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Mechanics

Mechanics. Chapter 3. Theorem of momentum and the law of momentum conservation 3.5 Applications of Newton’s laws. 3.5 Applications of Newton’s laws. Problems in dynamics can be divided into two classes:. Given the forces acting on an object, find motion or the equilibrium state of the object.

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Mechanics

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  1. Mechanics Chapter 3. Theorem of momentum and the law of momentum conservation 3.5 Applications of Newton’s laws Ch.3 Theorem of momentum and the law of momentum conservation

  2. 3.5 Applications of Newton’s laws Problems in dynamics can be divided into two classes: • Given the forces acting on an object, find motion or the equilibrium state of the object • Given the motion or equilibrium state of an object, find the forces acting on the object These problem can be solved by using Newton’s laws: F = ma F = -F´ • If a is known, find the forces acting on the object; • If F is given, find a • Mixture of above: only some of the forces and some of the acceleration components are known Ch.3 Theorem of momentum and the law of momentum conservation

  3. Mechanics 3.5 Applications of Newton’s laws 3.5.1 Force-Mass-Acceleration(FMA) method 3.5.2 Rectilinear motion 3.5.3 Rectilinear motion under variable force 3.5.4 Curvilinear motion 3.5.5 Equilibrium of particle 3.5.6 Summary Chapter 3. Theorem of momentum and the law of momentum conservation Ch.3 Theorem of momentum and the law of momentum conservation

  4. 3.5.1 Force-Mass-Acceleration(FMA) method The process of relating the forces to the acceleration of the particle using Newton’s laws is called theforce-mass-acceleration(FMA) method F = ma Resultant force vector Mass times accelerationinertial vector Free body diagram(FBD) Mass-acceleration diagram(MAD) Statically equivalent Ch.3 Theorem of momentum and the law of momentum conservation

  5. 3.5.1 Force-Mass-Acceleration(FMA) method Free Body Diagram • Free-body diagrams are essential to help identify the relevant forces. • A free-body diagram is a diagram showing the chosen body itself free of its surroundings, with vectors drawn to show the magnitudes and directions of each of the forces applied to the body by the various other bodies that interact with it. Mass-acceleration diagram Displays the inertia vector of the particle We will use these in all our examples we perform Ch.3 Theorem of momentum and the law of momentum conservation

  6. m2 m1 1 2 All surfaces frictionless N T2 m2 m2 2 m2a2 m2g Example of FBD and MAD 3.5.1 Force-Mass-Acceleration(FMA) method Mass m1 and mass m2 are connected by ropes of negligible mass that pass over a frictionless pulley. smooth peg FBD == MAD Ch.3 Theorem of momentum and the law of momentum conservation

  7. 3.5.1 Force-Mass-Acceleration(FMA) method Solving problems Step1: Draw the free-body diagram(FBD) of the particle that shows all forces acing on the particle Step 2: Use kinematics to analyze the acceleration of the particle. Step 3: Sketch the mass-acceleration diagram for the particle that displays the inertia vector ma, utilizing the results of step 2 Step 4: Referring to the FBD and MAD, relate the forces to acceleration, and choose co-ordinate system that simplifies subsequent calculations Step 5: Solve the equation to calculate unknown quantities. Ch.3 Theorem of momentum and the law of momentum conservation

  8. 2. Rectilinear motion Note • The FMA method only provides solutions for forces andaccelerations. If velocityorpositionhave to be found, kinematics equations are used once the acceleration is found from the equation of motion. • Any of the tools learned in Chapter 2 may be needed to solve a problem. Make sure you use consistent positive coordinate directions as used in the equation of motion part of the problem! Ch.3 Theorem of momentum and the law of momentum conservation

  9. Mechanics 3.5 Applications of Newton’s laws • Force-Mass-Acceleration(FMA) method • Rectilinear motion • Rectilinear motion under variable force • Curvilinear motion • Equilibrium of particle • Summary 3.5.1 Force-Mass-Acceleration(FMA) method 3.5.2 Rectilinear motion 3.5.3 Rectilinear motion under variable force 3.5.4 Curvilinear motion 3.5.5 Equilibrium of particle 3.5.6 Summary Chapter 3. Theorem of momentum and the law of momentum conservation Ch.3 Theorem of momentum and the law of momentum conservation

  10. 3.5.2 Rectilinear motion Newton’s second law of motion F = ma The scalar representation in rectangular coordinates Fx= maxFy= may Fz= maz If the particle is in rectilinear motion along the x-axis Fx= max • Automatically satisfied if all forces are in the direction of motion; • If not, used to calculate the unknown forces Fy=Fz= 0 Ch.3 Theorem of momentum and the law of momentum conservation

  11. 3.5.2 Rectilinear motion Ideal Pulley Force on person • Pulley spins freely without friction, neglect (rotational) inertia (mass) of pulley. • Pulley changes direction, not magnitude, of tension. Force On Box Ch.3 Theorem of momentum and the law of momentum conservation

  12. 3.5.2 Rectilinear motion Example 3.5-1: Pulley Systems - Atwood’s Machine: Massesm1andm2are attached to an ideal massless string and hung as shown around an ideal massless pulley. y • Find the accelerations,a1and a2, of the masses. • What is the tension in thestringT ? o j T1 T2 Atwood’s machineIf two blocks have similar mass, acceleration of system small, and can measure g. m1 y1 a1 m2 a2 y2 Ch.3 Theorem of momentum and the law of momentum conservation

  13. 3.5.2 Rectilinear motion Solution: • Draw free body diagrams for each object • Applying Newton’s Second Law: ( j-components) • T1 - m1g = m1a1 • T2 - m2g = m2a2 Free Body Diagrams T1 Since pulley is ideal: T1 = T2 = T T2 Since the masses are connected by the string which is inextensible j a1 a2 |y1| + |y2| +2R =L m2g m1g Differentiating the equation twice |a1| + |a2| = 0 |a1| = -|a2| = a • and a1 = -a2= -a. Ch.3 Theorem of momentum and the law of momentum conservation

  14. 3.5.2 Rectilinear motion T - m1g = -m1a(a) T - m2g = m2a(b) • Two equations & two unknowns • we can solve for both unknowns (T and a). Ch.3 Theorem of momentum and the law of momentum conservation

  15. m2 F m1  3.5.2 Rectilinear motion Example 3.5-2: Block on an Incline (Frictionless) GIVEN: A block of mass m2 is placed on a wedge (楔) of mass m1, which is supported by a horizontal surface. A horizontal forceFis applied to the wedge. All surfaces frictionless FIND:The angle  when there is no relative motion between m1 and m2 SOLUTION: Kinematics: when there is no relative motion between m1 and m2, they have the same horizontal acceleration a Ch.3 Theorem of momentum and the law of momentum conservation

  16. N2  m1 m2 F == == m1a m2a  N1 m2g m1g 3.5.2 Rectilinear motion Kinetics: draw the FBDs of the wedge and the block and apply Newton’s second law N Ch.3 Theorem of momentum and the law of momentum conservation

  17. 3.5.2 Rectilinear motion Define the x-y frame as fixed to the horizontal surface y x Ch.3 Theorem of momentum and the law of momentum conservation

  18. aA A 300 3.5.2 Rectilinear motion Example 3.5-3: Block on an Incline (Frictionless) GIVEN:The 4 kg block B starts from rest and slides on the 10kg wedge A, which is placed on a horizontal surface. Neglecting friction B A 300 FIND:(a) the acceleration of the wedge (b) The acceleration of the block relative to the wedge SOLUTION: Kinematics: the accelerations of the wedge and the block Wedge A: constrained to move on the horizontal surface. aA: horizontal. Assume that aA is directed to the right Ch.3 Theorem of momentum and the law of momentum conservation

  19. aA B 300 aB/A N1 mAaA wA=mAg 300 300 N2 3.5.2 Rectilinear motion Block B: aB = aA + aB/A aB/A :The acceleration of B relative to A, directed along the inclined surface of the wedge Kinetics: Draw the FBDs and apply Newton’s second law Wedge A: N1 + N2 + wA =mAaA N1cos300 +wA - N2 = 0 == N1 sin300 = mAaA 0.5N1 = mAaA Ch.3 Theorem of momentum and the law of momentum conservation

  20. y y wB=mBg 300 x 300 x mBaA mBaB/A N1 3.5.2 Rectilinear motion Block B: == N1 + wB =mBaB = mB(aA + aB/A) Using the coordinate axes shown in the fig. x component y component Ch.3 Theorem of momentum and the law of momentum conservation

  21. 3.5.2 Rectilinear motion We get three equations for three unknowns: Solve for aA and substituting the numerical data, we have Substituting the value obtained for aA into Eq.(2), we have Ch.3 Theorem of momentum and the law of momentum conservation

  22. Mechanics 3.5 Applications of Newton’s laws • Force-Mass-Acceleration(FMA) method • Rectilinear motion • Rectilinear motion under variable force • Curvilinear motion • Equilibrium of particle • Summary 3.5.1 Force-Mass-Acceleration(FMA) method 3.5.2 Rectilinear motion 3.5.3 Rectilinear motion under variable force 3.5.4 Curvilinear motion 3.5.5 Equilibrium of particle 3.5.6 Summary Chapter 3. Theorem of momentum and the law of momentum conservation Ch.3 Theorem of momentum and the law of momentum conservation

  23. 3.5.3 Rectilinear motion under variable force The forces acing on a particle are generally a function of the position and velocity of the particle, and time 1) Given kinematical equation of motion, find the force 2) Given the force, find the kinematical equation of motion Use the method given in ch. 2 Ch.3 Theorem of momentum and the law of momentum conservation

  24. f =-v o m y W=mg Example 3.5-4: Free fall with aerodynamic drag force 3.5.3 Rectilinear motion under variable force GIVEN:A ball of mass m is falling from high altitude. Aerodynamic draw face: fD = -v FIND: speed of the ball as a function of time t SOLUTION: • Draw free body diagram • Applying Newton’s Second Law: • Chose oy coordinate, positive direction downward Ch.3 Theorem of momentum and the law of momentum conservation

  25. 3.5.3 Rectilinear motion under variable force Initial condition: when t=0, v=0 Ch.3 Theorem of momentum and the law of momentum conservation

  26. Mechanics 3.5 Applications of Newton’s laws • Force-Mass-Acceleration(FMA) method • Rectilinear motion • Rectilinear motion under variable force • Curvilinear motion • Equilibrium of particle • Summary 3.5.1 Force-Mass-Acceleration(FMA) method 3.5.2 Rectilinear motion 3.5.3 Rectilinear motion under variable force 3.5.4 Curvilinear motion 3.5.5 Equilibrium of particle 3.5.6 Summary Chapter 3. Theorem of momentum and the law of momentum conservation Ch.3 Theorem of momentum and the law of momentum conservation

  27. 3.5.4 Curvilinear motion 1) Rectangular coordinate system Fx= maxFy= may Fz= maz 2) Path coordinate system Fn , Ft : the n-t components of the resultant force Special case: circular motion =R Fn :centripetal force Ch.3 Theorem of momentum and the law of momentum conservation

  28. F3 mat m  c man F2 F1 3.5.4 Curvilinear motion == MAD FBD 3) Cylindrical coordinate Ch.3 Theorem of momentum and the law of momentum conservation

  29. z z maz F3 r ma m mar F2 F1 y y  x x 3.5.4 Curvilinear motion == MAD FBD Ch.3 Theorem of momentum and the law of momentum conservation

  30. SOLVING PROBLEMS WITH n-t COORDINATES 3.5.4 Curvilinear motion •Use n-t coordinates when a particle is moving along a known,curvedpath. •Establish then-t coordinate systemon the particle. •Drawfree-bodyandmass-acceleration diagramsof the particle. Thenormal acceleration(an) always acts “inward” (the positive n-direction). Thetangential acceleration(at) may act in either the positive or negative t direction. •Apply theequations of motionand solve. •It may be necessary to employ thekinematic relations: at = dv/dt = v dv/ds an = v2/r Ch.3 Theorem of momentum and the law of momentum conservation

  31. 3.5.4 Curvilinear motion Example 3.5-5 GIVEN: A car travels around a curve which has a radius of 316 m. The curve is flat, not banked, and the coefficient of static friction between the tires and the road is 0.780. FIND:At what speed can the car travel around the curve without skidding(侧滑)? y SOLUTION:  FN FN r fs fs mg mg Ch.3 Theorem of momentum and the law of momentum conservation

  32. 3.5.4 Curvilinear motion • Now, the car will not skid as long as Fc is less than the maximum static frictional force Ch.3 Theorem of momentum and the law of momentum conservation

  33. y  FN  FN r r  fs fs mg  mg 3.5.4 Curvilinear motion Example 3.5-6: To reduce skidding, use a banked curve. Consider same conditions as previous example, but for a curve banked at the angle  Choose this coordinate system since ar is radial Since acceleration is radial only Ch.3 Theorem of momentum and the law of momentum conservation

  34. 3.5.4 Curvilinear motion • Since we want to know at what velocity the car will skid, this corresponds to the centripetal force being equal to the maximum static frictional force Substitute into previous equation Ch.3 Theorem of momentum and the law of momentum conservation

  35. 3.5.4 Curvilinear motion Substitute for FN and solve for v Ch.3 Theorem of momentum and the law of momentum conservation

  36. 3.5.4 Curvilinear motion • Adopt r = 316 m and  = 31°, and s=0.780 from earlier • If s=0,that is, there is no friction between the tire and road Ch.3 Theorem of momentum and the law of momentum conservation

  37. Mechanics 3.5 Applications of Newton’s laws • Force-Mass-Acceleration(FMA) method • Rectilinear motion • Rectilinear motion under variable force • Curvilinear motion • Equilibrium of particle • Summary 3.5.1 Force-Mass-Acceleration(FMA) method 3.5.2 Rectilinear motion 3.5.3 Rectilinear motion under variable force 3.5.4 Curvilinear motion 3.5.5 Equilibrium of particle 3.5.6 Summary Chapter 3. Theorem of momentum and the law of momentum conservation Ch.3 Theorem of momentum and the law of momentum conservation

  38. 3.5.5 Equilibrium of particle Particle is at rest or in uniform rectilinear motion  Equilibrium equation Fi = 0 Rectangular components: Fi x = Fi y = Fi z = 0 Ch.3 Theorem of momentum and the law of momentum conservation

  39. A B  T0 T 3.5.5 Equilibrium of particle Example 3.5-7: rope that passes over a peg GIVEN: A thin rope passes over a cylinder of radius r. The contact area is between A and B, which subtends a central angle . The coefficient of static friction between the rope and cylinder is 0 FIND: the relationship between the tension T0 and T Ch.3 Theorem of momentum and the law of momentum conservation

  40. Assumed direction of impending motion Distributed friction force A B  Distributed normal force T0 T 3.5.5 Equilibrium of particle SOLUTION: Neglect the weight of the rope. The cylinder exerts normal and friction forces on the rope, both of which are distributed along the contact area between A and B Direction of impending motion of the rope is clockwise, the direction of friction force is anticlockwise Equilibrium implies that T > T0 Ch.3 Theorem of momentum and the law of momentum conservation

  41. n N f0 t d T T´ 3.5.5 Equilibrium of particle Consider an infinitesimal length of the rope that subtends the differential angle d The forces acting on this differential element: • Tension on the left side of the element: T • Tension on the right side of the element: T´ = T+dT • The normal force: N • The friction: f0 Equilibrium condition: T + T´+ N + f0= 0 Ch.3 Theorem of momentum and the law of momentum conservation

  42. n N f0 t d T T´ 3.5.5 Equilibrium of particle Resolve the equation into n and t components, where n and t refer to the directions that a normal and tangent to the cylinder surface at the center of the element. Note: the angle between each tension and t-direction is d/2 Approximation: Ch.3 Theorem of momentum and the law of momentum conservation

  43. 3.5.5 Equilibrium of particle Ch.3 Theorem of momentum and the law of momentum conservation

  44. Mechanics 3.5 Applications of Newton’s laws • Force-Mass-Acceleration(FMA) method • Rectilinear motion • Rectilinear motion under variable force • Curvilinear motion • Equilibrium of particle • Summary 3.5.1 Force-Mass-Acceleration(FMA) method 3.5.2 Rectilinear motion 3.5.3 Rectilinear motion under variable force 3.5.4 Curvilinear motion 3.5.5 Equilibrium of particle 3.5.6 Summary Chapter 3. Theorem of momentum and the law of momentum conservation Ch.3 Theorem of momentum and the law of momentum conservation

  45. 3.5.6 Summary The basic equations for solving kinetic problems are Newton’s second and third laws of motion Rectangular coordinate Path coordinate Circular motion: Ch.3 Theorem of momentum and the law of momentum conservation

  46. 3.5.6 Summary Cylindrical coordinate Draw two diagrams: FBD and MAD Ch.3 Theorem of momentum and the law of momentum conservation

  47. 3.5.6 Summary Ch.3 Theorem of momentum and the law of momentum conservation

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