A wire is initially stress-free between rigid supports. It has Y = 2.0 × 1 0 11 Pa , area 0.50 mm 2 and α = 12 × 1 0 − 6 K − 1 . For a 40 K cooling, find tension to first order.

A = 0.50 × 1 0 − 6 m 2 . F = Y A α Δ T . F = ( 2 × 1 0 11 ) ( 0.5 × 1 0 − 6 ) ( 12 × 1 0 − 6 ) ( 40 ) = 48 N .

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PhysicsHeat & ThermoThermal ExpansionMedium2 minQB

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A wire is initially stress-free between rigid supports. It has $Y = 2.0 \times 1 0^{11} Pa$ , area $0.50 mm^{2}$ and $α = 12 \times 1 0^{- 6} K^{- 1}$ . For a 40 K cooling, find tension to first order.

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Answer:

Solution:

$A = 0.50 \times 1 0^{- 6} m^{2}$ .

$F = Y A α Δ T . F = (2 \times 1 0^{11}) (0.5 \times 1 0^{- 6}) (12 \times 1 0^{- 6}) (40) = 48 N .$

Subject:PhysicsTopic:Heat & ThermoSubtopic:Thermal Expansion

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