ME 141 Engineering Mechnics Lecture 10: Kinetics of prticles: Newton s nd Lw Ahmd Shhedi Shkil Lecturer, Dept. of Mechnicl Engg, BUET E-mil: sshkil@me.buet.c.bd, shkil6791@gmil.com Website: techer.buet.c.bd/sshkil Courtesy: Vector Mechnics for Engineers, Beer nd Johnston
Introduction m Newton s Second Lw of Motion 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. Must be expressed with respect to Newtonin (or inertil) frme of reference, i.e., one tht is not ccelerting or rotting. This form of the eqution is for constnt mss system
Liner Momentum of Prticle Replcing the ccelertion by the derivtive of the velocity yields dv m dt d dl mv dt dt L liner momentum of the prticle 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.
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 m 1kg1 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
Equtions of Motion Newton s second lw m Cn use sclr component equtions, e.g., for rectngulr components, i j k m i j k x x x m y mx x z y y m my y x y z z z m mz z
Dynmic Equilibrium Alternte expression of Newton s second lw, m 0 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. Inerti vectors re often clled inertil forces s they mesure the resistnce tht prticles offer to chnges in motion, i.e., chnges in speed or direction. Inertil forces my be conceptully useful but re not like the contct nd grvittionl forces found in sttics.
Smple Problem 1.1 SOLUTION: Resolve the eqution of motion for the block into two rectngulr component equtions. Unknowns consist of the pplied force P nd the norml rection N from the plne. The two equtions my be solved for these unknowns. A 00-lb block rests on horizontl plne. ind the mgnitude of the force P required to give the block n ccelertion of 10 ft/s to the right. The coefficient of kinetic friction between the block nd plne is m k 0.5.
y O Smple Problem 1.1 m 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. x m : y P cos30 0.5N 0 : N P sin 30 00lb 6.1lbs ft 10ft s 6.1lb 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 Psin 30 00lb P cos30 0.5 0 P sin 30 00lb 6.1lb P 151lb
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.
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. m x A A T1 100kg m : y y y B m B B B : A B B 300 kg 9.81m s T 300 kg T T g T m 1 xa 1 B m 940N - C C T 1 0 : 0 A B 300 kg B
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 B 1 xa B 1 1 100kg A 940N - 300 kg 940N - T T B 1 A 300 kg1 T 1 0 940 N B 150kg 100kg 0 A A A A 8.40 m 1 100 kg T 1 A s 4.0 m A 1680 N s 840 N
Prob # 1.19 Block A hs mss of 40 kg, nd block B hs mss of 8 kg. The coefficients of friction between ll surfces of contct re μ s = 0.0 nd μ k = 0.15. If P = 40 N, determine () the ccelertion of block B, (b) the tension in the cord
Prob # 1.35 Block B of mss 10 kg rests s shown on the upper surfce of -kg wedge A. Knowing tht the system is relesed from rest nd neglecting friction, determine () the ccelertion of B, (b) The velocity of B reltive to A t t = 0.5 s.
Kinetics: Norml nd Tngentil Coordintes Aircrft nd roller costers cn both experience lrge norml forces during turns.
Equtions of Motion Newton s second lw m or tngentil nd norml components, t t m t dv m dt n n m n v m
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.
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. : mg sin 30 m t m t n m n : t g sin 30.5mg n g t t mg cos30.5 cos30 n 4.9m m 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
Smple Problem 1.6 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. 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. Solve for the vehicle speed.
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. y 0 : R cosq W 0 W R cosq n m n : Rsinq W g W sinq cosq Solve for the vehicle speed. v g tnq n W g 3. ft s 400fttn18 v v 64.7ft s 44.1mi h
Prob # 1.45 During high-speed chse, 400-lb sports cr trveling t speed of 100 mi/h just loses contct with the rod s it reches the crest A of hill. () Determine the rdius of curvture r of the verticl profile of the rod t A. (b) Using the vlue of r found in prt, determine the force exerted on 160-lb driver by the set of his 3100-lb cr s the cr, trveling t constnt speed of 50 mi/h, psses through A.
Prob # 1.47 The roller-coster trck shown is contined in verticl plne. The portion of trck between A nd B is stright nd horizontl, while the portions to the left of A nd to the right of B hve rdii of curvture s indicted. A cr is trveling t speed of 7 km/h when the brkes re suddenly pplied, cusing the wheels of the cr to slide on the trck (μ k = 0.0). Determine the initil decelertion of the cr if the brkes re pplied s the cr () hs lmost reched A, (b) is trveling between A nd B, (c) hs just pssed B.
Prob # 1.49 A series of smll pckges, ech with mss of 0.5 kg, re dischrged from conveyor belt s shown. Knowing tht the coefficient of sttic friction between ech pckge nd the conveyor belt is 0.4, determine () the force exerted by the belt on the pckge just fter it hs pssed point A, (b) the ngle θ defining the point B where the pckges first slip reltive to the belt.