Transcription

Ch. 3: Kinetics of Particles3.0 Outline Introduction Newton’s Second Law Equations of Motion Rectilinear Motion Curvilinear Motion3.0 Outline

Ch. 3: Kinetics of Particles3.1 IntroductionKinetics is the study of the relations between the forcesand the motion. Here we will not seriously concernwhether the forces cause the motion or the motiongenerates the forces (causality).In this chapter, the focus is on the particles. That is thebody whose physical dimensions are so small comparedwith the radius of curvature of its path.There are at least 3 approaches toe the solution ofkinetic problems: (a) Newton’s second law (b) work andenergy method (c) impulse and momentum method.3.1 Introduction

Ch. 3: Kinetics of Particles3.2 Newton’s Second LawF mam mass (resistance to rate of change of velocity) of the particleF resultant force acting on the particlea resulting acceleration measured in a nonaccelerating frame of referenceFor most engineering problems on earth, the accelerationmeasured w.r.t. reference frame fixed to the earth’ssurface may be treated as absolute. And Newton’s 2ndlaw of motion holds.Newton’s 2nd law breaks when the velocities of the orderof the speed of light are involved Æ theory of relativity3.2 Newton’s Second Law

Ch. 3: Kinetics of Particles3.3 Equation of Motion and Solution of Problems F ma--- equation of motionscalar components decomposition according to a specified coordinateTwo problems of dynamics(1) specified kinematic conditions, find forces Æstraightforward application of Newton’s law asalgebraic equations(2) specified forces, find motion ÆDifficulty depends on the form of force function(t, s, v, a), as the solutions are found by solvinga system of differential equations.For simple functions, we can find closed form solutionsof motion as in rectilinear motion (sec. 2.2).3.3 Equation of Motion and Solution

Ch. 3: Kinetics of ParticlesUnconstrained motionMotion of the particle is determined by its initial motion andthe forces from external sources. It is free of constraintsand so has three degrees of freedom to specifythe position. Three scalar equations of motion wouldhave to be applied and integrated to obtain the motion.Constrained motionMotion of the particle is partially or totally determined byrestraining guides, other than its initial motion and theforces from external sources. Therefore, all forces, bothapplied and reactive, that act on the particle must beaccounted for in Newton’s law. The number of d.o.f. andequations are reduced regarding to the type of constraints.3.3 Equation of Motion and Solution

Ch. 3: Kinetics of ParticlesFree body diagramAll forces acting on the particle needed to be accountedin the equations of motion. Free body diagram unveilsevery force that acts on the isolated particle. Only afterthe FBD has been completed should the equations ofmotion be written. The appropriate coordinate axes anddirections should be indicated and consistently usedthroughout the problem.Treatment of the body as particle is valid when the forcesmay be treated as concurrent through the mass center.3.3 Equation of Motion and Solution

Ch. 3: Kinetics of Particles3.4 Rectilinear MotionIf the x-axis is the direction of the rectilinear motion, Fx ma x Fy 0 Fz 0If we are not free to choose a coordinate direction along the motion,the nonzero acceleration component will be shown up in all equations: Fx ma x Fy ma y Fz ma zOther coordinate system such as n-t or r-θa may be determined via the use of relative motionFor pure translating moving reference framea A a B a A/B3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/17 The coefficient of static friction between the flatbed of the truck and the crate it carries is 0.30.Determine the minimum stopping distance s thatthe truck can have from a speed of 70 km/h withconstant deceleration if the crate is not toslip forward.3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesmgP. 3/17 xF 0.3NNIf the crate is not to slip, crate and truck must have same acceleration.If the crate is not to slip, friction static friction at impending status.Minimum stopping distance when the deceleration is the max allowable value. Fx ma x 0.3mg ma x , a x 0.3g constant for minimum distance v 2 v o2 2a ( s s o ) 210 0 70 2 ( 0.3g ) s, s 64.2 m36 3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/18If the truck of Prob. 3/17 comes to stop from an initial forwardspeed of 70 km/h in a distance of 50 m with uniform deceleration,determine whether or not the crate strikes the wall at the forwardend of the flat bed. If the crate does strike the wall, calculate itsspeed relative to the truck as the impact occurs. Use the frictioncoefficients μs 0.3 and μk 0.25.3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/18stopping distance 50 m, which is less than minimum value 64.2 m the crate slips210 v v 2a ( s s o ) 0 70 2a truck 50, a truck 3.781 m/s 236 10 v v o a ( t t o ) 0 70 3.781 t, t stop 5.14 s36 22oFriction force: Fs 0.3mg 2.943m and Fk 0.25mg 2.45m xAssume crate and truck go together a truck a crate Fx ma x F m ( 3.781) required friction 3.781m Fs the crate slips and F Fk Fk ma crate , a crate 2.45 m/s[ a crate/truck a crate a truck ]F 0.3N2a crate/truck 2.45 ( 3.781) 1.331 m/s 2N the crate slips forward but will it strike the wall?1 22 s s o v o ( t t o ) 2 a ( t t o ) relative motion calculation13 1.331 t 2 , t strike 2.123 s t stop crate will strike the wall before the truck stops2 v v o a ( t t o ) relative motion calculationv crate/truck 0 1.331 2.123 2.826 m/s3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/23 If the coefficients of static and kinetic frictionbetween the 20-kg block A and the 100-kg cart Bare both essentially the same value of 0.50,determine the acceleration of each part for(a) P 60 N and (b) P 40 N.3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/2320g2P(a) N A 20g, Fmax 0.5N A 98.1 N 120 N block A moves forward relative to B Fx ma x 120 98.1 20a A , a A 1.095 m/s98.1 100a B , a B 0.981 m/s 22NAFFNA100g(b) Fmax 80 N block A does not move relative to B Fx ma x A & B move together80 120a, a 0.667 m/s 2NBFind developed friction by isolated FBD at A or B80 F 20a, F 66.67 N Fmax assumption is validF 100a, F 66.67 N3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/24 A simple pendulum is pivoted at O and is free toswing in the vertical plane of the plate. If theplate is given a constant acceleration a up theincline θ, write an expression for the steadyangle β assumed by the pendulum after allinitial start-up oscillations have ceased. Neglectthe mass of the slender supporting rod.3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/24yTβxθmg Fy 0 Tcosβ mgcosθ 0 Fx ma x Tsinβ mgsinθ ma a gsinθ β tan gcosθ 13.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/28 For the friction coefficients μs 0.25 andμk 0.20, calculate the acceleration of eachbody and the tension T in the cable.3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/28T60g2TFNs A 2s B c l a A 2a B 020gN 60gcos30, Fmax μs N 127.4 NAssume motion impends at block A F Fmax and equilibrium F 0 60gsin30 Fmax T 0, T 166.9 Nbut cylinder B will not be in equilibrium ( 20g 2T 0 move up )Assum block A slides down and block B moves up F ma 60gsin30 Fk T 60a A 120a B20g 2T 20a B , T 105.35 N, a B 0.725 m/s 2 , a A 1.45 m/s 23.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/35A bar of length l and negligible mass connects the cart of massM and the particle of mass m. If the cart is subjected to a constantacceleration a to the right, what is the resulting steady-stateangle θthat the freely pivoting bar makes with the vertical?Determine the net force P (not shown) that must be applied tothe cart to cause the specified acceleration.3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/35MgPyθTTNxmgFrom the given statements, pendulum and cart have same accelerationAt the pendulum, Fy 0 Tcosθ mg 0, T mg/cosθ Fx ma x a Tsinθ ma, θ tan 1 g At the cart, Fx ma x P Tsinθ Ma, P ( m M ) gtanθ3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/36 Determine the accelerations of bodies A and Band the tension in the cable due to theapplication of the 250 N force. Neglect all frictionand the masses of the pulleys.3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/3670g35g2TNAsA300 N3TsBNB2s A 3s B c l 2a A 3a B 0 Fx ma x 2T 70a A and 300 3T 35a Ba A 2.34 m/s 2 , a B 1.56 m/s 2 , T 81.8 N3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/44 The sliders A and B are connected by a lightrigid bar and move with negligible friction in theslots, both of which lie in a horizontal plane.For the position shown, the velocity of A is0.4 m/s to the right. Determine the accelerationof each slider and the force in the barat this instant.3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesNBP. 3/44TNATsBsA40 NKinematics: triangle OABs A s B and 0.5 s A cos15 s B cos15, s A s B 0.2588 ml 2 s 2A s B2 2s A s B cos150diff: 0 2s A v A 2s B v B 2 cos150 ( s A v B s B v A )given: v A 0.4 m/s v B 0.4 mdiff: 0 v 2A s A a A v B2 s Ba B cos150 ( s A a B s Ba A 2v A v B )0 0.04287 0.4829a A 0.4829a B(1)Kinetics: F ma 40 Tcos15 2a A and Tcos15 3a B into (1)a A 7.95 m/s 2 , a B 8.04 m/s 2 , T 25.0 N3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/46 With the blocks initially at rest, the force P isincreased slowly from zero to 260 N. Plot theaccelerations of both masses as functions of P.3.4 Rectilinear Motion

Ch. 3: Kinetics of Particles35gP. 3/46FAN A 35g, N B N A 42g 77gNANAFAP42gFAmax 0.2N A 68.67 N, FBmax 0.15N B 113.3 NFBFAk 0.15N A 51.5 N, FBk 0.10N B 75.54 NNBThree possible situations: no motion, B & A move together, and B & A move separately.Two impossible situations: B moves alone then FA will 0 A will move eventuallyand A moves alone P is applied at block B and force P is increased slowly from zero( not jump right to F )A max1) 0 P FBmax : FB will be developed to cancel with the applied P,FA will stay zero, and so there is no motion a A 0 & a B 03.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/462) assume both A and B go together in this phase FA FAmax and FB FBk F ma FA 35a & P FA FBk 42aat P Pmin FBmax ( increased slowly ) , a 0.49 m/s 2 and FA 17.16 N ( jumping )at FA FAmax( about to slip relative to each other ) , P 226.6 N and a 1.962 m/s 2between these extremum values, a FBmaxP FBk: linear function of P77 P 226.6 : a A a B which varies linearly from 0.49 to 1.962 m/s 23) A slide ( backward ) relative to B increasing P makes B accelerates more and moreP 226.6 N makes A slips FA FAk F ma FAk 35a A & P FA k FBk 42a BP 127.04: linear function of P42 226.6 P 260.0 : a A 1.47 m/s 2 constant and 2.37 a B 3.166 m/s 2a A 1.47 m/s 2 constant and a B ( jumping )3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/463.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/48 The system is released from rest in the positionshown. Calculate the tension T in the cord andthe acceleration a of the 30 kg block. The smallpulley attached to the block has negligible massand friction. (Suggestion: First establish thekinematic relationship between the accelerationsof the two bodies.)3.4 Rectilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/48 bKinematics: b 2 c 2 x 2 and b y lcdiff: bb xx and b y 0b 2 bb x 2 xx and b y 0(1) y xat this instant: x/b 4 / 5, x 0, b 0 ( initially rest )assume cylinder moves down, hence block moves to the leftT30gKinetics: for 30 kg block F ma T 3/5 T 30g N 0, N 30g 2T/5F T 4/5 30xassume the block moves F 0.25Nfor 15 kg cylinder15g T 15y 15bx 4 b ( T/15 g ) 30 , T 137.9 Nb 5 x7.5g 0.7T7.5g 0.7T 0.766 m/s 2x 30FTNTrecall (1) ,15g3.4 Rectilinear Motion

Ch. 3: Kinetics of Particles3.5 Curvilinear MotionChoose appropriate coordinate system (x-y, n-t, or r-θ)for the given problem. Determine the motion alongthose axes. Then set up the Newton’s lawalong those axes. The positive sense of the force andacceleration must be consistent. F mx F myn-t system: F m ( ρβ ) m ( v / ρ ) F mvr-θ system: F m ( r rθ ) Fθ m ( rθ 2rθ )x-y system:xy22nt2r3.5 Curvilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/55The member OA rotates about a horizontal axis through Owith a constant counterclockwise velocity ω 3 rad/s. As itpasses the position θ 0, a small block of mass m is placedon it at a radial distance r 450 mm. If the block is observedto slip at θ 50 , determine the coefficient of static frictionμs between the block and the member.3.5 Curvilinear Motion

Ch. 3: Kinetics of ParticlesP. 3/55mguse n-t coordinate systemgiven: ρ 0.45 m, β 50 , ρ 0 ( no slip until β 50 ) ,tβ 3 rad/s, β 0()mgsin50 F m ( ρβ ) Ft ma t N mgcos50 m ρβ ρβ , N mgcos50 Fn ma n 2nAt 50 , F Fs μs N and directs upward because gsin50 ρβ 2FNwhich means bar OA rotates too slow than required to keepthe block stays on the bar. The friction will develop to resist(the block from sliding down or to match F with ρβ ) .2If the bar rotates very very slow, friction force cannot make F to match ρβ ( F cannot be reduced any more, F ρβ ) .2