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Why do we use ramps (inclined planes)?

Why do we use ramps (inclined planes)?. How would you draw F W ?. Is this in the same direction as the motion?. Soooo what force makes the object slide down the plane?. Are there any other forces we have to consider?. F W

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Why do we use ramps (inclined planes)?

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  1. Why do we use ramps (inclined planes)?

  2. How would you draw FW? Is this in the same direction as the motion? Soooo what force makes the object slide down the plane? Are there any other forces we have to consider?

  3. FW must resolve intotwo components, one directed parallel (Fװ ) to surface and the other directed perpendicular (F┴ ) to the surface

  4. Fװis the component of of FW that makes the box slide down

  5. Where would you draw in the Normal Force?

  6. Remember FN is perpendicular to the surface

  7. If you knew angle and FW, how would you calculate F┴ and Fװ? To calculate F┴use:Fw cos ӨTo calculate Fװuse: Fw sin Ө

  8. F┴ is equal but opposite FNFװ = FNET if NO FRICTION

  9. How would you determine acceleration ? FNET = ma How would you determine acceleration if there was friction ? FNET = FA - FF How would you determine FF? FF = μFN

  10. Did you need to calculate F┴ ? You will when there is friction.

  11. If friction is present:

  12. Remember that FF= µFN

  13. Example 3 Given: 100kg cratecoefficient between crate and incline = 0.3

  14. Fw (100kg)(9.8m/s2) 980N • F┴ (100kg)(9.8m/s2)(cos30) 849N • Fװ (100kg)(9.8m/s2)(sin30) 490N • FN = F┴=849N • FF = µFN (0.3)(849N) = 255N

  15. Will the block move down the ramp? • Is there an unbalanced force parallel with the ramp? • What wins? • Fװ • By how much? • FNET = Fװ– FF 490N – 255N 235N

  16. If there is a net force, is there acceleration? • Yes • What is the rate of acceleration down the ramp? • a = FNET/m 235N /100kg 2.35 m/s2

  17. What else could I determine? • If the object started from rest and I knew the length of the ramp, how would I determine vf • Vf2 =vi2 + 2ad • Could I then determine time?

  18. What my family does for Thanksgiving……

  19. I caught an alligator!!!!! I need to get it in my “Critter Gitter” but I do not want to push it up the ramp. The gator might bite!!! I will attach a cable and use a winch to pull it up………………

  20. Example 15 • Given: 500 kg mass being pulled upward by a cable Inclined plane 12o Constant speed 3.0m/s upward FF = 460.0 N Make the free body diagram and determine the FNET

  21. I caught an alligator!!!!! Example 6 The alligator (500 kg) is pulled up a ramp into a “Critter Gitter” by a cable at a constant velocity of 3.0 m/s. The ramp makes an angle of 12.00 with the horizontal. The frictional force is 460.0 N.

  22. I caught an alligator!!!!! What direction is the frictional force of 460.0 N as the alligator moves up? Is there any other force working down the ramp?

  23. FT FN 500kg FF F II FW 12o

  24. Our ultimate question: What is the force applied by the cable and what would happen if the cable broke? I

  25. What is Fw (500 kg)(9.8 m/s2) 4900 N What is F┴ (4900 N)(cos12) 4790 N FN = F┴=4790 N What is Fװ (4900 N)(sin12) 1020 N What is FF 460 N (given) What is the total force opposing upward motion? Fװ+FF Is there a net force? No What is the force applied by winch? 1020 N + 460 N = 1480N

  26. If the cable breaks, what is the acceleration down the ramp? • Determine FNET • FNET = Fװ – FF • FNET = 1020 N - 460 N • FNET = 560 N • Determine a • a = FNET/m • a = 560 N/500 kg • a = 1.12m/s2 • FRICTION OPPOSES MOTION IN BOTH DIRECTIONS!!!

  27. What is the coefficient of friction between the ramp and the alligator? • FF/FN = 460 N/4790N = 0.096

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