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Inclined Planes

Inclined Planes. What is an Inclined Plane?. 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. Real World Applications of Inclined Planes.

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Inclined Planes

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  1. Inclined Planes

  2. What is an Inclined Plane? • 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

  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. Background Information • 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 Fgrav and Fnorm The normal force is always perpendicular to the inclined surface The gravitational force (WEIGHT) is always in downward direction

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

  6. HINTS… 1. F┴ is always equal and opposite Fnorm 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. 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 Example to Do Together! F║ = Ffrict = a = F┴ = Fnet =

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

  10. 2. This example has friction, therefore let’s solve for Ffrict next 849 Ffrict = µ•Fnorm Ffrict = 0.3•849 Ffrict = 255 N F ║ = 490 N F ┴ = 849 N Remember, Fnorm is ALWAYS equal and opposite F┴ 980

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

  12. 4. Lastly, we need to solve for the acceleration Fnet = ma a = Fnet ÷ m a = 235 ÷ 100 a = 2.35 m/s2 849 255 F ║ = 490 N F ┴ = 849 N 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┴ = Fnorm therefore Fnorm = 849 N

  14. Answers Continued 4. Ffrict = μFnorm Ffrict = .3(849) = 255 N 5. Fnet = F║ - Ffrict Fnet = 490 – 255 = 235 N 6. F = ma so a = F/m a = 235/100 = 2.35 m/s2

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