A piano wire 0.90 m long and 0.85 mm in diameter is fixed on one end....

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Physics

A piano wire 0.90 m long and 0.85 mm in diameter is fixed on oneend. The other end is wrapped around a tuning peg 3.5 mm indiameter. Initially the wire, whose Young's modulus is2.4Ã—1010N/m2,
has a tension of 30 N .
Part A
Find the tension in the wire after the tuning peg has beenturned through one complete revolution.
Express the tension in newtons to two significant figures.
T =
166.16
Â Â NÂ Â
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4.1 Ratings (503 Votes)

ans)

from above data that

given that

{Initial Length} = L₀ = 0.90 m

we know that
{Initial Cross Section Area} = A₀ = Ï€*D²/4
Â Â Â Â Â Â Â Â Â = Ï€*(0.85e-3 m)²/4
Â Â Â Â Â Â Â Â Â = 5.67 e-7m³

{Stretch Length} = Î”L = Ï€*(3.5e-3m) = 1.09 e-2m

{Î”Force} = Î”F = E*A₀*Î”L/L₀ =Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (<--- E = Young's Modulus)
Â Â Â Â Â Â Â Â Â Â Â Â Â = (2.24e+10 N/m²)*(5.67 e-7 m³ )*(1.09 e-2 m)/(0.90 m)
Â Â Â Â Â Â Â Â Â Â Â Â Â = 153.8N

{New Wire Tension}

= T_new = T_old + Î”F
Â Â Â Â Â Â Â Â Â = (30.0 N) +(153.8 N)
Â Â Â Â Â Â Â Â Â = 183.8Â Â N

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