distance can be varied. Determine, as a function of x, the components of the force exerted by the beam on the hinge.

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Textbook Question

Suppose a 65-kg person jumps from a height of 3.0 m down to the ground. Estimate the stress and determine if the tibia will break in a stiff-legged landing (d = 1.0 cm).

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Textbook Question

When a mass of 25 kg is hung from the middle of a fixed straight aluminum wire, the wire sags to make an angle of 12° with the horizontal as shown in Fig. 12–90. Determine the radius of the wire.

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Textbook Question

If 25 kg is the maximum mass m that a person can hold in a hand when the arm is positioned with a 105° angle at the elbow as shown in Fig. 12–102, what is the maximum force Fₘₐₓ that the biceps muscle exerts on the forearm? Assume the forearm and hand have a total mass of 2.0 kg with a cg that is 15 cm from the elbow, and that the biceps muscle attaches 5.0 cm from the elbow.

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Textbook Question

A 25-kg object is being lifted by two people pulling on the ends of a 1.15-mm-diameter nylon cord that goes over two 3.00-m-high poles that are 4.5 m apart, as shown in Fig. 12–93. How high above the floor will the object be when the cord breaks?

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