PH 221-3A Fall EQUILIBRIUM and. Lectures Chapter 12 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition)

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1 PH 221-3A Fall 2009 EQUILIBRIUM and ELASTICITY Lectues Chapte 12 (Halliday/Resnick/Walke, Fundamentals of Physics 8 th edition) 1

2 Chapte 12 Equilibium and Elasticity In this chapte we will define equilibium and find the conditions needed so that an object is at equilibium. We will then apply these conditions to a vaiety of pactical engineeing poblems of static equilibium. We will also examine how a igid body can be defomed by an extenal foce. In this section we will intoduce the following concepts: Stess and stain Young s modulus (in connection with tension and compession) Shea modulus (in connection with sheaing) Bulk modulus (in connection to hydaulic stess) 2

3 Equilibium We say that an object is in equilibium u when we the teoow following gtwocodtos conditions ae satisfied: 1. The linea momentum P of the cnte of mass is constant 2. The angula momentum L about the cente of mass o any othe point is a constant Ou concen in this chapte is with situtations in which P = 0 and L = 0 That is we ae inteested in objects that ae not moving in any way (this includes tanslational as well as otational motion) in the efeence fame fom which we obseve them. Such objects ae said to be in static equilibium In chapte 8 we diffeentiated between stable and unstable static equilibium If a body that is in static equilibium is displaced slightly fom this position the foces on it may etun it to its old position. In this case we say that the equilibium is stable. If the body does not etun to its old position then the equilibium is unstable. 3

4 A simple method fo the expeimental detemination of the cente of mass of a body of a complicated shape A igid body suppoted by an upwad foce acting at a point on the vetical line though its cente of mass is in equilibium The cente of mass is at the intesection of the new and old polongations of the sting. Stable, unstable and neutal equilibium A body suspended d fom a point above its cente of mass is in stable equilibium i 2) The foce of gavity and suppoting foce poduce a toque that tends to etun the body to the equilibium position. 4

5 If the body is suppoted by a foce applied at a point below the cente of mass it is in unstable equilibium A chai suppoted at a point diectly below the cente of mass. If we tun the chai slightly the chai tends to topple ove. A body suppoted at its cente of mass is in neutal equlibium If we tun such a body, it emains in equilibium in its new position and exhibits no tendency to etun to its oiginal position o to tun fathe away. 5

6 An example of unstable equilibium is shown in the figues In fig.a we balance a domino with the domino's centeofmass vetically above the suppoting edge. The toque of the gavitational foce F g about the suppoting edge is zeo because the line of action of F g passes though the edge. Thus the domino is in equilibium. i Even a slight foce on the domino ends the equilibium. As the line of action of Fg moves to one side of the suppoting edge (see fig.b) the toque due to F g is non-zeo and the domino otates in the clockwise diection away fom its equilibium position of fig.a. The domino in fig.a is in a position of unstable equilibium. The domino is fig.c is not quite as unstable. To topple the domino the applied foce would have to otate it though and beyond the position of fig.a. A flick of the finge against the domino can topple it. 6

7 The Conditions of equilibium In chapte 9 we calculated the ate of change fo the linea momentum of the cente dp of mass of an object. Fnet If an object is in tanslational equilibium then dt = dp P = constant and thus = 0 Fnet = 0 dt In chapte 11 we analyzed otational motion and saw that Newton's second law takes the dl fom: = τ Fo an object in otational ti equilibium i we have: net L = constant t dt dl = 0 τ net = 0 dt The two equiements fo a body to be in equilibium ae: 1. The vecto sum of all the extenal foces F net = 0 on the body must be zeo 2. The vecto sum of all the extenal toques that act on the body measued about any point must be zeo τ = 0 net 7

8 In component fom the conditions of equilibium ae: Balance of foces: F = 0 F = 0 F = 0 x y z Bl Balance of toques: τ = 0 τ = 0 τ = 0 x y z We shall simplify mattes by consideing only poblems in which all the foces that act on the body lie in the xy-plane. This means that the only toques geneated by these foces tend to cause otation about an axis paallel to the z-axis. With this assumption the conditions fo equilibium become: Balance of foces: F = 0 F = 0 Balance of toques: z x z = 0 y Hee τ is the net toque poduced τ by all extenal foces eithe about the z-axis o about any axis paellel to it. Finally fo static equilibium the linea momentum P of the cente of mass must be ze o: P = 0 τ F = 0 F = 0 x y z = 0 8

9 The cente of Gavity (cog) The gavitational i lfoce acting on an extended ddbd body is the vecto sum of fh the gavitational foces acting on the individual elements of the body. The gavitational foce F on a body effectively acts at a single point known as the cente of gavity g of the body. Hee "effectively" has the following meaning: If the individual gavitational foces on the elements of the body ae tuned off and eplaced by F g acting at the cente of gavity, then the net foce and the net toque about any point on the body does not chang e. We shall pove that if the acceleation of gavity g is the same fo all the elements of the body then the cente of gavity coicides with the cente of mass. This is a easonable appoximation fo objects nea the suface of the eath because g changes vey little. 9

10 Conside the extended object of mass M shown in fig.a. In fig.a we also show the i-th element of mass mi. The gavitational foce on mi is equal to mg i i whee gi is the acceleation of gavity in the vicinity of mi. The toque τ i on mi is equal to Fgixi. The net toque τnet = τi = Fgi xi (eqs.1) i i Conside now fig.b in which we have eplaced the foces Fgi by the net gavitational foce Fg acting at the cente of gavity. The net toque τ is equal to : τ = x F = x F (eqs.2) net net cog g cog gi i If we compae equation 1 with equation 2 we get: x F = Fx cog gi gi i i i We substitute mg fo F and we have: x mg = mgx i i gi cog i i i i i i i mx i i i If we set g = g fo all the elements x = = x m i i cog com i 10

11 Statics Poblem Recipe 1. Daw a foce diagam. (Label lth the axes) 2. Choose a convenient oigin O. A good choice is to have one of the unknown foces acting at O 3. Sign of the toque τ fo each foce: - If the foce induces clockwise (CW) otation + If the foce induces counte-clockwise (CCW) otation 4. Equilibium conditions: F = 0 F = 0 x y τ = z 0 5. Make sue that: numbes of unknowns = numbe of equations 11

12 Examples of Static Equilibium A locomotive of 90,000 kg is one thid of the way acoss a bidge 90 m long. The bidge consists of a unifom ion gide of 900,000 kg, which ests at two pieces. What is the load on each piece? Method: 1. Select the body which is to obey the equilibium condition 2. List all the extenal foces that act on this body and display them on a fee body diagam 3. Choose coodinate axes and esolve the foces into x and y components. (+ o with espect to positive diection that you have chosen) 4. Apply equilibium condition: ΣFx = 0 ΣFy = 0 5. Make a choice of axis otation. Choose positive diection of otation. Apply the static equilibium condition 6. Solve the equations in steps 4 and 5 fo the desied unknown quantities. 12

13 13

14 O Sample Poblem A unifom beam of length L and mass m = 1.8 kg is at est on two scales. A unifom block of mass M = 2.7 kg is at est on the beam at a distance L/4 fom its left end. Cal c u lat e the scales eadings F = F + F Mg mg = y l 0 (eqs.1) We choose to calculate the toque with epsect to an axis though h the left end of the beam (point O). L L τ z = ( mg ) ( Mg ) + ( L)( F ) = 0 (eqs.2) 4 2 Mg mg Fom equation 2 we get: F = + = + = N We solve quation 1 fo F F = Mg+ mg F = = N F 29 N l l l ( ) 14

15 Sample Poblem 12-2: A ladde of length L = 12 m and mass m = 45 kg leans against a fictionless wall. The ladde's uppe end is at a height h = 9.3 m above the pavement on which the lowe end ests. The com of the ladde is L/3 fom the lowe end. A fiefighte of mass M = 72 kg climbs half way up the ladde Find the foces exeted on the ladde by the wall 2 2 and the pavement. Distance a = L h = 7.58 m We take toques about an axis though point O. a a τ z = ( h)( Fw ) + ( mg) + ( Mg) = M m ga ( 72/ /3 ) Fw = = = 407 N 410 N h 9.3 F = F F = 0 F = F = 410 N x w px px w ( ) Fnet, y = Fpy Mg mg = 0 Fpy = Mg + mg = = N 1100 N 15

16 F = F T = 0 F = T = 6093 N x h c h c y v v ( ) ( ) = h + v = + Sample Poblem 12-3: A safe of mass M = 430 kg hangs by a ope fom a boom with dimensions a = 1.9 m and b = m. The beam of fthe boom has mass m = 85 kg Find the tension T in the cable and the magnitude of the c net foce F exeted on the beam by the hinge. We calculate the net toque about an axis nomal to the page that passes though point O. b τ z = ( a)( Tc ) ( b)( T ) ( mg ) = 0 2 m gb M /2 T 2.5( ) c = = 6100 N a 1.9 ( ) ( ) F = F mg T = 0 F = mg + T = g m + M = = 5047 N F F F N 16

17 O Sample Poblem 12-4: A 70 kg ock climbe hangs by the cimp hold of one hand. He feet touch the ock diectly below he finges. Assume that the foce fom the hoizontal ledge suppoting he finges is equally shaed by the fou finges. Calculate the hoizontal and vetical components F and F of the foce on each fingetip. F = F + 4F = 0 x N h mg F y = 4Fv mg = 0 Fv = = = N 170 N 4 4 h v We calculate the net toque about an axis that is pependicula to the page and passes though point O. ( ) F ( )( mg) ( )( F ) ( )( F ) τ = = 0 z N h v F h = = N 17 N

18 Poblem (static equilibium): The dawing shows an A-shaped ladde. Both sides of the ladde ae equal in length. This ladde standing on a fictionless hoizontal suface, and only the cossba (which has a negligible mass) of the A keeps the ladde fom collapsing. The ladde is unifom and dhas a mass of f200k 20.0 kg. Detemine the tension in the cossba of fthe ladde. 18

19 Indeteminate Stuctues. Fo the poblems in this chapte we have the following thee equations at ou disposal: F = 0 F = 0 τ = 0 x y z If the poblem has moe than thee unknowns we cannot solve it We can solve a statics poblem fo a table with thee legs but not fo one with fou legs. Poblems like these ae called indeteminatenate An example is given in the figue. A big elephant sits on a wobly table. If the table does not collapse it will defom so that all fou legs touch the floo. The upwad foces exeted on the legs by the floo assume definite and diffeent values. How can we calculate the values of these foces? To solve such an indeteminate equilibium poblem we must supplement the thee equilibium equations with some knowledge of elasticity, the banch of physics and engineeing that descibes how eal bodies defom when foces ae applied to them. 19

20 Elasticity Metallic solids consist of a lage numbe of atoms positioned on a egula thee-dimensional lattice as shown in the figue. The lattice is epetition of a patten (in the figue this patten is a cube) Each atom of the solid has a well-defined equilibium distance fom its neaest neighbos. The atoms ae held togethe by inteatomic foces that can be modeled as tiny spings. If we ty to change the inteatomic distance the esulting foce is popotional to the atom displacement fom the equilibium position. The sping constants ae lage and thus the lattice is emakably igid. Nevetheless all "igid" bodies ae to some extend elastic, which means that we can change thei dimensions slightly by pulling, pushing, twisting o compessing them. Fo example, if you suspend a subcompact ca fom a steel od 1 m long and 1 cm in diamete, the od will stetch by only 0.5 mm. The od will etun to its oiginal length of 1 m when the ca is emoved. If you suspend two cas fom the od the od will be pemanently defomed. If you suspend thee cas the od will beak. 20

21 stess = modulus stain In the thee figues above we show the thee ways in which a solid might change its dimensions unde the action of extenal defoming foces. In fig. a the cylinde is stetched by foces acting along the cylinde axis. In fig. b the cylinde is defomed by foces pependicula to its axis. In fig. c a solid placed in a fluid unde high pessue is compessed unifomly on all sides. All thee defomation types have stess in common (defined as defoming foce pe unit aea). These stesses ae known as tensile/compessive fo fig.a, sheaing fo fig.b, and hyda ulic fo fig.c. The application of stess on a solid esults in stain which take diffeent fom fo the thee types of stain. Stess is elated to stain via the equa tion: stess = modulus stain 21

22 F = A A E ΔL L F Tensile stess is defined as the atio whee A is the solid aea A Δ L Stain (symbol S) is defined as the atio whee ΔL L is the change in the length L of the cylindical solid. Stess is plotted vesus stain in the uppe figue. Fo a wide ange of applied stesses the stess-stain elation is linea and the solid etuns to its oiginal length when the stess is emoved. This is known as the elastic ange. If the stess is inceased beyond a maximum value known as the yield stength the cylinde becomes pemanently S y defomed. If the stess continues to incease the cylinde beaks at a stess value known as ultimate stength Su Fo stesses below S y (elastic ange) stess and stain ae connected via the equation F ΔL = E The constant E (modulus) is known as: Young's modulus A L Note: Young's modulus is almost the same fo tension and compession The ultimate stength S u maybe diffeent 22

23 A F = G A F A Δx L ΔV p = B V Sheaing: In the case of sheaing defomation stain Δx is defined d as the dimensionless i atio. The stess/stain t L equation has the fom: F Δ x G A = L The constant G is known as the shea modulus F Hydaulic Stess. The stess is this case is the pessue p = A the suounding fluid exets on the immesed object. Hee A is the aea of the object. In this case stain is defined as the ΔV dimesionless n atio whee V is the volume of the object V and ΔV the change in the volume due to the fluid pessue. The V stess/stain equation has the fom: p = B Δ The constant V B is known as the bulk modulus of the mateial 23

24 Poblem 43. A hoizontal aluminum od 4.8 cm in diamete pojects 5.3 cm fom a wall. A 1200 kg object is suspended fom the end of the od. The shea modulus of aluminum is x10 N/m. Neglecting the od's mass, find (a) the shea stess on the od and (b) the vetical deflection of the end of the od. ()Th (a) The shea stess is given by F/A, whee F is the magnitude of the foce applied paallel to one face of the aluminum od and A is the coss sectional aea of the od. In this case F is the weight of the object hung on the end: F = mg, whee m is the mass of the object. If is the adius of the od then A = π 2. Thus, the shea stess is F A 2 (1200 kg)(9.8m/s ) = mg 2 2 π = π(0.024m) = N/m. (b) The shea modulus G is given by G F / = A Δx/ L whee L is the potusion of the od and Δx is its vetical deflection at its end. Thus, 6 2 ( F/ A ) L ( N/m )(0.053m) ) Δ x = = = 10 2 G N/m m. 24