Newton's second law

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Newton's second law

  1. 1. Observations <ul><li>The acceleration of the given body is directly proportional to the applied force. This means that the ratio of the acceleration is always constant. </li></ul>— = — = — = constant F 1 F 2 F 3 a 1 a 2 a 3 The constant is a measure of how effective is the given force in producing acceleration. This ratio is the property of the body called mass . m = — F a A force of 1 newton (1N) is that resultant force that will give a 1-kg mass an acceleration of 1 m/s 2 . The newton (N) is the SI unit of force.
  2. 2. Newton’s 2 nd Law of Motion Applying a constant force of 12 N in succession to 1-, 2- and 3-kg masses will produce accelerations of 12 m/s 2 , 6 m/s 2 and 4m/s 2 , respectively. a = — — = —— = 12 m/s 2 F 12N m 1 1kg a = — — = —— = 6 m/s 2 a = — — = —— = 4 m/s 2 F 12N m 2 2 kg F 12N m 3 3kg
  3. 3. Newton’s Second Law of Motion : “Whenever an unbalanced force acts on a body, it produces in the direction of the force an acceleration that is directly proportional to the force and inversely proportional to the mass of the body.” Σ F = + F 1 + F 2 + F 3 + … Force (N) = mass (kg) x acceleration (m/s 2 ) Force (lb) = mass (slug) x acceleration (ft/s 2 ) 1 lb = 4.448 N 1 slug = 14.59 kg F net = ma
  4. 4. Illustrative Example 2 <ul><li>It is determined that a resultant force of 60 N will give a wagon an acceleration of 10 m/s 2 . What force is required to give the wagon an acceleration of 2 m/s 2 ? </li></ul>F = ma = (6kg)(2 m/s 2 ) m = — — = ——— = 6 kg F 60N a 10m/s 2 F = 12 N
  5. 5. Illustrative Example 3 <ul><li>A 1000kg car moving north at 100 km/h brakes to a stop in 50 m. What are the magnitude and direction of the force? </li></ul>Given: m = 1000kg; v i = 100 km/h = 27.8 m/s; v f = 0 Find: F net = ? Formula: F net = ma a = — —— v f – v i Δ t Δ X 50 m v 13.9 m/s Δ t = — — = ———— Δ t = 3.6 s a = — ——— 0 – 27.8 m/s 3.6 s a = – 7.72 m/s 2
  6. 6. F = ma = (1000kg)(- 7.72 m/s 2 ) F = 7720 N, South Therefore, we can summarize as follows: SI: W(N) = mkg) x g(9.8m/s 2 ) English: W(lb) = m(slug) x g(32 ft/s 2 )
  7. 7. Relationship Between Mass and Weight <ul><li>MASS is a universal constant equal to the ratio of the body’s weight to the gravitational acceleration due to gravity. </li></ul><ul><li>WEIGHT is the force of gravitational attraction and varies depending of the acceleration due to gravity. </li></ul>W = mg or m = — — W g <ul><li>The mass of a particle is equal to its weight divided by the acceleration due to gravity. </li></ul><ul><li>Weight has the same units as the unit of force. </li></ul><ul><li>The acceleration of gravity has the same units as acceleration. </li></ul>
  8. 8. Illustrative Examples <ul><li>What is the weight of a 4.8-kg mailbox? </li></ul><ul><li>What is the mass of a 40-N tank? </li></ul><ul><li>What is the mass of a 60-lb child? </li></ul><ul><li>What is the weight of a 7-slug man? </li></ul>W = mg = (4.8kg)(9.8 m/s 2 )= 47N m = F/g = (40N)/(9.8 m/s 2 )= 4.08 kg m = F/g = (60lb)/(32 ft/s 2 )= 1.9 slug W = mg = (7slug)(32 m/s 2 )= 224 lb
  9. 9. Illustrative Example 2 <ul><li>Find the weight of the body whose weight on Earth is 100 N. If this mass were taken to a distant planet where g = 2.0 m/s 2 , what would be its weight on that planet? </li></ul>Given: W E = 100N; g E = 9.80 m/s 2 g P = 2.0 m/s 2 Find: W P = ?
  10. 10. Solution <ul><li>Mass on Earth: </li></ul>m = — — = ———— = 10.2 kg W E 100 N g E 9.80 m/s 2 W P = mg P = (10.2 kg)(2 m/s 2 ) Weight on the planet: W P = 20.4 N Ans.: a = 1.63 m/s 2 ; m = 81.6 kg on both places
  11. 11. Try this! <ul><li>A woman weighs 800 N on Earth. When she walks on the moon, she weighs only 133 N. What is the acceleration due to gravity on the Moon, and what is her mass on the Moon? On the Earth? </li></ul>
  12. 12. <ul><li>A ball is accelerated from rest at a rate of 1.20 m/s 2 after a force of 20.0 n is applied. What is the mass of the ball? </li></ul>Practice Exercise (p.61-62) 2. A 15.0-kg box is pushed by two boys with forces of 15.0 N and 18.0 toward the right. What is the magnitude and direction of the acceleration of the box? 3. If the forces in number 2 are applied opposite each other, what is the magnitude and direction of the acceleration? Solution Solution Solution
  13. 13. Solution: <ul><li>A ball is accelerated from rest at a rate of 1.20 m/s 2 after a force of 20.0 n is applied. What is the mass of the ball? </li></ul>Given: F net = 20.0 N; Find: m = ? a = 1.20m/s 2 m = — — = ————— = 16.7 kg F a 20.0 N 1.20 m/s 2 F = ma
  14. 14. Solution: Given: m = 15.0 kg; F 1 = +15.0 N; F 2 = +18.0 N Find: a = ? a = — — = ———————— F net m 15.0 N + 18.0 N 15.0 kg F net = F 1 + F 2 = ma 2. A 15.0kg box is pushed by two boys with forces of 15.0 N and 18.0 toward the right. What is the magnitude and direction of the acceleration of the box? a = 2.20 m/s 2
  15. 15. Solution: Given: m = 15.0 kg; F 1 = +15.0 N; F 2 = +18.0 N Find: a = ? a = — — = ———————— F net m 15.0 N + 18.0 N 15.0 kg F net = F 1 + F 2 = ma 2. A 15.0kg box is pushed by two boys with forces of 15.0 N and 18.0 toward the right. What is the magnitude and direction of the acceleration of the box? a = 1.53 m/s 2 to the right
  16. 16. Solution: Given: m = 15.0 kg; F 1 = +15.0 N; F 2 = -18.0 N Find: a = ? a = — — = ———————— F net m 15.0 N - 18.0 N 15.0 kg F net = F 1 + F 2 = ma 3. If the forces in number 2 are applied opposite each other, what is the magnitude and direction of the acceleration? a = 0.2 m/s 2 to the left
  17. 17. The Atwood Machine m M T T M m T T Mg mg a a
  18. 18. The Atwood Machine M m T T Mg mg a a For mass M: T – Mg = - Ma Mg - T = Ma Eq. 1 For mass m: T – mg = ma Eq. 2 Conbining eq.1 & eq. 2: Mg - T = Ma T – mg = ma + Mg - mg = (M + m)a a = — —— g M – m M + m
  19. 19. Example: Atwood Machine <ul><li>A 2.0-kg body and a 5.0-kg body are each suspended at the end of a cord that passes over a frictionless pulley. (a) What is the acceleration of the system? (b) What is the tension on the cord? </li></ul>5.0 kg 2.0 kg a a T T
  20. 20. Vertical and horizontal problems m 1 m 2 A 2.00-kg hanging block pulls a 3.00-kg block along a frictionless table. Calculate for the acceleration of the system and the tension on the cord.
  21. 21. Vertical and horizontal problems m 1 m 2 m 1 m 2 T a w 2 a T F net = T = m 1 a - eq.1 F net = T- w 2 = -m 2 a or w 2 – T = m 2 a - eq.2 T T
  22. 22. <ul><li>Since the blocks are connected by a single cord, the tension T of the cord for both blocks is the same, thus similar rate of motion for both. </li></ul>T = m 1 a w 2 – T = m 2 a Combining the 2 equations gives: w 2 = m 1 a + m 2 a w 2 = (m 1 + m 2 )a a = — —————— w 2 (m 1 + m 2 ) Working equation ---- Eq. 1 ---- Eq. 2
  23. 23. a = — —————— w 2 (m 1 + m 2 ) a = — —————— (2.00kg)(9.80 m/s 2 ) (3.0 kg + 2.0 kg) a = 3.92 m/s 2 For acceleration: For the tension: from eq.1 T = m 1 a = (3.0 kg)(3.92m/s 2 )= 11.8 N Another Example
  24. 24. Example: Horizontal and Vertical motion <ul><li>A 100-g mass lies on a frictionless table and a cord is attached to it. The cord passes over a pulley at the edge of the table and at the free end, a 10-g mass is hung. Find the acceleration of the system and the tension on the cord. </li></ul>100 g 10 g
  25. 25. Solution a = — ————— w 2 (m 1 + m 2 ) a = — ——————— 10g (980cm/s 2 ) (100g + 10g) a = 89.1 cm/s 2 The acceleration: The tension: Isolating the 10-g mass, the tension T acts upward and the acceleration downward. The unbalanced force F = (9800 – T)dynes. Applying the second law: F = ma 9800-T = 10g(89.1cm/s 2 ) T = 9800dy – 891dy T = 8910 dynes <ul><li>A 100-g mass lies on a frictionless table and a cord is attached to it. The cord passes over a pulley at the edge of the table and at the free end, a 10-g mass is hung. Find the acceleration of the system and the tension on the cord. </li></ul>
  26. 26. m 1 m 2 m 3 T 1 T 2 T 2 T 1 Vertical and horizontal problems If a 100-g counter weight is attached on the left side of m 1 which is 300 g, and a 200-g mass on the right, what would be the acceleration of the system and the tensions on the left and right cords?
  27. 27. m 1 T 1 T 2 Free-body Diagrams m 2 m 3 T 2 T 1 w 3 w 3 F net = T 1 – T 2 = m 1 a F net = T 2 – W 2 = m 2 a F net = T 1 – W 3 = -m 3 a a a m 1 = 300 g m 2 = 100g m 3 = 200 g
  28. 28. <ul><li>Combining the 3 equations yields: </li></ul>T 1 – T 2 = m 1 a T 2 – W 2 = m 2 a W 3 – T 1 = m 3 a W 3 – W 2 = (m 1 + m 2 + m 3 )a a = — ————— W 3 – W 2 m 1 + m 2 + m 3 + + Working equation F net2 = T 2 – W 2 = m 2 a F net1 = T 1 – T 2 = m 1 a F net3 = T 1 – W 3 = -m 3 a
  29. 29. a = — ———— = ————————————— W 3 – W 2 m 1 + m 2 + m 3 a = 1.63 m/s 2 (0.2kg)(9.8m/s 2 ) – (0.1kg)(9.8m/s 2 ) 0.3 kg + 0.2 kg + 0.1 kg m 1 = 300 g m 2 = 100g m 3 = 200 g
  30. 30. m 1 = 300 g m 2 = 100g m 3 = 200 g T 1 – T 2 = m 1 a –Eq.1 T 2 – W 2 = m 2 a –Eq.2 W 3 – T 1 = m 3 a –Eq.3 For Tensions, T 1 and T 2 W 3 – T 1 = m 3 a m 3 g – T 1 = m 3 a T 1 = m 3 g – m 3 a T 1 = m 3 (g – a) T 1 = 0.2kg(9.8m/s 2 – 1.63m/s 2 ) Using Equation 3 T 1 = 1.63 N
  31. 31. Using Equation 2 T 2 – W 2 = m 2 a T 2 – m 2 g = m 2 a T 2 = m 2 a + m 2 g T 2 = m 2 (g – a) T 2 = 0.1kg(9.8m/s 2 + 1.63m/s 2 ) T 2 = 1.14 N
  32. 32. Using Equation 1 T 1 – T 2 = m 1 a 1.63N – T 2 = 0.3kg(1.63m/s 2 ) 1.63N – T 2 = 0.489 N T 2 = 1.63N – 0.489 N T 2 = 1.14 N T 1 = 1.63 N
  33. 33. Solution a = — —— g M – m M + m a = — ———— (9.8m/s 2 ) 5kg – 2kg 5kg + 2kg a = 4.2 m/s 2
  34. 34. Another Solution 5.0 kg 2.0 kg a a F = ma Unbalanced force F : F = W 2 + W 1 = 5kg(9.8m/s 2 ) – 2kg(9.8m/s 2 ) = 49 N – 19.6 N = 29.4N The moving masses : m = 2kg + 5kg = 7kg T T The acceleration: F = ma 29.4N = 7kg (a) a = 29.4 N / 7kg = 4.2 m/s 2 w 1 w 2
  35. 35. Solution: The tension on the cord Consider the 2.0-kg body as the moving part of the system. We can isolate it as shown in the free-body diagram on the right . Let T be the tension on the cord. The unbalanced force is T – W 1 . The acceleration is 4.2 m/s 2 . F = ma T – 19.6N = (2.0kg)(4.2m/s 2 ) 2.0 kg T a T = 8.4 N + 19.6 N = 28.0 N

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