A stainless steel orthodontic wire is applied to a tooth, as shown in the figure
ID: 1905389 • Letter: A
Question
A stainless steel orthodontic wire is applied to a tooth, as shown in the figure above. The wire has an unstretched length of 2.9 cm and a diameter of 0.23 mm. The wire is stretched 0.1 mm. Young's modulus for stainless steel is 1.80E+11 Pa. What is the tension in the wire? b.)What is the magnitude of the force on the tooth due to the wires? Disregard the width of the tooth
A stainless steel orthodontic wire is applied to a tooth, as shown in the figure above. The wire has an unstretched length of 2.9 cm and a diameter of 0.23 mm. The wire is stretched 0.1 mm. Young's modulus for stainless steel is 1.80E+11 Pa. What is the tension in the wire? b.)What is the magnitude of the force on the tooth due to the wires? Disregard the width of the toothExplanation / Answer
in the elastic (initial, linear) portion of the stress-strain curve: E = F/Ao ÷ ?L/Lo where E is the Young's modulus (modulus of elasticity) F is the force applied to the object; Ao is the original cross-sectional area through which the force is applied; ?L is the amount by which the length of the object changes; Lo is the original length of the object. The Young's modulus of a material can be used to calculate the force it exerts under specific strain. F = (E * Ao * ?L) ÷ Lo A stainless steel orthodontic wire is applied to a tooth, as shown in the figure above. The wire has an unstretched length of 2.9 cm and a diameter of 0.23 mm. The wire is stretched 0.1 mm. Young's modulus for stainless steel is 1.80E+11 Pa. What is the tension in the wire? E = 1.80 * 10^11 Pa = 1.80 * 10^11 N/m^2 Ao = p * r^2 r = ½ * diameter = 0.115 mm = 0.115 * 10^-3 m Ao = p * (0.115 * 10^-3)^2 = 4.15 * 10^-8 m^2 ?L = 0.1 mm = 1 * 10^-4 m Lo = 2.9 cm = 2.9 * 10^-2 m F = (E * Ao * ?L) ÷ Lo F = [(1.80 * 10^11) * (p * (0.115 * 10^-3)^2) * (1 * 10^-4)] ÷ (2.9 * 10^-2) F = 25.8 N
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