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Net Force Contents: What is the Net force Using Newton’s Second law with more than one 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.

jeremy-potts
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Net Force Contents: What is the Net force Using Newton’s Second law with more than one force

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  1. 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

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

  3. Net Force – Example 1 Finding acceleration 17.0 N 9.0 N 5.0 kg F = ma Making to the right + <+17.0 N – 9.0 N> = (5.0kg)a 8.0 N = (5.0kg)a a = (8.0 N)/(5.0kg) = 1.6 m/s/s TOC

  4. Net Force – Example 2 Finding an unknown force F = ?? 450. N 35.0 kg Some other force is acting on the block a = 9.0 m/s/s F = ma Making to the right + <+450. N + F> = (35.0kg)(+9.0 m/s/s) 450. N + F = 315 N F = 315 N - 450. N = -135 N (to the left) TOC

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

  6. Find the acceleration: 7.0 N 3.0 N 5.0 kg F = ma Making to the right + <7.0 N – 3.0 N> = (5.0kg)a 4.0 N = (5.0kg)a a = .80 m/s/s W .80 m/s/s

  7. Find the acceleration: 5.0 N 3.0 N 23.0 kg 6.0 N F = ma <5.0 N – 3.0 N – 6.0 N> = (23.0kg)a -4.0 N = (23.0kg)a a = -.1739 = -.17 m/s/s W -.17 m/s/s

  8. 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 -13 N

  9. 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) W -770 N

  10. Net Force – Example 3 Using Weight 35 N Find the acceleration (on Earth) 5.0 kg TOC

  11. 35 N -49 N Net Force – Example 3 Using Weight 5.0 kg Draw a Free Body Diagram: Don’t Forget the weight: F = ma = 5.0*9.8 = 49 N TOC

  12. 35 N -49 N Net Force – Example 3 Using Weight 5.0 kg F = ma 35 N – 49 N = (5.0 kg)a -14 N = (5.0 kg)a a = -2.8 m/s/s TOC

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

  14. Find the acceleration: F = ma, weight = (8.0 kg)(9.80 N/kg) = 78.4 N down Making up + <100. N - 78.4> = (8.0kg)a 21.6 N = (8.0kg)a a = 2.7 m/s/s 100. N 8.0 kg W 2.7 m/s/s

  15. Find the acceleration: F = ma, wt = (15.0 kg)(9.8 N/kg) = 147 N down <120. N - 147 N> = (15.0kg)a -27 N = (15.0kg)a a = -1.8 m/s/s It accelerates down 120. N 15.0 kg W -1.8 m/s/s

  16. 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) W 180 N

  17. 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) W 636 N

  18. 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 • 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 a = -4.50 m/s/s (DOWNWARD) W

  19. 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) W 1.57 kg

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