Transcription
Chapter 8 Equilibrium and ElasticityLecturePresentationChapter 8Equilibrium andElasticity 2015 Pearson Education, Inc.Chapter Goal: To learn about the static equilibrium ofextended objects, and the basic properties of springs andelastic materials. 2015 Pearson Education, Inc.Chapter 8 PreviewLooking AheadSlide 8-2Reading Question 8.1An object is in equilibrium ifA.B. 0C. Either A or BD. Both A and BText: p. 225 2015 Pearson Education, Inc.Slide 8-3 2015 Pearson Education, Inc.Slide 8-4
Reading Question 8.1Reading Question 8.2An object is in equilibrium ifIf you are performing the weightlifting exercise known asthe strict curl, the tension in your biceps tendon isA.B.A. Larger than the weight you are lifting.B. Equal to the weight you are lifting.C. Smaller than the weight you are lifting. 0C. Either A or BD. Both A and B 2015 Pearson Education, Inc.Slide 8-5 2015 Pearson Education, Inc.Reading Question 8.2Reading Question 8.3If you are performing the weightlifting exercise known asthe strict curl, the tension in your biceps tendon isAn object will be stable ifA.B.C.D.A. Larger than the weight you are lifting.B. Equal to the weight you are lifting.C. Smaller than the weight you are lifting. 2015 Pearson Education, Inc.Slide 8-7Slide 8-6Its center of gravity is below its highest point.Its center of gravity lies over its base of support.Its center of gravity lies outside its base of support.The height of its center of gravity is less than 1/2 its totalheight. 2015 Pearson Education, Inc.Slide 8-8
Reading Question 8.3Reading Question 8.4An object will be stable ifHooke’s law describes the force ofA.B.C.D.Its center of gravity is below its highest point.Its center of gravity lies over its base of support.Its center of gravity lies outside its base of support.The height of its center of gravity is less than 1/2 its totalheight. 2015 Pearson Education, Inc.Slide 8-9A.B.C.D.E.Gravity.A spring.Collisions.Tension.None of the above. 2015 Pearson Education, Inc.Slide 8-10Reading Question 8.4Reading Question 8.5Hooke’s law describes the force ofA very rigid material—one that stretches or compresses onlyslightly under large forces—has a large value ofA.B.C.D.E.Gravity.A spring.Collisions.Tension.None of the above. 2015 Pearson Education, Inc.A.B.C.D.Slide 8-11Tensile strength.Elastic limit.Density.Young’s modulus. 2015 Pearson Education, Inc.Slide 8-12
Reading Question 8.5A very rigid material—one that stretches or compresses onlyslightly under large forces—has a large value ofA.B.C.D.Tensile strength.Elastic limit.Density.Young’s modulus. 2015 Pearson Education, Inc.Section 8.1 Torque and Static EquilibriumSlide 8-13Torque and Static Equilibrium 2015 Pearson Education, Inc.Torque and Static Equilibrium For extended objectsthat can rotate, we mustconsider the net torque, too. An object at rest is in static equilibrium. As long as the object can be modeled as a particle, staticequilibrium is achieved when the net force on the particleis zero. 2015 Pearson Education, Inc.Slide 8-15 When the net force andthe net torque are zero,the block is in staticequilibrium. When the net force is zero,but the net torque is notzero, the object is notin static equilibrium. 2015 Pearson Education, Inc.Slide 8-16
Torque and Static EquilibriumQuickCheck 8.1 There are two conditions for static equilibrium on anextended object:Which object is in static equilibrium? The net force on the object must be zero. The net torque on the object must be zero. 2015 Pearson Education, Inc.Slide 8-17QuickCheck 8.1Slide 8-18Choosing the Pivot Point For an object in staticequilibrium, the nettorque about every pointmust be zero.Which object is in static equilibrium?D. 2015 Pearson Education, Inc. 2015 Pearson Education, Inc.Slide 8-19 You can choose any pointyou wish as a pivot pointfor calculating torque. 2015 Pearson Education, Inc.Slide 8-20
Choosing the Pivot PointChoosing the Pivot Point Any pivot point will work, but some pivot points cansimplify calculations. The solution is simplified ifyou choose the pivot point tobe the location where theforces are poorly specified. There is a “natural” axis of rotation for many situations. Anatural axis is an axis about which rotation would occur ifthe object were not in static equilibrium. 2015 Pearson Education, Inc.Slide 8-21Choosing the Pivot Point For the woman on the rockwall, the force of the wall onher feet is a mix of normaland frictional forces, and thedirection is not well known. 2015 Pearson Education, Inc.Slide 8-22Choosing the Pivot Point Choosing the point where thewoman’s foot contacts the wallas the pivot point eliminatesthe torque due to the force ofthe wall on her foot. The other forces and theirdirections are well known. Itis straightforward to calculatethe torque due to those forces(weight and tension).Text: p. 228 2015 Pearson Education, Inc.Slide 8-23 2015 Pearson Education, Inc.Slide 8-24
QuickCheck 8.2QuickCheck 8.2What does the scale read?What does the scale read?A. 500 NB. 1000 NC. 2000 ND. 4000 NA. 500 NB. 1000 NC. 2000 ND. 4000 NAnswering this requires reasoning, not calculating.Answering this requires reasoning, not calculating. 2015 Pearson Education, Inc.Slide 8-25 2015 Pearson Education, Inc.Example 8.4 Will the ladder slip?Example 8.4 Will the ladder slip? (cont.)A 3.0-m-long ladder leans against awall at an angle of 60 with respect tothe floor. What is the minimum valueof µs , the coefficient of static frictionwith the ground, that will prevent theladder from slipping? Assume thatfriction between the ladder and thewall is negligible.PREPARE The 2015 Pearson Education, Inc.Slide 8-26ladder is a rigid rod oflength L. To not slip, both the net forceand net torque on the ladder must bezero. FIGURE 8.9 on the next pageshows the ladder and the forces actingon it. We are asked to find thenecessary coefficient of static friction.Slide 8-27 2015 Pearson Education, Inc.Slide 8-28
Example 8.4 Will the ladder slip? (cont.)Example 8.4 Will the ladder slip? (cont.)First, we’ll solve for the magnitudesof the static friction force and thenormal force. Then we can use thesevalues to determine the necessaryvalue of the coefficient of friction.These forces both act at the bottomcorner of the ladder, so even thoughwe are interested in these forces, thisis a good choice for the pivot point because two of theforces that act provide no torque, which simplifies thesolution.With this choice of pivot, the weightof the ladder, acting at the center ofgravity, exerts torque d1w and theforce of the wall exerts torque d2n2.The signs are based on the observationthat would cause the ladder torotate counterclockwise, whilewould cause it to rotate clockwise. 2015 Pearson Education, Inc.Slide 8-29 2015 Pearson Education, Inc.Example 8.4 Will the ladder slip? (cont.)Example 8.4 Will the ladder slip? (cont.)SOLVE TheAltogether, we have three equationswith the three unknowns n1, n2, and fs.If we solve the third equation for n2,x- and y-components ofareThe torque about the bottom corner is 2015 Pearson Education, Inc.Slide 8-30we can then substitute this into thefirst equation to findSlide 8-31 2015 Pearson Education, Inc.Slide 8-32
Example 8.4 Will the ladder slip? (cont.)Example 8.4 Will the ladder slip? (cont.)Our model of static friction isfs fs max µsn1. We can find n1fromthe second equation: n1 Mg. Fromthis, the model of friction tells us thatASSESS Youknow from experiencethat you can lean a ladder or otherobject against a wall if the ground is“rough,” but it slips if the surface istoo smooth. 0.29 is a “medium” valuefor the coefficient of static friction,which is reasonable.Comparing these two expressions forfs, we see that µs must obeyThus the minimum value of the coefficient of static frictionis 0.29. 2015 Pearson Education, Inc.Slide 8-33 2015 Pearson Education, Inc.Slide 8-34Stability and Balance An extended object has a base of support on which it restswhen in static equilibrium. A wider base of support and/or a lower center ofgravity improves stability.Section 8.2 Stability and Balance 2015 Pearson Education, Inc. 2015 Pearson Education, Inc.Slide 8-36
Stability and Balance As long as the object’s center of gravity remains over thebase of support, torque due to gravity will rotate the objectback toward its stable equilibrium position. The object isstable. If the object’s center of gravity moves outside the base ofsupport, the object is unstable.Conceptual Example 8.5 How far to walk theplank?A cat walks along a plankthat extends out from a table.If the cat walks too far outon the plank, the plank willbegin to tilt. What determineswhen this happens?REASON Anobject is stableif its center of gravity lies over its base of support, andunstable otherwise. Let’s take the cat and the plank to beone combined object whose center of gravity lies along aline between the cat’s center of gravity and that of the plank. 2015 Pearson Education, Inc.Slide 8-37 2015 Pearson Education, Inc.Slide 8-38Conceptual Example 8.5 How far to walk theplank? (cont.)Conceptual Example 8.5 How far to walk theplank? (cont.)In FIGURE 8.12a, when thecat is near the left end of theplank, the combined centerof gravity is over the baseof support and the plankis stable. As the cat movesto the right, he reaches apoint where the combined center of gravity is directly overthe edge of the table, as shown in FIGURE 8.12b. If the cattakes one more step, the cat and plank will become unstableand the plank will begin to tilt.ASSESS 2015 Pearson Education, Inc.Slide 8-39Because the plank’scenter of gravity must be tothe left of the edge for it tobe stable by itself, the catcan actually walk a shortdistance out onto theunsupported part of theplank before it starts to tilt. The heavier the plank is, thefarther the cat can walk. 2015 Pearson Education, Inc.Slide 8-40
Example ProblemStability and Balance of the Human BodyA 2-m-long board weighing 50 N extends out over the edgeof a table, with 40% of the board’s length off the table. Howfar beyond the table edge can a 25-N cat walk before theboard begins to tilt? As a human moves, the body’s centerof gravity is constantly changing. 2015 Pearson Education, Inc.Slide 8-41 To maintain stability, peopleunconsciously adjust the positionsof their arms and legs to keep theircenter of gravity over their baseof support. 2015 Pearson Education, Inc.Slide 8-42Stability and Balance of the Human BodyTry It Yourself: Impossible Balance When the woman stands on hertiptoes, she leans forward, readjustingher center of gravity to be over theballs of her feet (her base of support).Stand facing a wall with your toes touching the base of thewall. Now rise onto your tiptoes. You will not be able to doso without falling backward. As we see from Figure 8.13b,your body has to lean forward to stand on tiptoes. With thewall in your way, you cannot lean enough to maintain yourbalance, and you will begin to topple backward. 2015 Pearson Education, Inc.Slide 8-43 2015 Pearson Education, Inc.Slide 8-44
Try It Yourself: Balancing a Soda CanTry to balance a soda can—full orempty—on the narrow bevel at thebottom. It can’t be done because,either full or empty, the center ofgravity is near the center of the can.If the can is tilted enough to sit onthe bevel, the center of gravity liesfar outside this small base of support.But if you put about 2 ounces (60 ml) of water in an emptycan, the center of gravity will be right over the bevel and thecan will balance. 2015 Pearson Education, Inc.Slide 8-45Section 8.3 Springs and Hooke’s Law 2015 Pearson Education, Inc.Springs and Hooke’s LawSprings and Hooke’s Law We have assumed that objectsin equilibrium maintain theirshapes as forces and torquesare applied to them. A restoring force is a force that restores a system to anequilibrium position. Systems that exhibit restoring forces are called elastic. Springs and rubber bands are basic examples of elasticity. This is an oversimplification;every solid object stretches,compresses, or deformswhen a force acts on it. 2015 Pearson Education, Inc.Slide 8-47 2015 Pearson Education, Inc.Slide 8-48
Springs and Hooke’s LawSprings and Hooke’s Law The spring force is proportional to the displacement ofthe end of the spring. The spring force and the displacement of the end of thespring have a linear relationship. The slope k of the line is called the spring constant andhas units of N/m.Slide 8-49 2015 Pearson Education, Inc. 2015 Pearson Education, Inc.Slide 8-50Springs and Hooke’s LawQuickCheck 8.3 Hooke’s law describes the most general form of therelationship between the restoring force and thedisplacement of the end of a spring.The restoring force of three springs is measured as they arestretched. Which spring has the largest spring constant? For motion in the vertical (y) direction, Hooke’s law is(Fsp)y –k y 2015 Pearson Education, Inc.Slide 8-51 2015 Pearson Education, Inc.Slide 8-52
QuickCheck 8.3Example 8.6 Weighing a fishThe restoring force of three springs is measured as they arestretched. Which spring has the largest spring constant?A scale used to weigh fish consists of a spring hung from asupport. The spring’s equilibrium length is 10.0 cm. When a4.0 kg fish is suspended from the end of the spring, itstretches to a length of 12.4 cm.a. What is the spring constant k for this spring?A.Steepest slope.Takes lots of force fora small displacement.b. If an 8.0 kg fish is suspended from the spring, what willbe the length of the spring?PREPARE Thevisual overview in FIGURE 8.15 shows thedetails for the first part of the problem. The fish hangs instatic equilibrium, so the net force in the y-direction and thenet torque must be zero. 2015 Pearson Education, Inc.Slide 8-53 2015 Pearson Education, Inc.Slide 8-54Example 8.6 Weighing a fish (cont.)Example 8.6 Weighing a fish (cont.)a. Because the fish isin static equilibrium, we haveFrom Figure 8.15, thedisplacement of the springfrom equilibrium is y yf – yi (–0.124 m) –(–0.100 m) –0.024 m. Thisdisplacement is negativebecause the fish moves in the –y-direction. We can nowsolve for the spring constant:SOLVEso that k –mg/ y. (The nettorque is zero because thefish’s center of gravity comes to rest directly under the pivotpoint of the hook.) 2015 Pearson Education, Inc.Slide 8-55 2015 Pearson Education, Inc.Slide 8-56
Example 8.6 Weighing a fish (cont.)Example 8.6 Weighing a fish (cont.)b. The restoring force isproportional to thedisplacement of thespring from its equilibriumlength. If we double themass (and thus the weight)of the fish, the displacementof the end of the spring will double as well, to y –0.048 m. Thus the spring will be 0.048 m longer,so its new length is 0.100 m 0.048 m 0.148 m 14.8 cm.ASSESS The 2015 Pearson Education, Inc.Slide 8-57spring doesn’t[Insert Figure 8.15stretch very much when a(repeated)]4.0 kg mass is hung from it.A large spring constant of1600 N/m thus seemsreasonable for this stiff spring. 2015 Pearson Education, Inc.Slide 8-58Example ProblemExample ProblemA 20-cm-long spring is attached to a wall. When pulledhorizontally with a force of 100 N, the spring stretches to alength of 22 cm. What is the value of the spring constant?A 20-cm-long spring is attached to a wall. When pulledhorizontally with a force of 100 N, the spring stretches to alength of 22 cm. The same spring is now suspended from ahook and a 10.2-kg block is attached to the bottom end.How long is the stretched spring? 2015 Pearson Education, Inc.Slide 8-59 2015 Pearson Education, Inc.Slide 8-60
Example ProblemA spring with spring constant k 125 N/m is used to pull a25 N wooden block horizontally across a tabletop. Thecoefficient of friction between the block and the table is µk 0.20. By how much does this spring stretch from itsequilibrium length?Section 8.4 Stretching andCompressing Materials 2015 Pearson Education, Inc.Slide 8-61 2015 Pearson Education, Inc.Stretching and Compressing MaterialsStretching and Compressing Materials We can model most solidmaterials as being made upof particle-like atoms connectedby spring-like bonds. Steel is elastic, but under normal forces it experiences onlysmall changes in dimension. Materials of this sort arecalled rigid. Rubber bands and other materials that can be stretchedeasily or show large deformations with small forces arecalled pliant. Pulling on a steel rod willslightly stretch the bondsbetween particles, and therod will stretch. 2015 Pearson Education, Inc.Slide 8-63 2015 Pearson Education, Inc.Slide 8-64
Stretching and Compressing MaterialsQuickCheck 8.4 For a rod, the springconstant depends on thecross-sectional area A,the length of the rod, L,and the material fromwhich it is made.Bars A and B are attached toa wall on the left and pulled withequal forces to the right. Bar B,with twice the radius, is stretchedhalf as far as bar A. Which has the larger value ofYoung’s modulus Y?A.B.C.D. The constant Y is called Young’s modulus and is aproperty of the material from which the rod is made. 2015 Pearson Education, Inc.Slide 8-65QuickCheck 8.4YA YBYA YBYA YBNot enough information to tell 2015 Pearson Education, Inc.Slide 8-66Stretching and Compressing MaterialsBars A and B are attached toa wall on the left and pulled withequal forces to the right. Bar B,with twice the radius, is stretchedhalf as far as bar A. Which has the larger value ofYoung’s modulus Y?A.B.C.D.YA YBArea of B increases by 4. If YB YA,YA YBF L stretch would be only L/4. Stretch ofY L/2 means B is “softer” than A.ALYA YBNot enough information to tell 2015 Pearson Education, Inc.Slide 8-67 2015 Pearson Education, Inc.Slide 8-68
Stretching and Compressing MaterialsExample 8.8 Finding the stretch of a wire The restoring force can be written in terms of the changein length L:A Foucault pendulum in a physicsdepartment (used to prove that theearth rotates) consists of a 120 kgsteel ball that swings at the end of a6.0-m-long steel cable. The cable hasa diameter of 2.5 mm. When the ballwas first hung from the cable, by howmuch did the cable stretch? It is useful to rearrange the equation in terms of two ratios,the stress and the strain: The unit of stress is N/m2 If stress is due to stretching, we call it tensile stress. 2015 Pearson Education, Inc.Slide 8-69 2015 Pearson Education, Inc.Example 8.8 Finding the stretch of a wire(cont.)Example 8.8 Finding the stretch of a wire(cont.)PREPARE TheSOLVEamount by which thecable stretches depends on theelasticity of the steel cable. Young’smodulus for steel is given in Table 8.1as Y 20 1010 N/m2.Slide 8-70Equation 8.6 relates the stretch of the cable L to therestoring force F and to the properties of the cable.Rearranging terms, we find that the cable stretches byThe cross-section area of the cable is 2015 Pearson Education, Inc.Slide 8-71 2015 Pearson Education, Inc.Slide 8-72
Example 8.8 Finding the stretch of a wire(cont.)Example 8.8 Finding the stretch of a wire(cont.)The restoring force of the cable is equal to the ball’s weight:ASSESSThe change in length is thus 2015 Pearson Education, Inc.Slide 8-73If you’ve ever strung a guitar with steel strings, youknow that the strings stretch several millimeters with theforce you can apply by turning the tuning pegs. So a stretchof 7 mm under a 120 kg load seems reasonable. 2015 Pearson Education, Inc.Beyond the Elastic LimitBeyond the Elastic Limit As long as the stretch stayswithin the linear region, asolid rod acts like a springand obeys Hooke’s law. For a rod or cable of a particular material, there is anultimate stress.Slide 8-74 The ultimate stress, or tensile strength, is the largeststress the material can sustain before breaking. As long as the stretch is lessthan the elastic limit, the rodreturns to its initial lengthwhen force is removed. The elastic limit is the end of the elastic region. 2015 Pearson Education, Inc.Slide 8-75 2015 Pearson Education, Inc.Slide 8-76
Biological MaterialsBiological Materials Most bones in your body aremade of two different kindsof bony material: dense andrigid cortical (or compactbone) on the outside, andporous, flexible cancellous(or spongy) bone on theinside. 2015 Pearson Education, Inc.Slide 8-77Biological MaterialsSlide 8-78 2015 Pearson Education, Inc.Summary: General PrinciplesText: p. 240 2015 Pearson Education, Inc.Slide 8-79 2015 Pearson Education, Inc.Slide 8-80
Summary: General PrinciplesSummary: Important ConceptsText: p. 240Text: p. 240Slide 8-81 2015 Pearson Education, Inc.Summary: Important ConceptsSummary: ApplicationsText: p. 240 2015 Pearson Education, Inc.Slide 8-82 2015 Pearson Education, Inc.Slide 8-83 2015 Pearson Education, Inc.Text: p. 240Slide 8-84
Summary: ApplicationsText: p. 240 2015 Pearson Education, Inc.Slide 8-85
Slide 8-9 Reading Question 8.3 An object will be stable if A. Its center of gravity is below its highest point. B. Its center of gravity lies over its base of support.
