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3 years ago

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A mineshaft has an ore elevator hung from a single braided cable of diameter 2.5 cm. Young's modulus of the cable is 10×10^10N/m

2. When the cable is fully extended, the end of the cable is 900 m below the support. How much does the fully extended cable stretch when 1500 kg of ore is loaded? Express your answer with the appropriate unit

1 answer:

antiseptic1488 [7]3 years ago

4 0

Answer:

ΔL = 0.2697 m = 26.97 cm

Explanation:

Given

∅ = 2.5 cm = 0.025 m

E = 10*10¹⁰ N/m²

L = 900 m

m = 1500 Kg

ΔL = ?

We get P = W as follows

W = m*g = (1500 Kg)(9.81 m/s²) = 14715 N

then we apply the equation of Hooke's Law

ΔL = P*L / (A*E)

⇒ ΔL = 14715 N*900 m / (π*0.25*(0.025 m)²*10*10¹⁰ N/m²)

⇒ ΔL = 0.2697 m = 26.97 cm

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