Questions: The drawing shows a wire attached to two back teeth and stretched across a front tooth. The purpose of this arrangement is to apply a force F to the front tooth. (The figure has been simplified by drawing the wire as if it ran straight from the front tooth to the back teeth) If the tension in the wire is 18.8 N, what are the magnitude and direction of the resultant force F applied to the front tooth in Newtons? Assume +y to be down and +x be to the right. Enter a positive answer if the resultant force is toward the back of the mouth. Note: If the picture below does not display, move to the next/previous question and come back to this one.

Solution Steps

The tension in the wire is given as 18.8 N. This tension acts along the wire, pulling the front tooth towards the back teeth. Due to the symmetry of the arrangement, the angles between the wire and the vertical are equal (37.5° each). We can decompose each tension force into its vertical and horizontal components.

The vertical component of each tension force is given by T*cos(θ), where T is the tension and θ is the angle between the wire and the vertical. In this case, θ = 37.5° and T = 18.8 N. So, the vertical component of each tension force is 18.8 * cos(37.5°) ≈ 14.94 N. Since there are two wires, the total vertical force is 2 * 14.94 N ≈ 29.88 N. This force acts downwards.

The horizontal component of each tension force is given by T*sin(θ), where T is the tension and θ is the angle between the wire and the vertical. So, the horizontal component of one tension force is 18.8 * sin(37.5°) ≈ 11.46 N to the right, and the other is 11.46 N to the left.

The horizontal components cancel each other out because they are equal in magnitude and opposite in direction. Therefore, the net force is equal to the total vertical component, acting downwards. Since the question asks for a positive answer if the force is directed towards the back of the mouth (which is downwards, the +y direction), we enter the positive value.

Final Answer: 29.9 N