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Applications & Examples of Newton’s Laws

Applications & Examples of Newton’s Laws. Forces are VECTORS !! Newton’s 2 nd Law: ∑ F = ma ∑ F = VECTOR SUM of all forces on mass m  Need VECTOR addition to add forces in the 2 nd Law! Forces add according to rules of VECTOR ADDITION ! (Ch. 3). Newton’s 2 nd Law problems:

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Applications & Examples of Newton’s Laws

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  1. Applications & Examples of Newton’s Laws

  2. Forces are VECTORS!! • Newton’s 2nd Law: ∑F = ma ∑F = VECTORSUMof all forces on mass m  Need VECTORaddition to add forces in the 2nd Law! • Forces add according to rules of VECTOR ADDITION! (Ch. 3)

  3. Newton’s 2nd Law problems: • STEP 1:Sketch the situation!! • Draw a “Free Body” diagram for EACHbody in problem & draw ALL forces acting on it. • Part of your grade on exam & quiz problems! • STEP 2:Resolve the forces on each body into components • Use a convenient choice of x,y axes • Use the rules for finding vector components from Ch. 3.

  4. STEP 3: Apply Newton’s 2nd Law to EACH BODY SEPARATELY: ∑F = ma • A SEPARATE equation like this for each body! • Resolved into components: ∑Fx = max ∑Fy = may Notice that this is theLASTstep, NOTthe first!

  5. Conceptual Example Moving at constant v, withNO friction, which free body diagram is correct?

  6. ExampleParticle in Equilibrium “Equilibrium” ≡The total force is zero. ∑F = 0 or ∑Fx = 0 & ∑Fy = 0 Example (a) Hanging lamp (massless chain). (b) Free body diagram for lamp. ∑Fy = 0  T – Fg = 0; T = Fg = mg (c) Free body diagram for chain. ∑Fy = 0  T – T´ = 0; T´ = T = mg

  7. ExampleParticle Under a Net Force Example (a) Crate being pulled to right across a floor. (b) Free body diagram for crate. ∑Fx = T = max ax= (T/m) ay= 0, because of no vertical motion. ∑Fy = 0  n – Fg = 0; n = Fg = mg

  8. ExampleNormal Force Again “Normal Force” ≡When a mass is in contact with a surface, the Normal Force n = force perpendicular to (normal to) the surface acting on the mass. Example Book on a table. Hand pushing down. Book free body diagram.  ay= 0, because of no vertical motion (equilibrium). ∑Fy = 0  n – Fg - F = 0  n = Fg + F = mg + F Showing again that the normal force is not always = & opposite to the weight!!

  9. Example A box of mass m = 10 kg is pulled by an attached cord along a horizontal smooth (frictionless!) surface of a table. The force exerted is FP = 40.0 N at a 30.0° angle as shown. Calculate: a. The acceleration of the box. b. The magnitude of the upward normal force FN exerted by the table on the box. Free Body Diagram The normal force, FN is NOT always equal & opposite to the weight!!

  10. Example Two boxes are connected by a lightweight (massless!) cord & are resting on a smooth (frictionless!) table. The masses are mA = 10 kg & mB = 12 kg. A horizontal force FP = 40 N is applied to mA. Calculate:a. The acceleration of the boxes. b. The tension in the cord connecting the boxes. Free Body Diagrams

  11. Example 5.4: Traffic Light at Equilibrium (a) Traffic Light, Fg = mg = 122 N hangs from a cable, fastened to a support. Upper cables are weaker than vertical one. Will break if tension exceeds 100 N. Does light fall or stay hanging? (b) Free body diagram for light. ay= 0, no vertical motion. ∑Fy = 0  T3 – Fg = 0 T3 = Fg = mg = 122 N (c) Free body diagram for cable junction (zero mass).T1x = -T1cos(37°), T1y = T1sin(37°) T2x = T2cos(53°), T2y = T2sin(53°), ax= ay= 0. Unknowns are T1 & T2. ∑Fx = 0  T1x + T2x = 0 or -T1cos(37°) + T2cos(53°) = 0 (1) ∑Fy = 0  T1y + T2y – T3 = 0 orT1sin(37°) + T2sin(53°) – 122 N = 0 (2) (1) & (2) are 2 equations, 2 unknowns.Algebrais required to solve for T1& T2! Solution: T1 = 73.4 N, T2 = 97.4 N

  12. Example 5.6: Runaway Car

  13. Example 5.7: One Block Pushes Another

  14. Example 5.8: Weighing a Fish in an Elevator

  15. Example 5.9: Atwood Machine

  16. Example 4-13 (“Atwood’s Machine”) Two masses suspended over a (massless frictionless) pulley by a flexible (massless) cable is an “Atwood’s machine”. Example: elevator & counterweight. Figure: Counterweight mC = 1000 kg. Elevator mE = 1150 kg. Calculate a. The elevator’s acceleration. b. The tension in the cable. a   a aE = - a Free Body Diagrams aC = a

  17. Conceptual Example Advantage of a Pulley A mover is trying to lift a piano (slowly) up to a second-story apartment. He uses a rope looped over 2 pulleys. What force must he exert on the rope to slowly lift the piano’s mg = 2000 N weight? mg = 2000 N Free Body Diagram

  18. Example: Accelerometer A small mass m hangs from a thin string & can swing like a pendulum. You attach it above the window of your car as shown. What angle does the string make a.When the car accelerates at a constanta = 1.20 m/s2? b.When the car moves at constant velocity,v = 90 km/h? Free Body Diagram

  19. Example = 300 N Free Body Diagram FT1x = -FTcosθ FT1y = -FTsinθ FT2x = FTcosθ FT2y = -FTsinθ

  20. Inclined Plane Problems The tilted coordinate System isconvenient, but not necessary. Engineers & scientists MUSTunderstand these! Understand∑F = ma& how to resolve it into x,y components in the tilted coordinate system!! a

  21. Example: Sliding Down An Incline A box of mass m is placed on a smooth (frictionless!) incline that makes an angle θ with the horizontal. Calculate: a. The normal force on the box. b. The box’s acceleration. c. Evaluate both for m = 10 kg & θ = 30º Free Body Diagram

  22. Example 5.10Inclined Plane, 2 Connected Objects

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