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Friction

Friction. Friction - the force needed to drag one object across another. (at a constant velocity). Depends on: How hard the surfaces are held together (demo) What type of surface it is (I.e. rough, smooth). Not supposed to depend on: Surface area (pressure) Speed (low speeds) (demo).

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Friction

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  1. Friction

  2. Friction - the force needed to drag one object across another. (at a constant velocity) • Depends on: • How hard the surfaces are held together (demo) • What type of surface it is (I.e. rough, smooth) • Not supposed to depend on: • Surface area (pressure) • Speed (low speeds) (demo)

  3. Force of Friction in N Normal Force - Force exerted by a surface to maintain its integrity Coefficient of Friction. 0 < < 1 (Specific to a surface) - in your book (Table 4-2) Usually the weight (level surfaces) FFr = FN

  4. Page 97 These are approximate What’s up with the two columns? How are they related (one bigger)

  5. Kinetic Friction - Force needed to keep it going at a constant velocity. • FFr = kFN • Always in opposition to velocity • (Demo, example calculation) • Static Friction - Force needed to start motion. • FFr<sFN • Keeps the object from moving if it can. • Only relevant when object is stationary. • Always in opposition to applied force. • Calculated value is a maximum • (Demo, example calculation, examples of less than maximum)

  6. Whiteboards: Friction 1 | 2 | 3 | 4 | 5 | 6 TOC

  7. What is the force needed to drag a 12 kg chunk of rubber at a constant velocity across dry concrete? F = ma, FFr = kFN FN = weight = mg = (12 kg)(9.8 N/kg) = 117.6 NFFr = kFN = (.8)(117.6 N) = 94.08 N = 90 N W 90 N

  8. What is the force needed to start a 150 kg cart sliding across wet concrete from rest if the wheels are locked up? F = ma, FFr<sFN FN = weight = mg = (150 kg)(9.8 N/kg) = 1470 NFFr = kFN = (.7)(1470) = 1029 N = 1000 N W 1000 N

  9. v 72 N FFr 8.5 kg s = .62, k = .48 What is the acceleration if there is a force of 72 N in the direction an 8.5 kg block is already sliding? FFr = kFN, FN = mg, FFr = kmg FFr = (.48)(8.5 kg)(9.8 N/kg) = 39.984 N F = ma <72 N - 39.984 N> = (8.5 kg)a, a = 3.77 = 3.8 ms-2 W 3.8 m/s/s

  10. v=12m/s FFr 22 kg s = .62, k = .48 A 22 kg block is sliding on a level surface initially at 12 m/s. What time to stop? FFr = kFN, FN = mg, FFr = kmg FFr = (.48)(22 kg)(9.8 N/kg) = 103.488 N F = ma < -103.488 N> = (22 kg)a, a = -4.704 ms-2 v = u + at, v = 0, u = 12, a = -4.704 ms-2, t = 2.55 s = 2.6 s W 2.6 s

  11. v a = 3.2 ms-2 F = ? FFr 6.5 kg s = .62, k = .48 A 6.5 kg box accelerates and moves to the right at 3.2 m/s/s, what force must be applied? FFr = kFN, m = 6.5 kg FFr = (.48)(6.5 kg)(9.8 N/kg) = 30.576 N F = ma < F - 30.576 N > = (6.5 kg)(3.2 ms-2), F = 51.376 = 51 N W 51 N

  12. v=12m/s F = ? FFr 22 kg s = .62, k = .48 A 22 kg block is sliding on a level surface initially at 12 m/s stops in 2.1 seconds. What external force is acting on it besides friction?? v = u + at, v = 0, u = 12 m/s, t = 2.1 s, a = -5.7143 ms-2 FFr = kFN, FN = mg, FFr = kmg FFr = (.48)(22 kg)(9.8 N/kg) = 103.488 N F = ma < -103.488 N + F> = (22 kg)(-5.7143 ms-2), F = -22 N (left) W -22 N (to the left)

  13. a = 1.2 ms-2 v 35 N FFr m s = .62, k = .48 A force of 35 N in the direction of motion accelerates a block at 1.2 m/s/s in the same direction What is the mass of the block? FFr = kFN, FN = mg, FFr = kmg FFr = (.48)m(9.8 ms-2) = m(4.704 ms-2) F = ma < 35 N - m(4.704 ms-2) > = m(1.2 ms-2) 35 N = m(4.704 ms-2) + m(1.2 ms-2) = m(4.704 ms-2 + 1.2 ms-2) m = (35 N)/(5.904 ms-2) = 5.928 kg = 5.9 kg W 5.9 kg

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