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Chapter 5 The Laws of Motion EXAMPLES

Chapter 5 The Laws of Motion EXAMPLES. Example 1: Action-Reaction. The force exerted BY object 1 ON object 2 is equal in magnitude and opposite in direction to exerted BY object 2 ON object 1. Example 2: Action-Reaction.

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Chapter 5 The Laws of Motion EXAMPLES

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  1. Chapter 5The Laws of Motion EXAMPLES

  2. Example 1: Action-Reaction • The force exerted BY object 1 ON object 2 is equal in magnitude and opposite in direction to • exerted BY object 2 ON object 1

  3. Example 2: Action-Reaction • The force Fhnexerted BY the hammer ON the nail is equal in magnitude and opposite in direction to Fnhexerted BY the nail ONthe hammer Fhn = – Fnh

  4. Example 3: Action-Reaction • We can walk forward because when one foot pushes backward against the ground, the ground pushes forward on the foot. – FPG≡ FGP

  5. Example 4: Normal Force (a) m = 10 kg  Weight: Fg = mg = 98.0N The normal forceis equal to the weight!! Only this case FN = mg = 98.0N (b) Pushing force = 40N FN = mg +40N= 138.0N (c) Pulling force = 40N FN = mg – 40N = 58.0N

  6. Example 5: Accelerating the box From Newton’s 2nd Law: FP – mg = 100N – 98N = ma ma= 2N The box accelerates upwards because FP > m g

  7. Example 6: A Traffic Light at Rest(Example 5.4 Text Book) • This is an equilibrium problem • No movement, so a = 0 • Upper cables are not strong as the lower cable. They will break if the tension exceeds 100N. • Will the light remain or will one of the cables break?

  8. Example 6: cont. • Find T3 from applying equilibrium in the y-direction to the light • Find x and y components for T1 and T2 : T2y T1y T2x -T1x

  9. Example 6: final. • Applying Newton’s 2nd Law to find the net force for each axis for a particle in equilibrium: • In equation (1) solve for T2 in terms of T1 : • Substituting (3) into (2) and solving for T1 : • As we can see Both values are less than 100N, so the cables will not break!!!

  10. Example 7: Weight Loss (Example 5.2 Text Book) • Apparent weight loss. The lady weights 65kg = 640N, the elevator descends with a = 0.2m/s2. What does the scale read (FN)? • From Newton’s 2nd law:∑F = ma FN – mg = – m a FN = mg – m a FN = 640N – 13N = 627N = 52kgUpwards! FN is the force the scale exerts on the person, and is equal and opposite to the force she exerts on the scale.

  11. Example 7: cont. • What does the scale read when the elevator descends at a constant speed of 2.0m/s? • From Newton’s 2nd law:∑F = 0 FN – mg = 0 FN = mg = 640N = 65kg The scale reads her true mass! NOTE: In the first case the scale reads an “apparent mass” but her mass does not change as a result of the acceleration: it stays at 65 kg

  12. Example 8: Normal Force m = 10.0 kg mg = 98.0N Find: ax ≠ 0? FN? if ay = 0 m=10kg FPy = FPsin(30) = 20.0N FPx = FPcos(30) = 34.6N ΣFx = FPx = m ax ax= 34.6N/10.0kg ax= 3.46m/s2 ΣFy = FN + FPy – mg = m ay FN + FPy – mg = 0  FN = mg – FPy= 98.0N – 20.0N FN = 78.0N FPy FPx

  13. Example 9: The Hockey Puck Moving at constant velocity, withNO friction. Which free-body diagram is correct? (b)

  14. Example 10: The Runaway Car (Example 5.6 Text Book) • Replace the force of gravity with its components: Fgx = mgsin Fgy = mgcos With: ay = 0 & ax ≠ 0 • (A). Findax Using Newton’s 2nd Law: • y-Direction ΣFy = n – mgcos = may = 0 n = mgcos

  15. Example 10: Final. • x-direction ΣFx = mgsin = max ax = gsin Independent of m!! (B) How long does it take the front of the car to reach the bottom? (C). What is the car’s speed at the bottom?

  16. Example 11: Two Boxes Connected by a Cord • Boxes A & B are connected by a cord (mass neglected). Boxes are resting on a frictionless table. • FP = 40.0 N • Find: • Acceleration (a) of each box • Tension(FT) in the cord connecting the boxes • There is only horizontal motion With: aA = aB = a • Apply Newton’s Laws for box A: ΣFx = FP –FT = mAa (1) • Apply Newton’s Laws for box B: ΣFx = FT = mBa (2) Substituting (2) into (1): FP –mBa = mAa FP = (mA + mB)a  a = FP/(mA + mB) = 1.82m/s2 Substituting a into (2)  FT = mBa = (12.0kg)(1.82m/s2)= 21.8N

  17. Example 12: Atwood’s Machine • Forces acting on the objects: • Tension (same for both objects, one string) • Gravitational force • Each object has the same acceleration since they are connected • Draw the free-body diagrams • Apply Newton’s Laws • Solve for the unknown(s)

  18. Example 12: cont. • Vary the masses and observe the values of the tension and acceleration • Note the acceleration is the same for both objects • The tension is the same on both sides of the pulley as long as you assume a mass-less, frictionless pulley • Apply Newton’s 2nd Law to each Mass. ΣFy = T– m1g = m1 a (1) ΣFy = T– m2g = – m2 a (2) • Then: T= m1g + m1a (3) T= m2g– m2a (4)

  19. Example 12: final. • Equating: (3) = (4) and Solving for a m1g + m1 a = m2g– m2 a  m1 a + m2 a = m2g – m1g a (m1 + m2) = (m2 – m1)g  (5) Substituting (5) into (3) or (4): T= m1g + m1a (3) 

  20. Example 13: Two Objects and Incline Plane • Draw the free-body diagram for each object • One cord, so tension is the same for both objects • Connected, so acceleration is the same for both objects • Apply Newton’s Laws • xy plane: ΣFx = 0 &ΣFy = m1 a T– m1g = m1 a T= m1g + m1 a(1) • x’y’ plane: ΣFx = m2 a&ΣFy = 0 m2gsinθ – T = m2 a(2) n – m2gcosθ = 0(3)

  21. Example 13: final. • Substituting (1)in (2) gives: m2gsinθ– (m1g + m1 a) = m2a m2gsinθ– m1g – m1 a = m2a  a (m1 + m2) = m2gsinθ – m1g • Substituting ain (1): T= m1g + m1 a

  22. Assume:mg = 98.0N  n = 98.0 N, s= 0.40, k = 0.30  ƒs,max= sn = 0.40(98N)= 39N Find Force of Friction if the force applied FA is: FA = 0ƒs=FA= 0ƒs=0 Box does not move!! FA = 10N FA < ƒs,max or (10N < 39N) ƒs–FA= 0ƒs=FA= 10N The box still does not move!! FA = 38N < ƒs,maxƒs–FA= 0 ƒs=FA= 38N Force is still not quite large enough to move the box!!! FA = 40N > ƒs,maxkinetic friction. This one will start moving the box!!! ƒk kn = 0.30(98N) = 29N. The net force on the box is: ∑F = max 40N – 29N = max 11N = max ax = 11 kg.m/s2/10kg = 1.10 m/s2 Example 14: Pulling Against Friction n ƒs,k

  23. Example 14: Final. ƒs,max= 39N ƒs,k ƒk= 29N ƒs µs n

  24. Example 15: To Push or Pull a Sled Similar to Quiz 5.7 • Will you exerts less force if you push or pull the girl? (θ is the same in both cases). • Newton’s 2nd Law: ∑F = ma x direction:∑Fx= maxFx – ƒs,max = max • Pushing y direction:∑Fy = 0 n – mg – Fy= 0  n = mg + Fy ƒs,max =μsn  ƒs,max =μs (mg + Fy ) FBD n Fx ƒs,max Fy Pushing

  25. Example 15: Final. • Pulling y direction:∑Fy = 0 n +Fy – mg= 0  n = mg – Fy ƒs,max =μsn  ƒs,max =μs (mg – Fy ) NOTE:ƒs,max (Pushing) > ƒs,max (Pulling) Friction Force would be less if you pull than push!!! FBD FBD Fy n ƒs,max Fx Pulling

  26. Example 16: Why Does the Sled Move? • To determine ifthehorse (sled)moves:consider onlythe horizontal forces exerted ONthehorse (sled), then apply 2nd Newton’s Law:ΣF = m a. • Horse: T : tension exerted by the sled. fhorse : reaction exerted by the Earth. • Sled: T : tension exerted by the horse. fsled : friction between sled and snow.

  27. Example 16: Final. • Horse: If fhorse > T , the horse accelerates to the right. • Sled: If T> fsled , the sled accelerates to the right. • The forces that accelerates the system (horse-sled) is the net force fhorse  fsled • If fhorse = fsled the system will move with constant velocity.

  28. Example 17:Experimental Determination of µs and µk • Tilted coordinate system: K-Trigonometry: Fgx= Fgsinθ = mgsinθ Fgy= Fgcosθ = – mgcosθ ∑F = m a , ƒk,s k,sn ax ≠ 0 ay = 0 • y direction: ∑Fy = 0  n – mgcosθ = 0  n = mgcosθ(1) • x direction: ∑Fx = max mgsinθ – ƒk,s = max(2) ax

  29. Example 17: Final. • The block is sliding down the plane, so friction acts up the plane • This setup can be used to experimentally determine the coefficient of friction • From Equations (1) and (2) and the fact that:k,sƒk,s /n µ = tan θ • For µs, use the angle where the block just slips • For µk, use the angle where the block slides down at a constant speed

  30. Example 18: Sliding Hockey PuckExample 5.12 (Text Book) • Draw the FBD, including the force of kinetic friction Given: vxi = 20.0 m/s vxf= 0, xi= 0, xf = 115 m Find μk? • y direction: (ay = 0) ∑Fy = 0  n – mg = 0 n =mg(1) • x direction: ∑Fx = max – μkn = max(2)

  31. Example 18: Final. • Substituting (1) in (2) : – μk(mg) = max ax = –μk g • To the left (slowing down) & independent of the mass!! • Replacing ax in the Equation: vf2 = vi2 + 2ax(xf – xi)  0 = (20.0m/s)2 + 2(–μk g)(115m)  μk 2(9.80m/s2)(115m) = 400(m2/s2)  μk =400(m2/s2) / (2254m2/s2)  μk = 0.177

  32. Example 19: Acceleration of Two Objects Connected with Friction Example 5.13 (Text Book) a • If: ƒk kn, find: a • Friction acts only on the object in contact with another surface. Draw the FBD Mass 1: (Block) • y direction: ∑Fy = 0,ay= 0 n + Fsinθ– m1g = 0  n = m1g – Fsinθ(1) • x direction: ∑Fx = m1a Fcosθ– T –ƒk = m1a Fcosθ– T –kn= m1a T = Fcosθ–kn–m1a(2) FBD

  33. Example 19:Final. FBD a Mass 2:(Ball) • y direction: ∑Fy = m2a T – m2g = m2aT = m2g + m2a (3) • x direction: ∑Fx = 0,ax= 0 • Substitute n = m1g – Fsinθ(1) into T = Fcosθ–kn–m1a(2) T = Fcosθ– k(m1g – Fsinθ) – m1a(4) • Equate: (3) = (4) and solve for a:

  34. Example 20: The Skier • If:FG= mg , ay = 0, Ffr kFN, & k= 0.10, then Find: ax • FG components: FGx= mgsin30o & FGy= – mgcos30o • Newton’s 2nd Law y direction: ∑Fy = 0  FN– mgcos30o = 0  FN = mgcos30o(1) x direction: ∑Fx = max mgsin30o – kFN = max(2) Replacing (1) in (2) mgsin30o – kFN = max (3) • Substituting (1) into (3) and solving for ax mgsin30o – kmgcos30o = max ax = g(0.5) – 0.10g(0.87) ax = 0.41g ax = 4.00m/s2 ax ax

  35. Material for the Midterm • Material from the book to Study!!! • Objective Questions: 5-6-10 • Conceptual Questions: 2-12-16 • Problems: 10-12-19-20-25-28-34-46-47-48

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