The drawing shows an elastic cord attatched to two back teeth and stretched across a front tooth. The purpose of the arrangement is to apply a force to the front tooth. If the tension in the cord is 1.2N what are the magnitude and direction of the force applied to the front tooth?
(The drawing is of the top of the mouth with and angle of 66 degrees split into to so 33 degrees and 33 degrees)
To find the magnitude and direction of the force applied to the front tooth, we need to analyze the given information and use the concept of vector components.
First, let's break down the tension force into its horizontal and vertical components. We'll assume that the direction of the horizontal component is towards the right (positive x-axis) and the vertical component is upwards (positive y-axis).
Given that the angle between the cord and the horizontal axis is 66 degrees, we can determine the magnitudes of the horizontal and vertical components of the tension force using trigonometric functions.
The horizontal component (Fx) can be found using the formula:
Fx = Tension * cos(angle)
Substituting the values:
Fx = 1.2N * cos(66)
Fx ≈ 0.497N
The vertical component (Fy) can be found using the formula:
Fy = Tension * sin(angle)
Substituting the values:
Fy = 1.2N * sin(66)
Fy ≈ 1.108N
Therefore, the magnitude of the force applied to the front tooth is the vector sum of the horizontal and vertical components:
F = sqrt(Fx^2 + Fy^2)
F = sqrt((0.497)^2 + (1.108)^2)
F ≈ 1.213N
The direction of the force can be determined using the tangent function:
θ = tan^(-1)(Fy / Fx)
θ = tan^(-1)(1.108 / 0.497)
θ ≈ 66.9 degrees
Therefore, the magnitude of the force applied to the front tooth is approximately 1.213N, and the direction is approximately 66.9 degrees relative to the horizontal axis.