Friction Friction the force needed to drag one

Friction

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

Ff = μR Force of Friction in N Coefficient of Friction. 0<μ<1 (Specific to a surface) - in your book (Table 4 -2) Reaction Force - Force exerted by a surface to maintain its integrity (FN) (R) Usually the weight (level surfaces)

IB calls kinetic “dynamic” These are approximate What’s up with the two columns? How are they related (one bigger = demo with force probe)

Dynamic Friction - Force needed to keep it going at a constant velocity. (AKA Kinetic friction) FF = μd. R Always in opposition to velocity (direction it is sliding) (Demo, example calculation) Static Friction - Force needed to start motion. F F < μs. R 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)

A 5. 00 kg block rests on a level table where there is a static coefficient of friction of 0. 470, and a dynamic of 0. 170. a) What are the dynamic and maximum static forces of friction? a) 23. 1 N, 8. 34 N; b) Nada; c) 5. 33 m/s/s; d) -3. 17 m/s/s, e) +3. 59 N, f) -2. 46 N

A 5. 00 kg block rests on a level table where there is a static coefficient of friction of 0. 470, and a dynamic of 0. 170. b) if it is at rest and you exert a force of 12. 0 N sideways on it what happens? a) 23. 1 N, 8. 34 N; b) Nada; c) 5. 33 m/s/s; d) -3. 17 m/s/s, e) +3. 59 N, f) -2. 46 N

A 5. 00 kg block rests on a level table where there is a static coefficient of friction of 0. 470, and a dynamic of 0. 170. c) if it is at rest and you exert a force of 35. 0 N to the right on it, what is the acceleration of the block? a) 23. 1 N, 8. 34 N; b) Nada; c) 5. 33 m/s/s; d) -3. 17 m/s/s, e) +3. 59 N, f) -2. 46 N

A 5. 00 kg block rests on a level table where there is a static coefficient of friction of 0. 470, and a dynamic of 0. 170. d) If it is sliding to the right and you exert a force of 7. 50 N to the left, what is the acceleration of the block? a) 23. 1 N, 8. 34 N; b) Nada; c) 5. 33 m/s/s; d) -3. 17 m/s/s, e) +3. 59 N, f) -2. 46 N

A 5. 00 kg block rests on a level table where there is a static coefficient of friction of 0. 470, and a dynamic of 0. 170. e) If it is sliding to the right, but decelerating at 0. 950 m/s/s, what force is acting on the block? a) 23. 1 N, 8. 34 N; b) Nada; c) 5. 33 m/s/s; d) -3. 17 m/s/s, e) +3. 59 N, f) -2. 46 N

A 5. 00 kg block rests on a level table where there is a static coefficient of friction of 0. 470, and a dynamic of 0. 170. f) It is sliding to the right at 7. 20 m/s and it slides to a halt in a distance of 12. 0 m. What other force besides friction is exerted on the block as it slides to a halt? a) 23. 1 N, 8. 34 N; b) Nada; c) 5. 33 m/s/s; d) -3. 17 m/s/s, e) +3. 59 N, f) -2. 46 N

Whiteboards: Friction a-k

A 6. 00 kg block rests on a level counter where the static coefficient of friction is 0. 620, and the dynamic is 0. 310. a) Calculate the dynamic and maximum static friction between the block and the counter. (Store these in your calculator for later use) 6. 00 kg 36. 4932 ≈ 36. 5 N, 18. 2466 ≈ 18. 2 N W

m = 6. 00 kg, μs = 0. 620, μd = 0. 310 (36. 4932, 18. 2466) b) if the block is initially at rest, and you exert a force of 30. 0 N to the right, does the block move? Draw a force diagram with four arrows – weight, R, applied force, and static friction force. Use longer arrows for bigger forces, R = 58. 9 N indicate congruencies. Static Friction (-30. 0 N) 6. 00 kg Applied (30. 0 N) Weight = -58. 9 N yup nothin to see here

m = 6. 00 kg, μs = 0. 620, μd = 0. 310 (36. 4932, 18. 2466) c) Suppose the block is sliding to the right, and there is a force of 25. 0 N to the right, what is the acceleration of the block? 6. 00 kg +1. 13 m/s/s W

m = 6. 00 kg, μs = 0. 620, μd = 0. 310 (36. 4932, 18. 2466) d) If the block is sliding to the right, and there is a force of 12. 0 N to the left, what is the acceleration of the block? 6. 00 kg -5. 04 m/s/s W

m = 6. 00 kg, μs = 0. 620, μd = 0. 310 (36. 4932, 18. 2466) e) If the block is sliding to the left and there is a force of 45. 0 N to the right, what is the acceleration? 6. 00 kg +10. 5 m/s/s W

m = 6. 00 kg, μs = 0. 620, μd = 0. 310 (36. 4932, 18. 2466) f) What applied force would make the block slide to the right and accelerate to the right at 5. 60 m/s/s? 6. 00 kg +51. 8 N W

m = 6. 00 kg, μs = 0. 620, μd = 0. 310 (36. 4932, 18. 2466) g) What force is being applied to the block if it is sliding to the right but decelerating at 2. 40 m/s/s? 6. 00 kg +3. 85 N W

m = 6. 00 kg, μs = 0. 620, μd = 0. 310 (36. 4932, 18. 2466) h) What force is being applied to the block if it is sliding to the left, but decelerating at 5. 20 m/s/s? 6. 00 kg +13. 0 N W

m = 6. 00 kg, μs = 0. 620, μd = 0. 310 (36. 4932, 18. 2466) i) Suppose the block is sliding freely to the right at 8. 90 m/s. In what time will the block be brought to a halt? 6. 00 kg 2. 93 s W

m = 6. 00 kg, μs = 0. 620, μd = 0. 310 (36. 4932, 18. 2466) j) If the block begins to slide to the right, and there is a force of 42. 0 N acting to the right, what is the displacement of the block when it reaches a speed of 15. 0 m/s? 6. 00 kg +28. 4 m W

m = 6. 00 kg, μs = 0. 620, μd = 0. 310 (36. 4932, 18. 2466) k) If the block is sliding to the right at 9. 00 m/s, and is brought to a halt in a distance of 18. 0 m, what force besides friction acted horizontally as it was stopping? 6. 00 kg +4. 75 N W

What is the force needed to drag a 12 kg chunk of rubber at a constant velocity across dry concrete? F = ma, FFr = k. FN FN = weight = mg = (12 kg)(9. 8 N/kg) = 117. 6 N FFr = k. FN = (. 8)(117. 6 N) = 94. 08 N = 90 N W

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 < μs. FN FN = weight = mg = (150 kg)(9. 8 N/kg) = 1470 N FFr = μk. FN = (. 7)(1470) = 1029 N = 1000 N W

μs =. 62, μk =. 48 A 22 kg block is sliding on a level surface initially at 12 m/s. What time to stop? v=12 m/s FFr 22 kg FFr = μk. FN, FN = mg, FFr = μkmg 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

μs =. 62, μk =. 48 A 6. 5 kg box accelerates and moves to the right at 3. 2 m/s/s, what force must be applied? v a = 3. 2 ms -2 FFr 6. 5 kg F=? FFr = μk. FN, 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

μs =. 62, μk =. 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=12 m/s FFr 22 kg F=? v = u + at, v = 0, u = 12 m/s, t = 2. 1 s, a = -5. 7143 ms-2 FFr = μk. FN, FN = mg, FFr = μkmg 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) -22 N (to the left) W

μs =. 62, μk =. 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 vis thea mass -2 the block? = 1. 2 msof FFr m 35 N FFr = μk. FN, FN = mg, FFr = μkmg 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) mkg= (35 N)/(5. 904 ms-2) = 5. 928 kg = 5. 9 kg 5. 9 W