Dyanmics Normal Force and Inclines

Normal Force The normal force is associated with surfaces, but how does a surface create this force. When you lie on a mattress your weight causes the springs to compress. As they compress a force develops in each spring that pushes the springs back toward their original position (up). When a box sits on the floor or on a table it compresses the intermolecular forces (forces between molecules) in a manner that is similar to compressing the springs in a mattress. The compressed intermolecular forces act like springs that push back against the weight of the box. The total upward force of all the compressed intermolecular forces acting together creates the normal force. Why is it called the normal force? In geometry a “normal line” is a line that is perpendicular to an object. The force created by surfaces is always perpendicular to the surface. As a result, this force lies along a normal line and is known as the normal force.

Determining Normal Force There is no specific formula that solves for the normal force. The normal force is situation dependent, and the equations that you will use to solve for it will vary from problem to problem. You must solve the normal force using the methods described in the previous PowerPoint presentations.

Example 1 Determine the normal force on a 10 kg mass at rest on a horizontal surface. Assess: Stationary, ΣFy = 0 Diagram: Sum forces: Solve: Fg N

Example 2 Determine the normal force on a 10 kg mass at rest on a horizontal surface while a rope pulls upward with 30 N. Assess: Stationary, ΣFy = 0 Diagram: Sum forces: Solve: Fg N T

Forces Acting on Inclines When an object is positioned on an incline its motion along the incline will always be at an angle to the traditional coordinate axis system. However, we control the placement and orientation of the coordinate axis. Since objects on inclines move parallel to the incline simply tilt the axis aligning the x-direction parallel to the incline. θ

Forces Acting on Inclines This orientation makes solving motion along the incline easier, but there is one draw back. The force of gravity, which has always been directed along the y-axis, is now at an angle to the coordinate axis system. As always the x- and y-vector quantities solve independently. Forces at an angle to the coordinate axis system MUST ALWAYS be split into components. The x-components must then be used with other x-variables to solve in the x-direction only. Likewise the y- components must be used in y- equations with other y-variables. Fg θ

Forces Acting on Inclines The right triangles formed by the components are similar triangles to the incline itself, and the angle of incline can be used to determine the magnitude of each component. The x-component is opposite the angle θ . Fg parallel to incline = Fg sin θ The y-component is adjacent to the angle θ . Fg perpendicular to incline = Fg cos θ Fg Fg sin θ Fg cos θ θ θ

Forces Acting on Inclines The x-component of the force of gravity, Fg sin θ , influences the object’s motion along the incline. Fg sin θ is ALWAYS present on inclines. When summing forces along the incline Fg sin θ is used in place of Fg , and must ALWAYS be included. The y-component of the force of gravity presses the object into the incline. When an object pushes against a surface the surface pushes back with a normal force. Since the object is not moving perpendicular to the incline the normal force is equal to the component of the force of gravity that is perpendicular to the incline. N =Fg cos θ Fg Fg sin θ Fg cos θ θ θ θ

Incline Diagrams Free Body Diagram Fg N Alternate Diagram Fg sin θ Includes only acting forces. As a result, this diagram is missing the important components that will solve the problem. Fg N Alternate Diagram This can include the components needed to solve the problem. Remember: Fg sin θ is always present on inclines. Fg sin θ N = Fg cos θ Fg cos θ This diagram explains the geometry of the forces, and their components, that act upon objects positioned on inclines. θ Fg Fg sin θ Fg cos θ N = Fg cos θ

Example 3 N N Fg sin θ Fg cos θ Fg A 5.0 kg mass is released from rest on a frictionless 30o incline. 5.0 kg Assess: Parallel to incline, accelerates, ΣFx = ma Perpendicular to incline, stationary, ΣFy = 0 Diagram: Free Body Diagram Alternate diagram Sum forces: Solve: Fg N Fg sin θ N Fg cos θ

Normal Force Differs in Each Problem In this presentation, and in previous presentations, we have seen that the magnitude of the normal force depends on the position of the object, the orientation of the surface, and on the other acting forces. These are only a few examples. The normal force must be determined in every problem. Fg N T