16 Newton s Laws #3: Components, Friction, Ramps, Pulleys, and Strings

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1 Chpter 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings When, in the cse of tilted coordinte system, you brek up the grvittionl force vector into its component vectors, mke sure the grvittionl force vector itself forms the hypotenuse of the right tringle in your vector component digrm. All too often, folks drw one of the components of the grvittionl force vector in such mnner tht it is bigger thn the grvittionl force vector it is supposed to be component of. The component of vector is never bigger thn the vector itself. Hving lerned how to use free body digrms, nd then hving lerned how to crete them, you re in pretty good position to solve huge number of Newton s 2 nd Lw problems. An understnding of the considertions in this chpter will enble to you to solve n even lrger clss of problems. Agin, we use exmples to convey the desired informtion. Exmple 16-1 A professor is pushing on desk with force of mgnitude t n cute ngle below the horizontl. The desk is on flt, horizontl tile floor nd it is not moving. or the desk, drw the free body digrm tht fcilittes the direct nd strightforwrd ppliction of Newton s 2 nd Lw of motion. Give the tble of forces. While not required prt of the solution, sketch often mkes it esier to come up with the correct free body digrm. Just mke sure you don t combine the sketch nd the free body digrm. In this problem, sketch helps clrify wht is ment by t n cute ngle below the horizontl. Pushing with force tht is directed t some cute ngle below the horizontl is pushing horizontlly nd downwrd t the sme time. 95

2 Chpter 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings Here is the initil free body digrm nd the corresponding tble of forces. N = 0 P sf g Tble of orces Symbol=? Nme Agent Victim Norml N orce The loor Desk Sttic sf riction orce The loor Desk g = mg Grvittionl The Erth s orce Grvittionl ield Desk orce of P Professor Hnds of Professor Desk Note tht there re no two mutully perpendiculr lines to which ll of the forces re prllel. The best choice of mutully perpendiculr lines would be verticl nd horizontl line. Three of the four forces lie long one or the other of such lines. But the force of the professor does not. We cnnot use this free body digrm directly. We re deling with cse which requires second free body digrm. Cses Requiring Second ree Body Digrm in Which One of More of the orces tht ws in the irst ree Body Digrm is Replced With its Components Estblish pir of mutully perpendiculr lines such tht most of the vectors lie long one or the other of the two lines. After hving done so, brek up ech of the other vectors, the ones tht lie long neither of the lines, (let s cll these the rogue vectors) into components long the two lines. (Breking up vectors into their components involves drwing vector component digrm.) Drw second free body digrm, identicl to the first, except with rogue vectors replced by their component vectors. In the new free body digrm, drw the component vectors in the direction in which they ctully point nd lbel them with their mgnitudes (no minus signs). 96

3 Chpter 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings or the cse t hnd, our rogue force is the force of the professor. We brek it up into components s follows: y Px Py x P Px = P cos Py P = sin Px = P cos Py = P sin Then we drw second free body digrm, the sme s the first, except with P replced by its component vectors: N = 0 Px sf Py g 97

4 Chpter 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings Exmple 16-2 A wooden block of mss m is sliding down flt metl incline ( flt metl rmp) tht mkes n cute ngle with the horizontl. The block is slowing down. Drw the directly-usble free body digrm of the block. Provide tble of forces. We choose to strt the solution to this problem with sketch. The sketch fcilittes the cretion of the free body digrm but in no wy replces it. m v Since the block is sliding in the down-the-incline direction, the frictionl force must be in the upthe-incline direction. Since the block s velocity is in the down-the-incline direction nd decresing, the ccelertion must be in the up-the-incline direction. k f g N Tble of orces Symbol=? Nme Agent Victim Norml The N The Rmp orce Block Kinetic kf = µ K The N riction The Rmp Block orce g =mg Grvittionl orce The Erth s Grvittionl ield The Block No mtter wht we choose for pir of coordinte xes, we cnnot mke it so tht ll the vectors in the free body digrm re prllel to one or the other of the two coordinte xes lines. At best, the pir of lines, one line prllel to the frictionl force nd the other perpendiculr to the rmp, leves one rogue vector, nmely the grvittionl force vector. Such coordinte system is tilted on the pge. 98

5 Chpter 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings Cses Involving Tilted Coordinte Systems or effective communiction purposes, students drwing digrms depicting phenomen occurring ner the surfce of the erth re required to use either the convention tht downwrd is towrd the bottom of the pge (corresponding to side view) or the convention tht downwrd is into the pge (corresponding to top view). If one wnts to depict coordinte system for cse in which the direction downwrd is prllel to neither coordinte xis line, the coordinte system must be drwn so tht it ppers tilted on the pge. In the cse of tilted-coordinte system problem requiring second free body digrm of the sme object, it is good ide to define the coordinte system on the first free body digrm. Use dshed lines so tht the coordinte xes do not look like force vectors. Here we redrw the first free body digrm. (When you get to this stge in problem, just dd the coordinte xes to your existing digrm.) kf y N g x Now we brek up g into its x, y component vectors. This clls for vector component digrm. 99

6 Chpter 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings y Horizontl Line gy 90 g x gx g x g = sin g y g = cos gx = gsin g y = gcos Next, we redrw the free body digrm with the grvittionl force vector replced by its component vectors. N kf g x g y 100

7 Chpter 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings Exmple 16-3 A solid brss cylinder of mss m is suspended by mssless string which is ttched to the top end of the cylinder. rom there, the string extends stright upwrd to mssless idel pulley. The string psses over the pulley nd extends downwrd, t n cute ngle thet to the verticl, to the hnd of person who is pulling on the string with force T. The pulley nd the entire string segment, from the cylinder to hnd, lie in one nd the sme plne. The cylinder is ccelerting upwrd. Provide both free body digrm nd tble of forces for the cylinder. A sketch comes in hndy for this one: To proceed with this one, we need some informtion on the effect of n idel mssless pulley on string tht psses over the pulley. Effect of n Idel Mssless Pulley The effect of n idel mssless pulley on string tht psses over the pulley is to chnge the direction in which the string extends, without chnging the tension in the string. By pulling on the end of the string with force of mgnitude T, the person cuses there to be tension T in the string. (The force pplied to the string by the hnd of the person, nd the tension force of the string pulling on the hnd of the person, re Newton s-3 rd -lw interction pir of forces. They re equl in mgnitude nd opposite in direction. We choose to use one nd the sme symbol T for the mgnitude of both of these forces. The directions re opposite ech other.) The tension is the sme throughout the string, so, where the string is ttched to the brss cylinder, the string exerts force of mgnitude T directed wy from the cylinder long the length of the string. Here is the free body digrm nd the tble of forces for the cylinder: 101

8 Chpter 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings T g Tble of orces Symbol=? Nme Agent Victim Tension The T The String orce Cylinder g = mg Grvittionl The Erth s The orce Grvittionl ield Cylinder Exmple 16-4 A crt of mss m C is on horizontl frictionless trck. One end of n idel mssless string segment is ttched to the middle of the front end of the crt. rom there the string extends horizontlly, forwrd nd wy from the crt, prllel to the centerline of the trck, to verticl pulley. The string psses over the pulley nd extends downwrd to solid metl block of mss m B. The string is ttched to the block. A person ws holding the crt in plce. The block ws suspended t rest, well bove the floor, by the string. The person hs relesed the crt. The crt is now ccelerting forwrd nd the block is ccelerting downwrd. Drw free body digrm for ech object. A sketch will help us to rrive t the correct nswer to this problem. m C m B Recll from the lst exmple tht there is only one tension in the string. Cll it T. Bsed on our knowledge of the force exerted on n object by string, viewed so tht the pprtus ppers s it does in the sketch, the string exerts rightwrd force T on the crt, nd n upwrd force of mgnitude T on the block. 102

9 Chpter 16 Newton s Lws #3: Components, riction, Rmps, Pulleys, nd Strings There is reltionship between ech of severl vribles of motion of one object ttched by tut string, which remins tut throughout the motion of the object, nd the corresponding vribles of motion of the second object. The reltionships re so simple tht you might consider them to be trivil, but they re criticl to the solution of problems involving objects connected by tut spring. The Reltionships Among the Vribles of Motion or Two Objects, One t One End nd the Other t the Other End, of n lwys-tut, Unstretchble String Consider the following digrm. 2 1 Becuse they re connected together by string of fixed length, if object 1 goes downwrd 5 cm, then object 2 must go rightwrd 5 cm. So if object 1 goes downwrd t 5 cm/s then object 2 must go rightwrd t 5 cm/s. In fct, no mtter how fst object 1 goes downwrd, object 2 must go rightwrd t the exct sme speed (s long s the string does not brek, stretch, or go slck). In order for the speeds to lwys be the sme, the ccelertions hve to be the sme s ech other. So if object 1 is picking up speed in the downwrd direction t, for instnce, 5 cm/s 2, then object 2 must be picking up speed in the rightwrd direction t 5 cm/s 2. The mgnitudes of the ccelertions re identicl. The wy to del with this is to use one nd the sme symbol for the mgnitude of the ccelertion of ech of the objects. The ides relevnt to this simple exmple pply to ny cse involving two objects, one on ech end of n inextensible string, s long s ech object moves only long line colliner with the string segment to which the object is ttched. Let s return to the exmple problem involving the crt nd the block. The two free body digrms follow: N T m c T m B gc gb Note tht the use of the sme symbol T in both digrms is importnt, s is the use of the sme symbol in both digrms. 103

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