a stainless steel orthodontic wire is applied to a tooth.the wire has an upstretched length of 3.1cm and a radius of 0.11mm.if the wire is stretched 0.10mm,find the magnitude and direction of the force on the tooth.disregard the width of the tooth assume young modulus for stainless steel is 18*10*10pa
To find the magnitude and direction of the force on the tooth, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. In this case, the stainless steel wire can be treated as a spring.
Hooke's Law equation for a spring is: F = -k * Δx
Where:
F is the force applied to the spring in Newtons (N),
k is the spring constant in Newtons per meter (N/m),
Δx is the displacement from the equilibrium position in meters (m).
First, let's calculate the spring constant (k) using the formula:
k = (E * A) / L
Where:
E represents Young's modulus (18 * 10^10 Pa),
A is the cross-sectional area of the wire in square meters,
L is the original length of the wire in meters.
The cross-sectional area (A) can be calculated using the formula:
A = π * r^2
Where:
r is the radius of the wire in meters.
Converting the given values:
Original length (L) = 3.1 cm = 0.031 m
Radius (r) = 0.11 mm = 0.00011 m
Young's modulus (E) = 18 * 10^10 Pa
Now we can calculate the spring constant (k):
A = π * (0.00011 m)^2
A = 3.801*10^(-8) m^2
k = (18 * 10^10 Pa * 3.801*10^(-8) m^2) / 0.031 m
k = 2206451.6129032258 N/m
Next, let's calculate the displacement (Δx). We are given that the wire is stretched by 0.10 mm, which is equal to 0.0001 m:
Δx = 0.0001 m
Now we can calculate the force (F) using Hooke's Law:
F = -k * Δx
F = -2206451.6129032258 N/m * 0.0001 m
F = -220.6451612903226 N
The negative sign indicates that the force is acting in the opposite direction of the displacement. Since the wire is stretched, the force should be pulling the tooth inward.
Therefore, the magnitude of the force on the tooth is approximately 220.645 N, and the direction of the force is inward.