Question

Example: 6.1 A force of 2500 N affected a metal wire 10 m long and 3.5 mm in diameter, extending by 0.5 cm find: a. Stress. b. Strain. c. Young modulus Solution F Stress = \frac{F}{A} 2500 = \frac{2500}{\pi \times (1.75 \times 10^{-3})^2} = 2.6 \times 10^8 \frac{N}{m^2} \Delta L Strain = \frac{\Delta L}{L} = \frac{0.5 \times 10^{-2}}{10} = 5 \times 10^{-4} Stress Y = \frac{Stress}{Strain} = \frac{2.6 \times 10^8}{5 \times 10^{-4}} = 5.2 \times 10^{11} \frac{N}{m^2}

          Example: 6.1
A force of 2500 N affected a metal wire 10 m long and 3.5 mm in diameter,
extending by 0.5 cm find:
a. Stress.
b. Strain.
c. Young modulus
Solution
F
Stress = \frac{F}{A}
2500
= \frac{2500}{\pi \times (1.75 \times 10^{-3})^2} = 2.6 \times 10^8 \frac{N}{m^2}
\Delta L
Strain = \frac{\Delta L}{L}
= \frac{0.5 \times 10^{-2}}{10} = 5 \times 10^{-4}
Stress
Y = \frac{Stress}{Strain}
= \frac{2.6 \times 10^8}{5 \times 10^{-4}} = 5.2 \times 10^{11} \frac{N}{m^2}

Example: 6.1
A force of 2500 N affected a metal wire 10 m long and 3.5 mm in diameter,
extending by 0.5 cm find:
a. Stress.
b. Strain.
c. Young modulus
Solution
F
Stress = (F)/(A)
2500
= (2500)/(π×(1.75 ×10^-3)^2) = 2.6 ×10^8 (N)/(m^2)
ΔL
Strain = (ΔL)/(L)
= (0.5 ×10^-2)/(10) = 5 ×10^-4
Stress
Y = (Stress)/(Strain)
= (2.6 ×10^8)/(5 ×10^-4) = 5.2 ×10^11 (N)/(m^2)

Added by Nuria B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Ameer Said

Step 1

5 cm and the original length is 2.1 cm. We are also given that the force applied is 2500 N and the area is 2 m^2. Show more…

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Could you please explain this question? How did we get 1.75? I need all the steps. Example: 6.1 Extending by 0.5 cm, find: a. Stress b. Strain c. Young modulus Solution: a. Stress = Force/Area Stress = 2500 N / (2 m^2) Stress = 1250 N/m^2 b. Strain = Change in length / Original length Strain = (0.5 cm) / (2.1 cm) Strain = 0.2381 c. Young modulus = Stress / Strain Young modulus = 1250 N/m^2 / 0.2381 Young modulus = 5246.03 N/m^2