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- Physics Homework #11 (SOLUTIONS reuploaded)

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4/18/09 2:19 AMMasteringPhysics Page 1 of 16http://session.masteringphysics.com/myct [ Assignment View ] Course PHY121(TTH) HW #11 Due at 11:59pm on Sunday, April 19, 2009 View Grading Details Unwinding Cylinder--Dynamics A cylinder with a moment of inertia (about its axis of symmetry), mass , and radius has a massless string wrapped around it which is tied to the ceiling . At time the cylinder is released from rest at height above the ground. Use for the magnitude of the acceleration of gravity. Assume that the string does not slip on the cylinder. Let represent the instantaneous velocity of the center of mass of the cylinder, and let represent the instantaneous angular velocity of the cylinder about its center of mass. Note that there are no horizontal forces present, so for this problem and . Part A The string constrains the rotational and translational motion of the falling cylinder, given that it doesn't slip. What is the relationship between the magnitude of the angular velocity and that of the velocity of the center of mass of the cylinder? Hint A.1Relating changes in the translational and rotational displacements Hint not displayed Express in terms of and . ANSWER: = Part B Let's look at some limiting cases as a way to build your intuition and also to check your answers. If you can't answer these questions now, work through other parts of this problem first and then consider these special cases using your final analytic answer. In the limit that the moment of inertia while the mass remains finite, what magnitudes would you expect for the tension in the vertical section of string and the downward acceleration of the center of mass? Note: This is a hypothetical cylinder with all its mass concentrated along its axis. The rest of the cylinder (i.e. the bulk) is massless. Hint B.1Torque considerations Hint not displayed Hint B.2Force considerations Hint not displayed Hint B.3Kinetic energy of rotation Hint not displayed Choose the option that best describes the limiting values of and under the conditions given. ANSWER: and and and and and and The string provides the torque that transfers gavitational potential energy into rotational kinetic energy. In the limit the kinetic energy of rotation as well, so in this case the string has no work to do. 4/18/09 2:19 AMMasteringPhysics Page 2 of 16http://session.masteringphysics.com/myct Part C Imagine that the string is wound around the center axle of a yo-yo; the axle radius is , but the yo-yo casing has a radius and moment of inertia . In the limit the moment of inertia of the yo-yo and the mass of the yo-yo remains finite, what magnitudes would you expect for the tension in the vertical section of string and the downward acceleration of the center of mass? Choose the option that best describes the limiting values of and under the conditions given. ANSWER: and and and and and and As the yo-yo becomes increasingly unresponsive (has a smaller angular acceleration) to any applied torque; it will seem like it cannot begin rotating. If it can't rotate, the string can't unwind, so the center of mass can't fall, and the net force on the yo-yo must be zero. Part D Using Newton's 2nd law, complete the equation of motion in the vertical direction that describes the translational motion of the cylinder. Express your answer in terms of the tension in the vertical section of string, , and ; a positive answer indicates an upward acceleration. ANSWER: Part E Using the equation of rotational motion and the definition of torque , complete the equation of rotational motion of the cylinder about its center of mass. Your answer should include the tension in the vertical section of string and the radius . A positive answer indicates a counterclockwise torque about the center of mass (in the direction). ANSWER: Part F In other parts of this problem expressions have been found for the vertical acceleration of the cylinder and the angular acceleration of the cylinder in the direction; both expressions include an unknown variable, namely, the tension in the vertical section of string The string constrains the rotational and vertical motions, providing a third equation relating and . Solve these three equations to find the vertical acceleration, , of the center of mass of the cylinder. Hint F.1 Find the relationship between linear and angular acceleration The string does not slip against the cylinder, so any vertical displacement corresponds to unwinding some string (a rotation displacement). Find the cylinder's vertical acceleration in terms of the angular acceleration . Hint F.1.1What equation to use Hint not displayed Express your answer in terms of and . ANSWER: = Hint F.2 Eliminate the angular acceleration algebraically. Use your answer from Part E and the previous hint to write an expression for that does not depend on . Your answer should depend on , , and . ANSWER: = Hint F.3 Putting it all together Now eliminate from the two equations containing and (One in the previous hint and one from the answer to part D). 4/18/09 2:19 AMMasteringPhysics Page 3 of 16http://session.masteringphysics.com/myct Express in terms of , , , and ; a positive answer indicates upward acceleration. ANSWER: = Part G Solve for the tension in the vertical section of string. Express in terms of the known variables , , , and . ANSWER: = Exercise 10.41: The spinning figure skater The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center . When his hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin-walled hollow cylinder. His hands and arms have a combined mass 9.0 . When outstretched, they span 1.8 ; when wrapped, they form a cylinder of radius 26 . The moment of inertia about the rotation axis of the remainder of his body is constant and equal to . Part A If his original angular speed is 0.40 , what is his final angular speed? Express your answer using two significant figures. ANSWER: = 1.1 Exercise 10.35 A 3.40 rock has a horizontal velocity of magnitude 12.0 when it is at point in the figure . Part A At this instant, what is the magnitude of its angular momentum relative to point ? ANSWER: = 196 Part B What is the direction of the momentum in part (A)? ANSWER: into the page out of the page 4/18/09 2:19 AMMasteringPhysics Page 4 of 16http://session.masteringphysics.com/myct Part C If the only force acting on the rock is its weight, what is the magnitude of the rate of change of its angular momentum at this instant? ANSWER: = 213 Part D What is the direction of the rate in part (A)? ANSWER: into the page out of the page A Rolling Hollow Sphere A hollow spherical shell with mass 1.80 rolls without slipping down a slope that makes an angle of 35.0 with the horizontal. Part A Find the magnitude of the acceleration of the center of mass of the spherical shell. Hint A.1How to approach the problem Hint not displayed Hint A.2Translational motion in the x direction Hint not displayed Hint A.3Torque on the spherical shell Hint not displayed Hint A.4Moment of inertia Hint not displayed Hint A.5Relation between the translational and angular accelerations Hint not displayed Take the free-fall acceleration to be = 9.80 . ANSWER: = 3.37 Part B Find the magnitude of the frictional force acting on the spherical shell. Hint B.1How to approach the problem Hint not displayed Take the free-fall acceleration to be = 9.80 . ANSWER: = 4.05 The frictional force keeps the spherical shell stuck to the surface of the slope, so that there is no slipping as it rolls down. If there were no friction, the shell would simply slide down the slope, as a rectangular box might do on an inclined (frictionless) surface. Part C Find the minimum coefficient of friction needed to prevent the spherical shell from slipping as it rolls down the slope. Hint C.1How to approach the problem Hint not displayed ANSWER: = 0.280 Exercise 10.40 A small block on a frictionless, horizontal surface has a mass of 2.30×10 ?2 . It is attached to a massless cord passing through a hole in the surface (the figure ). The block is originally revolving at a distance of 0.300 from the hole with an angular speed of 4/18/09 2:19 AMMasteringPhysics Page 5 of 16http://session.masteringphysics.com/myct 1.65 . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 . Model the block as a particle. Part A Is angular momentum of the block conserved? ANSWER: yes no Part B Why or why not? ANSWER: Answer not displayed Part C What is the new angular speed? ANSWER: = 6.60 Part D Find the change in kinetic energy of the block. ANSWER: = 8.45×10 ?3 Part E How much work was done in pulling the cord? ANSWER: 8.45×10 ?3 Problem 10.62 The mechanism shown in the figure is used to raise a crate of supplies from a ship's hold. The crate has total mass 53 . A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius 0.30 and a moment of inertia = 2.5 about the axle. The crate is suspended from the free end of the rope. One end of the axle pivots on frictionless bearings; a crank handle is attached to the other end. When the crank is turned, the end of the handle rotates about the axle in a vertical circle of radius 0.12 , the cylinder turns, and the crate is raised. Part A What magnitude of the force applied tangentially to the rotating crank is required to raise the crate with an acceleration of 0.85 ? (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.) Express your answer using two significant figures. ANSWER: = 1.5 Introduction to Static Equilibrium 4/18/09 2:19 AMMasteringPhysics Page 6 of 16http://session.masteringphysics.com/myct Learning Goal: To understand the conditions necessary for static equilibrium. Look around you, and you see a world at rest. The monitor, desk, and chair?and the building that contains them?are in a state described as static equilibrium. Indeed, it is the fundamental objective of many branches of engineering to maintain this state in spite of the presence of obvious forces imposed by gravity and static loads or the more unpredictable forces from wind and earthquakes. The condition of static equilibrium is equivalent to the statement that the bodies involved have neither linear nor angular acceleration. Hence static mechanical equilibrium (as opposed to thermal or electrical equilibrium) requires that the forces acting on a body simultaneously satisfy two conditions: and ; that is, both external forces and torques sum to zero. You have the freedom to choose any point as the origin about which to take torques. Each of these equations is a vector equation, so each represents three independent equations for a total of six. Thus to keep a table static requires not only that it neither slides across the floor nor lifts off from it, but also that it doesn't tilt about either the x or y axis, nor can it rotate about its vertical axis. Part A Frequently, attention in an equilibrium situation is confined to a plane. An example would be a ladder leaning against a wall, which is in danger of slipping only in the plane perpendicular to the ground and wall. By orienting a Cartesian coordinate system so that the x and y axes are in this plane, choose which of the following sets of quantities must be zero to maintain static equilibrium in this plane. Hint A.1Simplifying the equations Hint not displayed ANSWER: and and and and and and and and and and and Part B As an example, consider the case of a board of length and negligible mass. Take the x axis to be the horizontal axis along the board and the y axis to be the vertical axis perpendicular to the board. A mass of weight is strapped to the board a distance from the left-hand end. This is a static equilibrium problem, and a good first step is to write down the equation for the sum of all the forces in the y direction since the only nonzero forces of that exist are in the y direction. What is ? Your equation for the net force in the y direction on the board should contain all the forces acting vertically on the board. Express your answer in terms of the weight and the tensions in the two vertical ropes at the left and right ends and . Recall that positive forces point upward. ANSWER: The only relevant component of the torques is the z component; however, you must choose your pivot point before writing the equations. This point could be anywhere; in fact, the pivot point does not even have to be at a point on the body. You should choose this point to your advantage. Generally, the best place to locate the pivot point is where some unknown force acts; this will eliminate that force from the resulting torque equation. Part C What is the equation that results from choosing the pivot point to be the point from which the mass hangs (where acts)? Express your answer in terms of the unknown quantities and and the known lengths and . Recall that counterclockwise torque is positive. ANSWER: This gives us one equation involving two unknowns, and . We can use this result and to solve for and . Part D 4/18/09 2:19 AMMasteringPhysics Page 7 of 16http://session.masteringphysics.com/myct What is the equation that results from choosing the pivot point to be the left end of the plank (where acts)? Express your answer in terms of , , , and the dimensions and . Not all of these variables may show up in the solution. ANSWER: Part E What is the equation that results from choosing the pivot point to be the right end of the plank (where acts)? Express your answer in terms of , , , and the dimensions and . Not all of these variables may show up in the solution. ANSWER: Part F Solve for , the tension in the right rope. Hint F.1 Choose the correct equation Hint not displayed Express your answer in terms of and the dimensions and . Not all of these variables may show up in the solution. ANSWER: = Part G Solve for , the tension in the left rope. Hint G.1Choose the correct equation Hint not displayed Express your answer in terms of and the dimensions and . Not all of these variables may show up in the solution. ANSWER: = Part H Solve for the tension in the left rope, , in the special case that . Be sure the result checks with your intuition. Express your answer in terms of and the dimensions and . Not all of these variables may show up in the solution. ANSWER: = Only one set of forces, exactly balanced, produces static equilibrium. From this perspective it might seem puzzling that so much of the world is static. One must realize, however, that many forces?like those of the tensions in the ropes here or those between the floor and an object resting on it?increase very quickly as the object moves. If there is a slight imbalance of the forces, the object accelerates so that its position changes until the object has adjusted itself to restore the force balance. It then oscillates about this point until friction or some other dissipative mechanism causes it to become stationary at the exact equilibrium point. The Center of It All Learning Goal: To learn the definition of the center of mass for systems of particles and for extended objects and be able to locate it. Imagine throwing a rock upward and away from you. With negligible air resistance, the rock will follow a parabolic path before hitting the ground. Now imagine throwing a stick (or any other extended object). More likely than not, the stick will rotate throughout its flight; the motion of each point of the stick will be fairly complex. However, there will be one point that will follow the parabolic path, namely, the point about which the stick is rotating. Such a special point can be located experimentally, for instance, by videotaping the stick in flight and then analyzing its motion. The exciting part is that no matter how you throw the stick, such a special point will always exist, and will always be located at the same spot within the stick! The motion of the entire stick can then be described as a combination of the motion of that point, as if the entire mass of the stick were concentrated there, and the rotation of the stick about that point. Such a point, it turns out, exists for any rigid object or system of massive particles. It is called the center of mass. 4/18/09 2:19 AMMasteringPhysics Page 8 of 16http://session.masteringphysics.com/myct Locating the center of mass of an object seems a worthy task, because it helps in analyzing the motion of the object. However, it would be annoying and sometimes infeasible to videotape each object in motion in order to locate its center of mass. Fortunately, the location of the center of mass of an object or system can be calculated using the following set of formulas (note that we will usually consider cases in which the z coordinate is irrelevant): For a system of massive point particles that have coordinates and masses : , . If we replace the quantity by the total mass of the system , these formulas can be rewritten as follows: , In this problem, you will practice locating the center of mass for various systems of particles and for some simple extended objects. Part A Two particles of masses and ( ) are located 10 meters apart. Where is the center of mass of the system located? ANSWER: less than 5 meters from the particle of mass exactly 5 meters from the particle of mass more than 5 meters but less than 10 meters from the particle of mass more than 10 meters from the particle of mass Part B For the system of three particles shown, which have masses , , and as indicated, where is the center of mass located? ANSWER: to the left of the particle of mass between the particle of mass and the particle of mass between the particle of mass and the particle of mass to the right of the particle of mass Perhaps you could "feel" that the center of mass should be located to the right of the particle of mass , since the particle to the right of it has greater mass than the particle to the left. A calculation, however, allows one to pinpoint the exact location of the center of mass. Part C For the system of particles described in the Part B, find the x coordinate of the center of mass. Assume that the particle of mass is at the origin and the positive x axis is directed to the right. Express your answer in terms of . ANSWER: = 4/18/09 2:19 AMMasteringPhysics Page 9 of 16http://session.masteringphysics.com/myct Part D Let us now consider a two-dimensional case. The system includes three particles of equal mass located at the vertices of an isosceles triangle as shown in the figure. Which arrow best shows the location of the center of mass of the system? Do not calculate. ANSWER: 1 2 3 4 5 Part E What is the x coordinate of the center of mass of the system described in Part D? Express your answer in terms of . ANSWER: = From the symmetry of the situation, you can see that the x and y coordinates of the center of mass are the same. Part F A system of four buckets forms a square as shown in the figure. Initially, the buckets have different masses (it is not known how these masses are related). A student begins to add water gradually to the bucket located at the origin. As a result, what happens to the coordinates of the center of mass of the system of buckets? Hint F.1 The movement of the center of mass Hint not displayed ANSWER: The x coordinate stays the same; the y coordinate increases. The x coordinate stays the same; the y coordinate decreases. The x coordinate increases; the y coordinate stays the same. The x coordinate decreases; the y coordinate stays the same. The x coordinate increases; the y coordinate increases. The x coordinate decreases; the y coordinate decreases. The x coordinate stays the same; the y coordinate stays the same. There is not enough information to answer the question. Part G Find the x coordinate of the center of mass of the system of particles shown in the figure. 4/18/09 2:19 AMMasteringPhysics Page 10 of 16http://session.masteringphysics.com/myct Express your answer in meters to two significant figures. ANSWER: = 0.56 Part H Find the y coordinate of the center of mass of the system of particles described in the previous part. Express your answer in meters to two significant figures. ANSWER: = 0.40 For an extended object of mass , the sum in parentheses is replaced by an integral, and the formulas look like this: , , where is the infinitesimal mass element (the mass of a small area or volume in the limit as the area or volume goes to zero). For instance, if the density of an object is , then the infinitesimal mass element would be , where is an infinitesimal volume element and is evaluated at the point where is located. Frequently, the calculations are aided by the very powerful idea of symmetry, which plays a major role in all areas of physics. For instance, the center of mass of a uniform circle is at its center; that of a uniform straight rod is at the midpoint; etc. Note that in dealing with extended objects of irregular shape, it is often useful to use the idea of symmetry to locate the centers of mass of the symmetrical subparts and then treat these subparts as point particles located at their respective centers of mass. Part I Let us now consider an extended object. A wire is bent at its midpoint through a right angle as shown in the figure. Which arrow best shows the location of the center of mass of the system? Do not calculate. ANSWER: 1 2 3 4 5 For an extended object, the center of mass does not have to be within the object itself. This wire is one example; one can think of many others including bagels. The exact location of the center of mass can be found using the symmetry approach: It is located halfway between the midpoints of the two straight segments of the wire (can you see why?). Part J A straight rod has one end at the origin and the other end at the point and a linear density given by , where is a known constant and is the x coordinate. Since this wire is not uniform, you will have to use integrtation to solve this part. 4/18/09 2:19 AMMasteringPhysics Page 11 of 16http://session.masteringphysics.com/myct Use to find the total mass . Find for this rod. Hint J.1 How to approach the problem Hint not displayed Hint J.2 What is ? Hint not displayed Express your answer in terms of one or both of and . ANSWER: = There is another concept closely related to the center of mass, the center of gravity. This concept is useful, since gravity is a "distributed force": It is applied to each particle making up an object. Gravity is also the most ubiquitous force of all, which means that the net gravitational force almost always plays a role in determining an object's motion. The center of gravity is defined as the point at which the net force of gravity acting on the object is applied. The good news is that as long as the gravitational field is uniform, the locations of the center of mass and the center of gravity coincide. In practical terms, thsi means that the center of gravity and the center of mass coincide as long as the object's size is much less than that of Earth (or whatever planet or star provides the force of gravity in a particular situation). A Bar Suspended by Two Vertical Strings A rigid, uniform, horizontal bar of mass and length is supported by two identical massless strings. Both strings are vertical. String A is attached at a distance from the left end of the bar and is connected to the ceiling; string B is attached to the left end of the bar and is connected to the floor. A small block of mass is supported against gravity by the bar at a distance from the left end of the bar, as shown in the figure. Throughout this problem positive torque is that which spins an object counterclockwise. Use for the magnitude of the acceleration due to gravity. Part A Find , the tension in string A. Hint A.1Choosing an axis Hint not displayed Hint A.2Find the torque around the best axis Hint not displayed Hint A.3Summing the torques Hint not displayed Express the tension in string A in terms of , , , , , and . ANSWER: = Part B Find , the magnitude of the tension in string B. Hint B.1Two different methods to find Hint not displayed Hint B.2Direction of forces Hint not displayed Express the magnitude of the tension in string B in terms of , , , and . ANSWER: 4/18/09 2:19 AMMasteringPhysics Page 12 of 16http://session.masteringphysics.com/myct ANSWER: = Part C If the bar and block are too heavy the strings may break. Which of the two identical strings will break first? ANSWER: string A string B Part D If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of such that the bar remains stable (call it )? Hint D.1Nature of the unstable motion Hint not displayed Hint D.2Tension in string B at the critical point Hint not displayed Hint D.3Calculate the torques Hint not displayed Express your answer for in terms of , , , and . ANSWER: = Part E Note that since , as computed in the previous part, is not necessarily positive. If , the bar will be stable no matter where the block of mass is placed on it. Assuming that , , and are held fixed, what is the maximum block mass for which the bar will always be stable? In other words, what is the maximum block mass such that ? Hint E.1Requirement of stability If is calculated to be less than zero, the solution is unphysical. (The bar does not extend there to support it!) The minimum value that can have is obviously zero. If is less than the mass that would give then the bar will be stable for any physical value of . Answer in terms of , , and . ANSWER: = Answer not displayed Forces on a Bridge A bridge, constructed of 11 beams of equal length and negligible mass, supports an object of mass as shown. Real bridges of this sort have steel rockers at the ends (check one out sometime), these assure that the support forces on the bridge are vertical even when it expands or contracts thermally. Throughout this problem, use for the magnitude of the acceleration due to gravity. Part A Find , the vertical force that pier P exerts on the left end of the bridge. Hint A.1How to approach the problem Hint not displayed 4/18/09 2:19 AMMasteringPhysics Page 13 of 16http://session.masteringphysics.com/myct Hint A.2A suggested origin for determining torques Hint not displayed Hint A.3Find the sum of the torques Hint not displayed Express the vertical force at P in terms of and . ANSWER: = Answer not displayed Part B Part not displayed Part C Assuming that the bridge segments are free to pivot at each intersection point, what is the tension in the horizontal segment directly above the point where the object is attached? If you find that the horizontal segment directly above the point where the object is attached is being stretched, indicate this with a positive value for . If the segment is being compressed, indicate this with a negative value for . Hint C.1One way to start Hint not displayed Hint C.2Find the torque about a special point Hint not displayed Hint C.3How to find from the torque Hint not displayed Express the tension in terms of and . ANSWER: = Answer not displayed Young's Modulus Learning Goal: To understand the meaning of Young's modulus, to perform some real-life calculations related to stretching steel, a common construction material, and to introduce the concept of breaking stress. Hooke's law states that for springs and other "elastic" objects , where is the magnitude of the stretching force, is the corresponding elongation of the spring from equilibrium, and is a constant that depends on the geometry and the material of the spring. If the deformations are small enough, most materials, in fact, behave like springs: Their deformation is directly proportional to the external force. Therefore, it may be useful to operate with an expression that is similar to Hooke's law but describes the properties of various materials, as opposed to objects such as springs. Such an expression does exist. Consider, for instance, a bar of initial length and cross-sectional area stressed by a force of magnitude . As a result, the bar stretches by . Let us define two new terms: Tensile stress is the ratio of the stretching force to the cross- sectional area: . Tensile strain is the ratio of the elongation of the rod to the initial length of the bar: . 4/18/09 2:19 AMMasteringPhysics Page 14 of 16http://session.masteringphysics.com/myct It turns out that the ratio of the tensile stress to the tensile strain is a constant as long as the tensile stress is not too large. That constant, which is an inherent property of a material, is called Young's modulus and is given by Part A What is the SI unit of Young's modulus? Hint A.1Look at the dimensions Hint not displayed ANSWER: Part B Consider a metal bar of initial length and cross-sectional area . The Young's modulus of the material of the bar is . Find the "spring constant" of such a bar for low values of tensile strain. Hint B.1Use the definition of Young's modulus Hint not displayed Express your answer in terms of , , and . ANSWER: = Part C Ten identical steel wires have equal lengths and equal "spring constants" . The wires are connected end to end, so that the resultant wire has length . What is the "spring constant" of the resulting wire? Hint C.1The spring constant Hint not displayed ANSWER: Part D Ten identical steel wires have equal lengths and equal "spring constants" . The wires are slightly twisted together, so that the resultant wire has length and its cross-sectional area is ten times that of the individual wire. What is the "spring constant" of the resulting wire? Hint D.1The spring constant Hint not displayed ANSWER: Part E Ten identical steel wires have equal lengths and equal "spring constants" . The Young's modulus of each wire is . The wires are connected end to end, so that the resultant wire has length . What is the Young's modulus of the resulting wire? ANSWER: Part F Ten identical steel wires have equal lengths and equal "spring constants" . The Young's modulus of each wire is . The 4/18/09 2:19 AMMasteringPhysics Page 15 of 16http://session.masteringphysics.com/myct wires are slightly twisted together, so that the resultant wire has length and is ten times as thick as the individual wire. What is the Young's modulus of the resulting wire? ANSWER: By rearranging the wires, we create a new object with new mechanical properties. However, Young's modulus depends on the material, which remains unchanged. To change the Young's modulus, one would have to change the properties of the material itself, for instance by heating or cooling it. Part G Consider a steel guitar string of initial length meter and cross-sectional area square millimeters. The Young's modulus of the steel is pascals. How far ( ) would such a string stretch under a tension of 1500 newtons? Use two significant figures in your answer. Express your answer in millimeters. ANSWER: = 15 Steel is a very strong material. For these numeric values, you may assume that Hooke's law holds. However, for greater values of tensile strain, the material no longer behaves elastically. If the strain and stress are large enough, the material deteriorates. The final part of this problem illustrates this point and gives you a sense of the "stretching limit" of steel. Part H Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earth has proven much more difficult. The deepest mines ever drilled are only about 10 miles deep. To illustrate the difficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubic meter, and its breaking stress, defined as the maximum stress the material can bear without deteriorating, is about pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that the magnitude of the acceleration due to gravity remains constant at 9.8 meters per second per second. Hint H.1Why does the cable break? Hint not displayed Hint H.2Find the stress in the cable Hint not displayed Use two significant figures in your answer, expressed in kilometers. ANSWER: 26 This is only about 16 miles, and we have assumed that no extra load is attached. By the way, this length is small enough to justify the assumption of virtually constant acceleration due to gravity. When making such assumptions, one should always check their validity after obtaining a result. Stress on a Square Steel Plate A square steel plate is 11.0 on a side and 0.500 thick. Assume that the plate is oriented with the square faces horizontal to the ground. Part A Find the shear strain that results if a force of magnitude 9.00×10 5 is applied to the top square face of the plate, parallel to the side, and a force of equal magnitude is applied in the opposite direction to the bottom face of the plate as shown . Note that the shear modulus of steel in pascals is 7.50×10 10 . Hint A.1How to approach the problem Hint not displayed 4/18/09 2:19 AMMasteringPhysics Page 16 of 16http://session.masteringphysics.com/myct Hint A.2Calculate the shear stress Hint not displayed Express the strain numerically to three significant figures. ANSWER: Shear strain = 9.92×10 ?4 Part B Find the displacement in centimeters. Hint B.1Definition of shear strain Hint not displayed Express your answer in centimeters to three significant figures. ANSWER: = 4.96×10 ?4 Summary10 of 12 items complete (84.28% avg. score) 12.12 of 15 points clockwork MasteringPhysics

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