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Forces in 2 Dimensions

Forces in 2 Dimensions. There are two types of problems One type is when the applied force is at an angle with the surface The other type is when the object is on an inclined plane (“ramp problems”. Forces in 2 dimensions. F f. F A. 1. This is what we did last chapter. F A. 2. F f. θ.

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Forces in 2 Dimensions

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  1. Forces in 2 Dimensions • There are two types of problems • One type is when the applied force is at an angle with the surface • The other type is when the object is on an inclined plane (“ramp problems”

  2. Forces in 2 dimensions Ff FA 1. This is what we did last chapter FA 2. Ff θ Now the applied force is at an angle

  3. Forces in 2 dimensions FA Fv Ff Fh θ

  4. Forces in 2 Dimensions • The applied Force is at an angle • Only the horizontal component is used to make it move forward • If the object is being pulled, the Normal force is decreased by the vertical component. • If the object is being pushed, the Normal force is increased by the vertical component.

  5. Forces in 2 dimensions FN Fv Fv FW The Normal force is equal to the weight minus the vertical component

  6. Forces in 2 dimensions FN Fv FN Fv Pulling FW FW FN = FW FN = FW - FV

  7. Forces in 2 dimensions FN FN Fv Pushing Fv FW FW FN = FW FN = FW + FV

  8. Forces in 2 dimensions You need to find both the horizontal and vertical component by using the following formulas: Cos θ = Fh Sin θ = Fv FA FA FA Fv Ff θ Fh

  9. Forces in 2 dimensions A sled is pulled with a force of 45 N and makes an angle of 35º with the surface. If the sled has a mass 10 kg, what is it’s acceleration if the µ is 0.30 FA FV Ff θ Fh

  10. Forces in 2 dimensions First you have to determine the horizontal and vertical components. Cos 35° = Fh Sin 35° = Fv 45 N 45N Fh = 37 N Fv = 26 N

  11. Forces in 2 dimensions Next you need to calculate the force of friction Ff = µFN FN = FW - FV = (10 kg)(9. 8m/s2) – 26 N = 72 N (If the box was just sitting there, the normal force would equal 98N) Ff = µFN Ff = (0.3)(72N) Ff = 21.6 N

  12. Forces in 2 dimensions Finally, you put it all together into a net force equation: Fnet = Fh - Ff (m)(a)= 37 N - 21.6 N (10 kg)(a) = 37 N - 21.6 N a = 1.54 m/s2

  13. Forces in 2 Dimensions The second type of problem is the inclined plane or “Ramp” problems. It easy to confuse these because they tend to look alike.

  14. Forces in 2 dimensions Inclined Plane or “The Ramp” - when box is sliding down ramp. Ff FN F II F┴ FW

  15. Forces in 2 dimensions Inclined Plane or “The Ramp” - when box is being pushed up the ramp. FA FN F 11 Ff F┴ FW

  16. Forces in 2 dimensions The forces on an inclined plane are: FF Force of Friction- always opposes FA F11 Parallel Force- Force that causes box to slide down the ramp. It always going down the ramp F ┴ Perpendicular Force – It is the force that the ramp pushes up with. It is equal but opposite to Normal Force FN Normal Force – it is not equal to the weight - it is equal to the perpendicular force Fw Weight force – always points straight down FA Applied force – when box is being slid up the ramp

  17. Forces in 2 dimensions If the object is stationary or moving at a constant speed, the following is true: FN = F ┴ (always) F11 = Ff If F11 > Ff object will accelerate F11 = Sin θ • FW F ┴ = Cos θ • FW Ff = µ F┴ (or Ff = µ FN )

  18. “The Skier” A skier goes down a mountain that makes a 30° angle with the horizontal. If he weighs 562 N and the snow has a coefficient of friction of 0.20, how fast will he accelerate? Ff FN F11 F┴ Fw

  19. “The Skier” A skier goes down a mountain that makes a 30° angle with the horizontal. If he weighs 562 N and the snow has a coefficient of friction of 0.20, how fast will he accelerate? Ff FN F11 F┴ 30° Fw = 562N

  20. First, you have to redraw the parallel force vector. Ff FN F11 30° F┴ 30° Fw = 562N F11

  21. Second, find F11 and F ┴ F11 = (Sin 30°)(562N) F ┴ = (Cos30)(562N) F11 = 281 N F ┴ = 487N Ff FN F11 30° F┴ 30° Fw = 562N F11

  22. Third, find the Force of friction Ff = µF ┴ Ff = (0.2)(487N) FF = 97.4 N Ff FN F11 30° F┴ 30° Fw = 562N F11

  23. Next, you need to know the mass. Fw = mg 562 N = m (9.8 m/s2) m = 57.3 kg Ff FN F11 30° F┴ 30° Fw = 562N F11

  24. Finally, use the net Force equation: Fnet = F11 – Ff (m)(a) = 281 N - 97.4 N 57.3 kg (a) = 183.6 N a = 3.2 m/s2 Ff FN F11 30° F┴ 30° Fw = 562N F11

  25. Forces in 2 Dimensions • There are several more types of problems dealing with this concept. • They are called “Sign” problems • The Sign problems.ppt

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