Net Force Contents What is the Net force

Net Force Contents: • What is the Net force • Using Newton’s Second law with more than one force • Whiteboard Net Force • Applying weight • Whiteboards with weight

Net Force In F = ma m = mass a = acceleration F = <The vector sum of all the forces> TOC

Net Force – Example 1 Finding acceleration 9. 0 N 17. 0 N 5. 0 kg F = ma Making to the right + <+17. 0 N – 9. 0 N> = (5. 0 kg)a 8. 0 N = (5. 0 kg)a a = (8. 0 N)/(5. 0 kg) = 1. 6 m/s/s TOC

Net Force – Example 2 Finding an unknown force Some other force is acting on the block F = ? ? 450. N 35. 0 kg a = 9. 0 m/s/s F = ma Making to the right + <+450. N + F> = (35. 0 kg)(+9. 0 m/s/s) 450. N + F = 315 N - 450. N = -135 N (to the left) TOC

Whiteboards: Net Force 1 1|2|3|4 TOC

Find the acceleration: 3. 0 N 7. 0 N 5. 0 kg F = ma Making to the right + <7. 0 N – 3. 0 N> = (5. 0 kg)a 4. 0 N = (5. 0 kg)a a =. 80 m/s/s W

Find the acceleration: 3. 0 N 5. 0 N 23. 0 kg 6. 0 N F = ma <5. 0 N – 3. 0 N – 6. 0 N> = (23. 0 kg)a -4. 0 N = (23. 0 kg)a a = -. 1739 = -. 17 m/s/s W

Find the other force: F = ? ? 452 kg 67. 3 N a =. 12 m/s/s F = ma <67. 3 N + F> = (452 kg)(. 12 m/s/s) <67. 3 N + F> = 54. 24 N F = 54. 24 N - 67. 3 N F = -13. 06 = -13 N W

Find the other force: F ? ? ? 2100 kg 580 N 125 N a =. 15 m/s/s F = ma <580 N - 125 N + F> = (2100 kg)(-. 15 m/s/s) 455 N + F = -315 N F = -770 N (To the LEFT) -770 N W

Net Force – Example 3 Using Weight 35 N Find the acceleration (on Earth) 5. 0 kg TOC

Net Force – Example 3 Using Weight Draw a Free Body Diagram: Don’t Forget the weight: F = ma = 5. 0*9. 8 = 49 N 35 N 5. 0 kg -49 N TOC

Net Force – Example 3 Using Weight F = ma 35 N – 49 N = (5. 0 kg)a -14 N = (5. 0 kg)a a = -2. 8 m/s/s 35 N 5. 0 kg -49 N TOC

Whiteboards: Using Weight 1|2|3|4|5 TOC

Find the acceleration: F = ma, weight = (8. 0 kg)(9. 80 N/kg) = 78. 4 N down Making up + <100. N - 78. 4> = (8. 0 kg)a 21. 6 N = (8. 0 kg)a a = 2. 7 m/s/s 100. N 8. 0 kg W

Find the acceleration: F = ma, wt = (15. 0 kg)(9. 8 N/kg) = 147 N down <120. N - 147 N> = (15. 0 kg)a -27 N = (15. 0 kg)a a = -1. 8 m/s/s It accelerates down -1. 8 m/s/s 120. N 15. 0 kg W

Find the force: F = ma, wt = (16 kg)(9. 8 N/kg) = 156. 8 N down <F – 156. 8 N> = (16. 0 kg)(+1. 5 m/s/s) F – 156. 8 N = 24 N F = 180. 8 N = 180 N F 16 kg a = 1. 5 m/s/s (upward) 180 N W

Find the force: F = ma, wt = 1176 N downward <F – 1176 N> = (120. kg)(-4. 50 m/s/s) F – 1176 N = -540 N F = 636 N F 120. kg a = -4. 50 m/s/s (DOWNWARD) 636 N W

A falling 52. 0 kg rock climber hits the end of the rope going 13. 5 m/s, and is stopped in a distance of 4. 20 m. What was the average force exerted to stop them? F = ma, a = +21. 696 m/s/s (from kinematics) wt = 509. 6 N downward <F – 509. 6 N> = (52. 0 kg)(+21. 696 m/s/s) F = 1637. 8 N +1640 N F 52. 0 kg W

A 65. 0 kg dumbwaiter is going up at 5. 80 m/s and is brought to rest in a distance of 1. 90 m What is the tension in the cable supporting it as it is stopping? F = ma, a = -8. 853 m/s/s (from kinematics) wt = 637 downward <F – 637 N> = (65. 0 kg)(-8. 853 m/s/s) F = 61. 58 N +61. 6 N F 65. 0 kg W

Find the force: F = ma, wt = 1176 N downward <F – 1176 N> = (120. kg)(-4. 50 m/s/s) F – 1176 N = -540 N F = 636 N • Relationship between tension, weight and acceleration • Accelerating up = more than weight (demo, elevators) • Accelerating down = less than weight (demo, elevators, acceleration vs velocity) • Climbing ropes F 120. kg a = -4. 50 m/s/s (DOWNWARD) W

Find the mass: F = ma, wt = m(9. 80 m/s/s) downward <13. 6 – m(9. 80 m/s/s)> = m(-1. 12 m/s/s) 13. 6 N = m(9. 80 m/s/s) - m(1. 12 m/s/s) 13. 6 N = m(9. 80 m/s/s-1. 12 m/s/s) 13. 6 N = m(8. 68 m/s/s) m = 1. 5668 kg = 1. 57 kg 13. 6 N m a = 1. 12 m/s/s (downward) 1. 57 kg W