1. Inclined Planes

2. An inclined plane is a type of simple machine An inclined plane is a large and flat object that is tilted so that one end is higher than the other What is an Inclined Plane?

3. Real World Applications of Inclined Planes Hadley Canal in Massachusetts used Inclined Plane engineering around 1800’s to raise and lower boats over Great Falls

4. The greater the angle of the inclined surface, the faster an object will slide down the incline There are always at least 2 forces acting on an object on an inclined plane F grav and F norm Background Information The normal force is always perpendicular to the inclined surface The gravitational force (WEIGHT) is always in downward direction

5. Solving for the Forces 1. Break down F grav into its x and y components F ║ = mgsinΘ and F ┴ = mgcosΘ

6. HINTS… 1. F ┴ is always equal and opposite F norm 2. When there is NO FRICTION, F ║ is the net force

7. 2. Deduce the net force and solve for other unknowns including acceleration and µ What is the net force for the Inclined Plane diagram to the right?

8. Example to Do Together! Solve for all unknowns listed. The crate has a mass of 100 kg and the coefficient of friction between the crate and the incline is 0.3 F ║ = F frict = a = F ┴ = F net =

9. 1. Break down F grav into its components F ║ = mgsinΘ F ║ = (100)(9.8)sin30° F ║ = 490 N F ┴ = mgcosΘ F ┴ = (100)(9.8)cos30° F ┴ = 849 N F grav = mg F grav = (100)(9.8) F grav = 980 N

10. 2. This example has friction, therefore let’s solve for F frict next Ffrict = µF norm Ffrict = 0.3849 Ffrict = 255 N F ║ = 490 N F ┴ = 849 N Remember, Fnorm is ALWAYS equal and opposite F ┴ 849 980

11. 3. Now, from our results we can deduce F net F net = F ║ - F frict F net = 490 – 255 F net = 235 N F ║ = 490 N F ┴ = 849 N 849 255 980 Remember, when there is NO FRICTION Fnet = F ║

12. 4. Lastly, we need to solve for the acceleration F net = ma a = F net ÷ m a = 235 ÷ 100 a = 2.35 m/s 2 F ║ = 490 N F ┴ = 849 N 849 255 980 Fnet = 235 N

13. Answers 1. F ║ = mgsinΘ F ║ = (100)(9.8)sin30° F ║ = 490 N 2. F ┴ = mgcosΘ F ┴ = (100)(9.8)cos30° F ┴ = 849 N 3. F ┴ = F norm therefore F norm = 849 N

14. Answers Continued 4. F frict = μF norm F frict =.3(849) = 255 N 5. F net = F ║ - F frict F net = 490 – 255 = 235 N 6. F = ma so a = F/m a = 235/100 = 2.35 m/s 2