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Newton’s Third Law

Newton’s Third Law Chapter 7 The 3 rd Law “For every force, there is an equal and opposite force.” The 3 rd Law Runner example: Does the runner push on the earth? Why does the runner move more? Does the earth move at all? Backward force for the earth Forward force for the runner

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Newton’s Third Law

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  1. Newton’s Third Law Chapter 7

  2. The 3rd Law “For every force, there is an equal and opposite force.”

  3. The 3rd Law Runner example: • Does the runner push on the earth? • Why does the runner move more? • Does the earth move at all? Backward force for the earth Forward force for the runner

  4. The 3rd Law Rocket Example: • Equal force up and down • Does the rocket move because it pushes against the earth? • If that is so, why does a rocket move in space? Fforward Fgases

  5. The 3rd Law Sled of bricks on Ice: • Would the sled move? ICE

  6. The 3rd Law Why would Cyclops be in trouble?

  7. Should Green Lantern be able to float still in space as he pushes back his opponent?

  8. Tension • Flexible cord (can only pull) • Exerts a force Ft on the object it pulls • Usually neglect the mass of the cord

  9. Tension: Example 1 Two boxes are connected by a cord as shown. They are then pulled by another short cord. Find the acceleration of each box and the tension in the cord between the boxes. 12.0 kg 10.0 kg Fp= 40.0 N

  10. First let’s find the acceleration (consider the boxes as one mass) SF = Fp (this is the only horizontal force) ma = Fp a= Fp/m = 40.0 N/(12.0 kg + 10.0 kg) a=1.82 m/s2

  11. To find the tension, let’s deal with each box one at a time SF = Fp – FT m1a= Fp – FT FT = m1a – Fp FT= [(10 kg)(1.82 m/s2) – 40 N FT= -21.8 N 10.0 kg FT FN Fp= 40.0 N m1g

  12. The second box only has a horizontal pull from the tension. SF = FT m2a2 = FT FT = (12.0 kg)(1.82 m/s2) FT = 21.8 N 12.0 kg FN FT m2g Note that the sign of the tension varies depending on which box you consider.

  13. Tension: Example 2 Calculate the acceleration of the elevator and the tension in the cable.

  14. Draw free-body diagrams for both the elevator and counterweight

  15. Set up the force equations: SF = m1a1 = FT – m1g SF = m2a2 = FT – m2g Or m1a1 = FT – m1g m2a2 = FT – m2g We have two equations, but three unknowns (FT, a1, and a2)

  16. However, since the elevator will drop the counterweight will rise: a1 = -a2 m1a1 = FT – m1g m2a2 = FT – m2g Three equations, three unknowns (FT, a1, and a2)

  17. m1a1 = FT – m1g Substitute a1 = -a2 -m1a2 = FT – m1g Solve for FT FT = m1g - m1a2 m2a2 = FT – m2g Substitute for FT m2a2 = m1g - m1a2 – m2g m2a2 + m1a2 = m1g– m2g a2(m2+m1)=g(m1-m2) a2 = g(m1-m2) = 9.8(1150-1000) = 0.68 m/s2 (m2+m1) (1000+1150)

  18. Since we have solved for a2, it makes most sense to use this equation to find FT m2a2 = FT – m2g FT = m2a2 + m2g FT = (1000 kg)(0.68m/s2) +(1000 kg)(9.8 m/s2) FT =10500 N

  19. Tension: Example 3 Mr. Fredericks uses a pulley to lift a 200 kg piano at a constant velocity. How much tension does he need to put on the rope?

  20. SF = 2FT –mg = ma a = 0 (constant vel.) 0 = 2FT –mg Rearrange FT = mg/2 Substitute FT =(200kg)(9.8m/s2)/2 FT =980 N (Note how the pulley doubles my effort force.)

  21. Tension: Example 4 A physics student gets stuck in the mud. In order to get out, she ties a rope to a tree and pushes at the midpoint (Fpush=300 N). If the care begins to budge at an angle of 5o, calculate the force of the rope pulling on the car.

  22. Note that the tension is always along the direction of the rope, and provided by the tree and the car. Since the car is just starting to budge, we will assume the sum of all the Forces is zero.

  23. SFx = 0 = FT1x – FT2x SFy = 0 = Fp - FT1y – FT2y 0 = FT1x – FT2x 0 = Fp - FT1y – FT2y Use trigonometry 0 = FT1cos5o – FT2cos 50 0 = 300N - FT1sin5o – FT2sin5o

  24. FT1cos5o = FT2cos 5o Rearrange FT1cos5o = FT2cos 5o FT1 = FT2 Substitute 0 = 300N – FT2sin5o – FT2sin5o 300N = 2FT2sin5o FT2= 300N/2sin5o = 1700 N (Note that she magnified her force almost 6 times!!!!)

  25. Friction: Example 3 In the following setup, the coefficient of kinetic friction between the box and the table is 0.20. What is the acceleration of the system? m1=5.0 kg m2=2.0 kg

  26. FN Ffr FT Box one does not move vertically, so we can just worry about the horizontal forces. Ff r= (0.20)(5.0 kg)(9.8m/s2) = 9.8 N SF=FT – Ffr m1a = FT - 9.8 N m1g

  27. FT In this case, the downward direction is considered positive since that is the direction of motion. That way a is the same for both boxes. Box two only moves vertically. SF=m2g - FT m22 = (2.0 kg)(9.8 m/s2) - FT m2a = 19.6 N - FT m2g

  28. m1a = FT - 9.8 N Two eqns, two m2a = 19.6 N – FT unknowns FT = m1a + 9.8 N m2a = 19.6 N – m1a - 9.8 N m2a + m1a = 9.8 N a = 9.8 N/(m2+m1) = 9.8 N/7.0 kg = 1.4 m/s2

  29. We can also calculate the tension: FT = m1a + 9.8 N FT = (5.0 kg)(1.4 m/s2) + 9.8 N FT = 17 N

  30. Tension: Ex. 4 A 90.0 kg mountain climber climbs from the ropes as shown. The maximum tension that rope 3 can hold is 1500 N before it breaks. Calculate the maximum angle of q. Rope 1 Rope 3 q Rope 2

  31. Tension: Ex. 5 A 200 kg stage set is lifted down by a 100 kg stagehand as shown. Calculate the stagehand’s acceleration. Ans: 3.27 m/s2 200 kg 100 kg

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