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Dynamics: Intro & Application of Newton’s Laws

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Presentation on theme: "Dynamics: Intro & Application of Newton’s Laws"— Presentation transcript:

1 Dynamics: Intro & Application of Newton’s Laws

2

3 Drawing Free-Body-Diagrams
AP Physics Development Committee May 2010 – New Wording for free-body or force diagrams. (p. 149 C&E) Students will be directed to "draw and label the forces (not components) that act on the [object]," where [object] is replaced by a reference specific to the question, such as "the car when it reaches the top of the hill." Any components that are included in the diagram will be scored in the same way as incorrect or extraneous forces.

4 Drawing Free-Body-Diagrams
In addition, in any subsequent part asking for a solution that would typically make use of the diagram, the following will be included. "If you need to draw anything other than what you have shown in part [x] to assist in your solution, use the space below. Do NOT add anything to the figure in part [x]." This will give students the opportunity to construct a working diagram showing any components that are appropriate to the solution of the problem. This second diagram will not be graded.

5 Drawing Free-Body-Diagrams
Binder pp

6 Force = Interaction Force: interaction between an agent and an object causing a push or pull Two kinds of forces 1. Contact 2. “Non-Contact” (long-range field forces due to gravitational, magnetic, and/or electric fields)

7 Force Types Contact Supportive (normal or ⊥) N Tension (rope/chain) T
Friction or Drag (always oppose motion) f Other push or pull P Non-Contact Gravitational G

8 Identify the interactions
System Schema block Identify the interactions table earth

9 label the interaction types N G
System Schema block label the interaction types N G table earth G

10 Dot around the system of interest N G
System Schema block Dot around the system of interest N G table earth G

11 You are only interested in the forces that cross the dotted line! G
System Schema block N You are only interested in the forces that cross the dotted line! G table earth G

12 Agent/Object Notation
Type of force Agent that produces the force. “Dealer” Object the force acts on. “Feeler” If the agent can't be identified, the force doesn't exist!

13 Constant Velocity FN T/B block N G FG E/B table earth G

14 Note: the velocity vector does NOT touch the dot.
Constant Velocity FN T/B When the object is moving, include a velocity vector off to the side v FG E/B Note: the velocity vector does NOT touch the dot.

15 Changing velocity FN T/B v block Ff T/B N f G FG E/B table earth G

16 Note: the acceleration vector does NOT touch the dot.
Changing velocity Ff T/B v FG E/B FN T/B a Note: the acceleration vector does NOT touch the dot.

17 Non Perpendicular Forces
Object slides without friction FN R/B v block N G FG E/B ramp earth G

18 Non Perpendicular Forces
Ff Ramp ll/B = f FN Ramp ⊥/B = FN FG E/B= Wt Another form of A/O notation Components should not appear on the FBD!!

19 Unambiguous Force Labeling
FT Rope1/B = T1 FG E/B= Wt FT Rope2/B = T2

20 Forces – Relative lengths
FG E/B = mg FT R/B= T θ Ff Tll/B = f FN T⊥/B = FN v

21 Ambiguity in HW is OK Ff A/B= D FG E/B= mg v FG E/B= mg v OR

22 Object at rest or moving with constant velocity. ΣF = 0 (Equilibrium)
NEWTON'S LAWS FIRST LAW Object at rest or moving with constant velocity. ΣF = 0 (Equilibrium) Isaac Newton ( ) Vectors should be written in component form: ΣFx = 0 ΣFy = 0

23 2005B2 2005 B2. A simple pendulum consists of a bob of mass 1.8 kg attached to a string of length 2.3 m. The pendulum is held at an angle of 30° from the vertical by a light horizontal string attached to a wall, as shown. a. Draw a free‑body diagram labeling the forces on the bob in the position shown.

24 FT s2/B=T2 FT s1/B = T1 FG E/B= mg

25 ΣFH = T2 – T1 cos 60º = 0 ΣFV = T1 sin 60º - mg = 0 T2 = 10.18 N
b. Calculate the tension in the horizontal string. T2 T1 mg 60° ΣFH = T2 – T1 cos 60º = 0 ΣFV = T1 sin 60º - mg = 0 T2 = N

26 c. The horizontal string is now cut close to the bob, and the pendulum swings down.
Calculate the speed of the bob at its lowest position. L h h = L - Lcosθ = 2.5 m/s

27 The End (for now) 27


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