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5.7 Some Applications of Newton’s Law, cont Multiple Objects When two or more objects are connected or in contact, Newton’s laws may be applied to the system as a whole and/or to each individual object Whichever you use to solve the problem, the other approach can be used as a check Example 5.16 Multiple Objects First treat the system as a whole: F x msystemax Apply Newton’s Laws to the individual blocks Solve for unknown(s) Check: |P21| = |P12| Example 5.17 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 Tension (FT) in the each box cord connecting the boxes Example 5.17 Two Boxes Connected by a Cord, final 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 Example 5.18 The 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) Example 5.18 The Atwood’s Machine, 2 The Atwood’s Machine: Find: a and T Apply Newton’s 2nd Law to each Mass. ΣFy = T – m1g = m1 a (1) ΣFy = T – m2g = – m2 a (2) Then: T = m1g + m1 a (3) T= m g – m a (4) Example 5.18 The Atwood’s Machine, 3 The Atwood’s Machine: 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 m2 m1 a g m1 m2 (5) Example 5.18 The Atwood’s Machine, final The Atwood’s Machine: Substituting (5) into (3) or (4): T = m1g + m1 a (3) m m1 T m1 g m1 2 g m1 m2 m1m1 g m1m2 g m1m2 g m1m1 g m1m2 g m1m1 g T m1 g T m1 m2 m1 m2 m1m2 T 2 g m1 m2 Active Figure 5.14 Example 5.19 Two Objects and Incline Plane Find: a and T One cord: so tension is the same for both objects Connected: so acceleration is the same for both objects Apply Newton’s Laws Solve for the unknown(s) Example 5.19 Two Objects and Incline Plane, 2 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) Example 5.19 Two Objects and Incline Plane, Final Substituting (1) in (2) gives: m2gsinθ – (m1g + m1 a) = m2 a m2gsinθ – m1g – m1 a = m2 a a (m1 + m2) = m2gsinθ – m1g Substituting a in (1) we’ll get: T = m1g + m1 a (1) m sin m1 T m1 g m1 2 g m1 m2 m2 sin m1 a g m1 m2 m1m1 g m2 m1 g m1m2 g sin m1m1 g T m1 m2 m2 m1 g m1m2 g sin m1m2 g sin 1 T m1 m2 m1 m2 Problem-Solving Hints Newton’s Laws Conceptualize the problem – draw a diagram Categorize the problem Equilibrium (SF = 0) or Newton’s Second Law (SF = m a) Analyze Draw free-body diagrams for each object Include only forces acting on the object Problem-Solving Hints Newton’s Laws, cont Analyze, cont. Establish coordinate system Be sure units are consistent Apply the appropriate equation(s) in component form Solve for the unknowns. This always requires Kinder Garden Algebra (KGA). Like solving two linear equations with two unknowns Finalize Check your results for consistency with your free- body diagram Check extreme values 5.8 Forces of Friction When an object is in motion on a surface or through a viscous medium, there will be a resistance to the motion This is due to the interactions between the object and its environment This resistance is called the Force of Friction Forces of Friction, 2 Friction exists between any 2 sliding surfaces. Two types of friction: Static (no motion) friction Kinetic (motion) friction The size of the friction force depends on: The microscopic details of 2 sliding surfaces. The materials they are made of Are the surfaces smooth or rough? Are they wet or dry? Forces of Friction, 3 Friction is proportional to the normal force ƒs µs n (5.8) and ƒk = µk n (5.9) These equations relate the magnitudes of the forces, THEY ARE NOT vector equations The force of static friction (maximum) is generally greater than the force of kinetic friction ƒs > ƒk The coefficients of friction (µk,s) depends on the surfaces in contact Forces of Friction, final The direction of the frictional force is opposite the direction of motion and parallel to the surfaces in contact The coefficients of friction (µk,s) are nearly independent of the area of contact Static Friction Static friction acts to keep the object from moving: ƒs = F If F increases, so does ƒs If F decreases, so does ƒs ƒs µs n where the equality holds when the surfaces are on the verge of slipping Called impending motion Static Friction, cont Experiments determine the relation used to compute friction forces. The friction force ƒs exists ║ to the surfaces, even if there is no motion. Consider the applied force F ∑F = ma = 0 & also v = 0 There must be a friction force ƒs to oppose F F – ƒs = 0 ƒs = F Kinetic Friction The force of kinetic friction (ƒk ) acts when the object is in motion Friction force ƒk is proportional to the magnitude of the normal force n between 2 sliding surfaces. ƒk n ƒk k n (magnitudes) k Coefficient of kinetic friction k : depends on the surfaces & their conditions k : is dimensionless & < 1 Static & Kinetic Friction Experiments find that the Maximum Static Friction Force ƒs,max is proportional to the magnitude (size) of the normal force n between the 2 surfaces. DIRECTIONS: ƒs,max n Then: ƒs,max = sn (magnitudes) Static & Kinetic Friction s Coefficient of static friction s : depends on the surfaces & their conditions s : is dimensionless & < 1 Always: ƒs,max > ƒk s n > k n (Cancel n) s > k ƒs ƒs,max = µsn ƒs µs n Some Coefficients of Friction Active Figure 5.16 Friction in Newton’s Laws Problems Friction is a force, so it simply is included in the Net Force (SF ) in Newton’s Laws The rules of friction allow you to determine the direction and magnitude of the force of friction Example 5.20 Pulling Against Friction 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: A. FA = 0 ƒs = FA = 0 ƒs = 0 Box does not move!! n ƒs,k B. FA = 10N FA < ƒs,max or (10N < 39N) ƒs – FA = 0 ƒs = FA = 10N The box still does not move!! Example 5.20 Pulling Against Friction, 2 C. FA = 38N < ƒs,max ƒs – FA = 0 ƒs = FA = 38N This force is still not quite large enough to move the box!!! n ƒs,k D. FA = 40N > ƒs,max kinetic friction. This one will start moving the box!!! ƒk k n = 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 ƒs,k Example 5.20 Pulling Against Friction, final ƒs,max = 39N ƒk= 29N Example 5.21 To Push or Pull a Sled Similar to Quiz 5.14 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 Pushing Pulling Example 5.21 To Push or Pull a Sled, 2 x direction: ∑Fx = max Fx – ƒs,max = max n Pushing y direction: ∑Fy = 0 n – mg – F y = 0 n = m g + Fy Fx ƒs,max Fy ƒs,max = μsn ƒs,max = μs (mg + Fy ) Pushing Example 5.21 To Push or Pull a Sled, final Pulling y direction: ∑Fy = 0 n + Fy – mg = 0 n = mg – Fy ƒs,max = μsn ƒs,max = μs (mg – Fy ) n Fy ƒs,max Fx NOTE: ƒs,max (Pushing) > ƒs,max (Pulling) Friction Force would be less if you pull than push!!! Pulling Conceptual Example 5.22 Why Does the Sled Move? (Example 5.11 Text Book) To determine if the horse (sled) moves: consider only the horizontal forces exerted ON the horse (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. Conceptual Example 5.22 Why Does the Sled Move? 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. Example 5.23 Sliding Hockey Puck Example 5.13 (Text Book) Draw the free-body diagram, including the force of kinetic friction Opposes the motion Is parallel to the surfaces in contact Continue with the solution as with any Newton’s Law problem Example 5.23 Sliding Hockey Puck, 2 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) Substituting (1) in (2) : – μk(mg) = max ax = – μk g Example 5.23 Sliding Hockey Puck, final 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 Example 5.24 Two Objects Connected with Friction Example 5.13 (Text Book) Known: ƒk kn Find: a 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) Example 5.24 Two Objects Connected with Friction, 2 Mass 2: (Ball) y direction: ∑Fy = m2a T – m2g = m2a T = m2g + m2a (3) x direction: ∑Fx = 0, ax = 0 n = m1g – Fsinθ (1) T = Fcosθ – kn – m1a (2) Substitute (1) into (2): T = Fcosθ – k(m1g – Fsinθ ) – m1a (4) Equate: (3) = (4) and solve for a: Example 5.24 Two Objects Connected with Friction, final m2 g m2 a F cos k (m1 g F sin ) m1a m1a m2 a F cos k m1 g k F sin m2 g m1 m2 a F cos k sin k m1 m2 g F cosθ μ k sinθ μ k m1 m2 g a m1 m2 Inclined Plane Problems Tilted coordinate system: Convenient, but not necessary. K-Trigonometry: Fgx= Fgsinθ = mgsinθ Fgy= Fgcosθ = – mgcosθ Understand: ∑F = m a , ƒk k n ax ≠ 0 ay = 0 y direction: ∑Fy = 0 n – mgcosθ = 0 n = mgcosθ (1) x direction: ∑Fx = max mgsinθ – ƒ = max (2) ax Is the normal force n equal & opposite to the weight Fg ? NO!!!! Experimental Determination of µs and µk 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 µs,k µs,k = tan s,k For µs use the angle where the block just slips For µk use the angle where the block slides down at a constant speed Active Figure 5.19 Example 5.25 The Skier Assuming: FG = mg , ay = 0 ƒk kn, k = 0.10 Find: ax Components: FGx= FGsin30o = mgsin30o FGy= FGcos30o = – mgcos30o Newton’s 2nd Law y direction: ∑Fy = 0 n – mgcos30o = 0 n = mgcos30o (1) x direction: ∑Fx = max mgsin30o – ƒk = max (2) ax Example 5.25 The Skier n = mgcos30o (1) mgsin30o – ƒk = max (2) Replacing ƒk kn in (2) mgsin30o – kn = max (3) Substituting (1) into (3) mgsin30o – kmgcos30o = max ax = gsin30o – μkgcos 30o ax = g(0.5) – 0.10g(0.87) ax = 0.41g ax = 4.00m/s2 ax Material for the Midterm Examples to Read!!! Example 5.2 Example 5.3 Example 5.12 (Page 120) (Page 122) (Page 134) Homework to be solved in Class!!! NONE