Astainless-steel orthodontic wire is applied to a tooth, as in the feu e bele .

Question

Astainless-steel orthodontic wire is applied to a tooth, as in the feu e bele . The wire has an unst ethed length of 3 24 cm and·radius of 0 13 mnrthe wire force on the tooth. Disregard the width of the tooth and assume Young's modulus for stainless steel is 18 x 10 Pa. magnitude stretched onmn, mnd the-agnitude and dinserofy - Need Help? Reedn A high-speed lifting mechanism supports a(n) 900-kg object with a steel cable that is 40.0 m long and 4.00 cm2 in cress-sectional area (a) Determine the elongation of the cable. (Enter your answer to at least two decimal places.) (b) By what additional amount does the cable increase in length if the object is accelerated upwards at a rate of 3.1 m/s (c) what is the greatest mass that can be accelerated upward at 3.1 m/s if the stress in the cable is not to exceed the elastic limit of the cable, which is 2.2 × 108 Pa, kg Need Help? Rases

Explanation / Answer

Young's modulus Y = FL/(A*e)

Y = 18*10^10 Pa

F = force aplied

L = length of the wire = 3.24*10^-2 m

A = area of cross section = pi*r^2 = pi*(0.13*10^-3)^2

e= stretch = 0.12*10^-3 m

F = Y*A*e/L

F = 18*10^10*pi*(0.13*10^-3)^2* 0.12*10^-3/(3.24*10^-2 )

F = 35.4 N

net force 2*F*sin30

Fnet = 2*35.4*sin30 = 35.4 N <<<<-----------ANSWER

direction downward <<----------------ANSWER

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Tension in wire T = mg

Y = T*L/(A*e)

Y = m*g*L/(A*e)

elongation e = m*g*L/(A*Y)

e = 900*9.8*40/(4*10^-4*18*10^10)

e = 4.9 mm    <<----------------ANSWER

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part(b)

from newtons second law

Fnet = T - mg = m*a

tension T = mg + ma = 900*(9.8+3.1) = 11610 N

elongation e = T*L/(A*Y)

e = 11610*40/(4*10^-4*18*10^10)

e = 6.45 mm <<<----------------ANSWER

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