Skip to main content
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
Capitolo 12, Problema 42

A 15-cm-long tendon was found to stretch 3.7 mm by a force of 13.4 N. The tendon was approximately round with an average diameter of 8.5 mm. Calculate Young’s modulus of this tendon.

Guida verificata passo dopo passo
1
Step 1: Recall the formula for Young's modulus, which is defined as \( E = \frac{\sigma}{\epsilon} \), where \( \sigma \) is the stress and \( \epsilon \) is the strain. Stress is given by \( \sigma = \frac{F}{A} \), and strain is given by \( \epsilon = \frac{\Delta L}{L_0} \).
Step 2: Calculate the cross-sectional area \( A \) of the tendon, assuming it is approximately circular. The formula for the area of a circle is \( A = \pi r^2 \), where \( r \) is the radius. Convert the diameter of 8.5 mm to meters (\( r = \frac{8.5}{2} \times 10^{-3} \) m) and substitute into the formula.
Step 3: Compute the strain \( \epsilon \) using the formula \( \epsilon = \frac{\Delta L}{L_0} \). Here, \( \Delta L \) is the elongation (3.7 mm converted to meters) and \( L_0 \) is the original length of the tendon (15 cm converted to meters).
Step 4: Calculate the stress \( \sigma \) using the formula \( \sigma = \frac{F}{A} \), where \( F \) is the applied force (13.4 N) and \( A \) is the cross-sectional area calculated in Step 2.
Step 5: Substitute the values of \( \sigma \) and \( \epsilon \) into the formula for Young's modulus \( E = \frac{\sigma}{\epsilon} \) to determine the modulus of elasticity for the tendon.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
4m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Young's Modulus

Young's modulus is a measure of the stiffness of a material, defined as the ratio of tensile stress to tensile strain. It quantifies how much a material will deform under a given load, providing insight into its elastic properties. A higher Young's modulus indicates a stiffer material that deforms less under stress.
Video consigliato:
Percorso guidato
12:00
Young's Double Slit Experiment

Tensile Stress

Tensile stress is the force applied per unit area of a material, typically measured in pascals (Pa). It is calculated by dividing the applied force by the cross-sectional area of the material. In the context of the tendon, it helps determine how much force is exerted on the tendon relative to its size.
Video consigliato:
Percorso guidato
07:49
Ray Diagrams for Convex Mirrors

Tensile Strain

Tensile strain is the measure of deformation representing the displacement between particles in a material when subjected to tensile stress. It is a dimensionless quantity calculated as the change in length divided by the original length. In this case, it indicates how much the tendon stretches relative to its original length when a force is applied.
Pratica correlata
Domanda del libro di testo

The Leaning Tower of Pisa is 55 m tall and about 7.7 m in radius. The top is 4.5 m off center. Is the tower in stable equilibrium? If so, how much farther can it lean before it becomes unstable? Assume the tower is of uniform composition.

1564
views
Domanda del libro di testo

Assume the supports of the uniform cantilever shown in Fig. 12–79 (m = 2900 kg) are made of wood. Calculate the minimum cross-sectional area required of each, assuming a safety factor of 9.0.

1755
views
Domanda del libro di testo

A refrigerator is approximately a uniform rectangular solid 1.9 m tall, 1.0 m wide, and 0.75 m deep. If it sits upright on a truck with its 1.0-m dimension in the direction of travel, and if the refrigerator cannot slide on the truck, how rapidly can the truck accelerate without tipping the refrigerator over? [Hint: The normal force would act at one corner.]

1563
views
Domanda del libro di testo

A steel cable is to support an elevator whose total (loaded) mass is not to exceed 3100 kg. If the maximum acceleration of the elevator is 1.8 m/s² , calculate the diameter of cable required. Assume a safety factor of 8.0.

1535
views
Domanda del libro di testo

A heavy load Mg = 62.0 kN hangs at point E of the single cantilever truss shown in Fig. 12–81. Use a torque equation for the truss as a whole to determine the tension FT in the support cable, and then determine the force FA→\(\overrightarrow{F_{A}\)} on the truss at pin A. Neglect the weight of the trusses, which is small compared to the load.

1353
views
1
rank
Domanda del libro di testo

A marble column of cross-sectional area 1.4m² supports a mass of 22,000 kg. By how much is the column shortened if it is 8.6 m high?

1795
views