Vector Mechnics for Engineers: Dynmics nnouncement Reminders Wednesdy s clss will strt t 1:00PM. Summry of the chpter 11 ws posted on website nd ws sent you by emil. For the students, who needs hrdcopy, plese come to the front tble. Homework 1.9 (7th) or 1.10 (8th) 1.17 (7th) or 1.18 (8th) 1. (7th) or 1.4 (8th) 1.5 (7th) or 1.7 (8th) 1.36 (7th) or 1.38 (8th) Due, next Wednesdy, 07/19/006! 1-30
Vector Mechnics for Engineers: Dynmics Smple Problem 11.1 SOLUTION: Evlute time t for q = 30 o. Evlute rdil nd ngulr positions, nd first nd second derivtives t time t. Rottion of the rm bout O is defined by q = 0.15t where q is in rdins nd t in seconds. Collr slides long the rm such tht r = 0.9-0.1t where r is in meters. Clculte velocity nd ccelertion in cylindricl coordintes. Evlute ccelertion with respect to rm. fter the rm hs rotted through 30 o, determine () the totl velocity of the collr, (b) the totl ccelertion of the collr, nd (c) the reltive ccelertion of the collr with respect to the rm. - 30
Vector Mechnics for Engineers: Dynmics Review of lst clss nd introduction of this clss ---- Rod Mp Finished Dynmics Unfinished Prticles Dels with Rigid body Curviliner motion Rectngulr coord. Curviliner Coord. Polr/Cylindr. Coord. Kinemtics Determintion of motion Sole prticle & three bsic problems Reltive Motion (P) Dependent Motion Rectiliner motion Kinetics Newton s lw of motion Dynmic Trnsltionl motion equilibrium Eq. Rottion Chpter 1 3-30
CHPTER 11 VECTOR MECHNICS FOR ENGINEERS: DYNMICS Ferdinnd P. eer E. Russell Johnston, Jr. Lecture Notes: Dr. Gngyi Zhou UC Irvine Jul 10th, 006 Kinetics of Prticles Prt : Force method
Vector Mechnics for Engineers: Dynmics Newton s Second Lw of Motion ---moving objects with constnt mss Newton s Second Lw: If the resultnt force cting on prticle is not zero, the prticle will hve n ccelertion proportionl to the mgnitude of resultnt nd in the direction of the resultnt. When prticle of mss m is cted upon by force the ccelertion of the prticle must stisfy F m (Mthemtic expression ) Remrks: If force cting on prticle is zero, prticle will not ccelerte, i.e., it will remin sttionry or continue on stright line t constnt velocity. Cution: ccelertion must be evluted with respect to Newtonin frme of reference, i.e., one tht is not ccelerting or rotting. F, 5-30
Vector Mechnics for Engineers: Dynmics Newton s Second Lw of Motion ---moving objects with vrible mss Consider how Rocket lunches: (1) Rockets propel outwrds high pressure gses; () The mss of Rockets decreses (3) Rockets re moving object with decresing mss Replcing the ccelertion by the derivtive of the velocity yields d dl F m v dt dt L liner momentum of the prticle generl form Liner Momentum Conservtion Principle: If the resultnt force on prticle is zero, the liner momentum of the prticle remins constnt in both mgnitude nd direction. 6-30
Vector Mechnics for Engineers: Dynmics Systems of Units Of the units for the four primry dimensions (force, mss, length, nd time), three my be chosen rbitrrily. The fourth must be comptible with Newton s nd Lw. Interntionl System of Units (SI Units): bse units re the units of length (m), mss (kg), nd time (second). The unit of force is derived, 1N 1kg m 1 s kg m 1 s U.S. Customry Units: bse units re the units of force (lb), length (m), nd time (second). The unit of mss is derived, 1lb 1lbm 3.ft s 1lb 1slug 1ft s lbs 1 ft 7-30
Vector Mechnics for Engineers: Dynmics Equtions of Motion Rectngulr Coord. Newton s second lw provides F m Solution for prticle motion is fcilitted by resolving vector eqution into sclr component equtions, e.g., for rectngulr components, Fxi Fy j Fzk m xi yj zk Fx mx Fy my Fz d x d y Fx m Fy m Fz dt For tngentil nd norml components, F F t t m t dv m dt F F n n m dt n v m m m z d z dt Curviliner Coord. 8-30
Vector Mechnics for Engineers: Dynmics Dynmic Equilibrium lternte expression of Newton s second lw, virtul F m 0 force m inertil vector With the inclusion of the inertil vector, the system of forces cting on the prticle is equivlent to zero. The prticle is in dynmic equilibrium. Methods developed for prticles in sttic equilibrium my be pplied, e.g., coplnr forces my be represented with closed vector polygon. 9-30
Vector Mechnics for Engineers: Dynmics Summry on Newton s nd lw Two expressions d Constnt mss: F=m; Generl form: mv F dt dv m v dt In Coordintes The components of force vectors cn be fully determined by the ssocite components of ccelertion vectors through Newton s lw. dm dt Force to induce velocity chnge Converting dynmic problem to sttic problem: Only by Inertil force ( virtul force opposite to ccelertions) Physicl mening: Force is the key to chnge motion. Force to induce mss chnge 10-30
Vector Mechnics for Engineers: Dynmics Steps to solve dynmic problems though Newton s nd lw Step 1: How mny prticles? Step : Setup coordintes Step 3: Determintion of motion for ech prticles 3.1, Kinemtics 3., Decomposition of forces on ech prticle long ech xis of coordintes 3.3, Dynmic equtions for ech prticle long ech xis of coordintes Step 4: Look up the solutions from the bove determined motions. 11-30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.1 SOLUTION: Resolve the eqution of motion for the block into two rectngulr component equtions. 00-lb block rests on horizontl plne. Find the mgnitude of the force P required to give the block n ccelertion or 10 ft/s to the right. The coefficient of kinetic friction between the block nd plne is m k 0.5. Unknowns consist of the pplied force P nd the norml rection N from the plne. The two equtions my be solved for these unknowns. 1-30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.1 y O m F x W g lbs 6.1 ft m N k 0.5N 00lb 3.ft s SOLUTION: Resolve the eqution of motion for the block into two rectngulr component equtions. F x m : F y P cos30 0.5N 0 : 6.1lbs ft 10ft s 6.1lb N Psin30 00lb 0 Unknowns consist of the pplied force P nd the norml rection N from the plne. The two equtions my be solved for these unknowns. N Psin30 00lb P cos30 0.5 Psin30 00lb 6.1lb P 151lb 13-30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.3 SOLUTION: The two blocks shown strt from rest. The horizontl plne nd the pulley re frictionless, nd the pulley is ssumed to be of negligible mss. Determine the ccelertion of ech block nd the tension in the cord. Write the kinemtic reltionships for the dependent motions nd ccelertions of the blocks. Write the equtions of motion for the blocks nd pulley. Combine the kinemtic reltionships with the equtions of motion to solve for the ccelertions nd cord tension. 14-30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.3 O y x SOLUTION: Write the kinemtic reltionships for the dependent motions nd ccelertions of the blocks. Write equtions of motion for blocks nd pulley. F m x y T 1 F m F y y m : 100kg : 300kg 9.81m s T 300kg T T 1 x g T m m 940N - C C T 1 0 : 0 1 300kg 15-30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.3 O y x Combine kinemtic reltionships with equtions of motion to solve for ccelertions nd cord tension. y T T T 1 1 x 100kg 940N - 940N - 300kg 1 T T 1 300kg 1 T 1 0 940 N 150kg 100kg 0 8.40m 1 100kg T 1 s 4.0m 1680 N s 840 N 16-30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.4 SOLUTION: The block is constrined to slide down the wedge. Therefore, their motions re dependent. Express the ccelertion of block s the ccelertion of wedge plus the ccelertion of the block reltive to the wedge. The 1-lb block strts from rest nd slides on the 30-lb wedge, which is supported by horizontl surfce. Neglecting friction, determine () the ccelertion of the wedge, nd (b) the ccelertion of the block reltive to the wedge. Write the equtions of motion for the wedge nd block. Solve for the ccelertions. 17-30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.4 y x SOLUTION: The block is constrined to slide down the wedge. Therefore, their motions re dependent. Write equtions of motion for wedge nd block. F m : F F x x y N 1 0.5N sin30 m W N m 1 x W m m sin30 y g cos30 : W g cos30 cos30 g sin30 m sin30: W cos30 30 W g sin 1 18-30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.4 Solve for the ccelertions. 0.5N1 W g N 1 W cos30 W g sin30 W g W cos30 W g gw cos30 W W sin30 3. ft s 1lbcos30 30lb 1lbsin30 5.07ft sin30 s cos30 g sin30 5.07 ft s cos30 3. ft s sin30 0.5ft s 19-30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.5 The bob of -m pendulum describes n rc of circle in verticl plne. If the tension in the cord is.5 times the weight of the bob for the position shown, find the velocity nd ccelertion of the bob in tht position. SOLUTION: Resolve the eqution of motion for the bob into tngentil nd norml components. Solve the component equtions for the norml nd tngentil ccelertions. Solve for the velocity in terms of the norml ccelertion. 0-30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.5 SOLUTION: Resolve the eqution of motion for the bob into tngentil nd norml components. Solve the component equtions for the norml nd tngentil ccelertions. F : mg sin30 m t m t Fn m n : t g sin30.5mg n g t t.5 cos30 4.9m mg cos30 m n n s 16.03m Solve for velocity in terms of norml ccelertion. n v v n m 16.03m s v 5.66m s s 1-30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.6 SOLUTION: The cr trvels in horizontl circulr pth with norml component of ccelertion directed towrd the center of the pth.the forces cting on the cr re its weight nd norml rection from the rod surfce. Determine the rted speed of highwy curve of rdius = 400 ft bnked through n ngle q = 18 o. The rted speed of bnked highwy curve is the speed t which cr should trvel if no lterl friction force is to be exerted t its wheels. Resolve the eqution of motion for the cr into verticl nd norml components. Solve for the vehicle speed. - 30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.6 SOLUTION: The cr trvels in horizontl circulr pth with norml component of ccelertion directed towrd the center of the pth.the forces cting on the cr re its weight nd norml rection from the rod surfce. Resolve the eqution of motion for the cr into verticl nd norml components. F y 0 : R cosq W 0 W R cosq Fn m n : Rsinq W g W sinq cosq Solve for the vehicle speed. v g tnq n W g 3. ft s 400ft tn18 v v 64.7ft s 44.1mi h 3-30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.15 lock weights 80lb, nd block weights 16lb. The coefficients of friction between ll surfces of contct re u s =0. nd u k =0.15. Knowing tht P=0, determine () the ccelertion of block, (b) the tension in the cord. 4-30
Vector Mechnics for Engineers: Dynmics Smple Problem 1.40 The 0.5kg flyblls of centrifugl governor revolve t constnt speed v in the horizontl circle of 150mm rdius shown. Neglecting the mss of links,c,d, nd DE nd requiring tht the links support only tensile forces, determine the rnge of the llowble vlues of v so tht the mgnetitudes of the forces in the links do not exceed 75N 5-30