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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: 9780137488179Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoGiancoli Douglas 5th editionCh. 12 - Static Equilibrium; Elasticity and FractureProblema 75b
Capitolo 12, Problema 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.]

Guida verificata passo dopo passo
1
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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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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The Normal Force

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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Intro To Elastic Collisions
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