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A force of 5000 N is applied outwardly to each end of a 5.0-m long rod with a radius of 34.0 mm and a Young's modulus of 125 x 108 N/m^2. The elongation of the rod is:

a. 0.55 mm
b. 1.42 mm
c. 0.0040 mm
d. 0.11 mm
e. 0.022 mm

Respuesta :

Answer:

0.55 mm

Explanation:

E = σ/ε

E = (F/A) / (ΔL/L₀)

ΔL/L₀ = F/EA

ΔL = FL₀/EA

ΔL = 5000(5.0) / (125e8(π(0.034²)))

ΔL = 0.00055070914... m

With that Young's modulus our shaft may be made of strong wood such as oak or douglas fir, or possibly lead.

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