cos(θ)sin(θ) Alternative Exercise Correct Correct θ = 0 skiladæmi 10 Part A Part B Part C Due: 11:59pm on Wednesday, November 11, 2015

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1 skiladæmi 10 Due: 11:59pm on Wednesday, November 11, 015 You will receive no credit for items you complete after the assignment is due Grading Policy Alternative Exercise 1115 A bar with cross sectional area A is subjected to equal and opposite tensile forces F at its ends Consider a plane through the bar making an angle θ with a plane at right angles to the bar (the figure ) Part A What is the tensile (normal) stress at this plane in terms of F, A, and θ? F cos (θ) A Part B What is the shear (tangential) stress at the plane in terms of F, A, and θ? F cos(θ)sin(θ) A Part C For what value of θ is the tensile stress a maximum? θ = 0
2 Part D For what value of θ is the shear stress a maximum? θ = 45 Alternative Exercise A moonshiner produces pure ethanol (ethyl alcohol) late at night and stores it in a stainless steel tank in the form of a cylinder 0310 m in diameter with a tight fitting piston at the top The total volume of the tank is 60 L In an attempt to squeeze a little more into the tank, the moonshiner piles lead breaks of total mass 1430 on top of the piston kg Part A What additional volume of ethanol can the moonshiner squeeze into the tank? (Assume that the wall of the tank is perfectly rigid) V = L ± A Wire under Stress A steel wire of length 01 with circular cross section must stretch no more than 000 when a tensile (stretching) force of 380 is applied to each end of the wire N m cm Part A What minimum diameter dmin is required for the wire? Express your answer in millimeters Take Young's modulus for steel to be Y = Pa Hint 1 How to approach the problem Recall that Young's modulus is defined as Y = tensile stress tensile strain Compute the strain in terms of quantities given in the problem introduction, and write the stress in terms of given quantities and the unknown wire diameter Use these along with the given value of Young's modulus for steel to solve for the diameter of the wire Hint Calculate the tensile strain
3 Calculate the tensile strain on the wire Hint 1 Definition of tensile strain Tensile strain is defined as the ratio of the elongation ΔL to the original length L0 of a material that is under stress: tensile strain= ΔL L0 tensile strain = Hint 3 Definition of tensile stress Tensile stress is defined as the force perpendicular to the surface of a material divided by the cross sectional area of the surface: tensile stress = F A Hint 4 Relation between the area and the diameter The relation between the area A and the diameter d of a circle is A = π d 4 dmin = 156 mm Note that you were asked for the minimum diameter Where does this figure? The extension is directly proportional to the stress, ie, the force per unit area One way to decrease the stress is to increase the surface area over which the stretching force is applied So any diameter (and so area) greater than the one you calculated would serve to keep the extension within the tolerance specified (ie, the maximum allowable extension) ± 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 F = kδx, where F is the magnitude of the stretching force, Δx is the corresponding elongation of the spring from equilibrium, and k 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
4 opposed to objects such as springs Such an expression does exist Consider, for instance, a bar of initial length L and cross sectional area A stressed by a force of magnitude F As a result, the bar stretches by ΔL Let us define two new terms: Tensile stress is the ratio of the stretching force to the cross sectional area: stress= F A Tensile strain is the ratio of the elongation of the rod to the initial length of the bar: strain= ΔL L 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 F/A Y = ΔL/L Part A What is the SI unit of Young's modulus? Hint 1 Look at the dimensions If you look at the dimensions of Young's modulus, you will see that they are equivalent to the dimension of pressure Use the SI unit of pressure Pa Part B Consider a metal bar of initial length L and cross sectional area A The Young's modulus of the material of the bar is Y Find the "spring constant" k of such a bar for low values of tensile strain Express your answer in terms of Y, L, and A Hint 1 Use the definition of Young's modulus Consider the equation defining Y Then isolate F and compare the result with Hooke's law: F = kδx k = Y A L Part C
5 Ten identical steel wires have equal lengths L and equal "spring constants" k The wires are connected end to end, so that the resultant wire has length What is the "spring constant" of the resulting wire? 10L Hint 1 The spring constant Use the expression for the spring constant determined in Part B From the expression derived in the Part B, you can determine what happens to the spring constant when the length of the spring increases 01k k 10k 100k Part D Ten identical steel wires have equal lengths L and equal "spring constants" k The wires are slightly twisted together, so that the resultant wire has length L and its cross sectional area is ten times that of the individual wire What is the "spring constant" of the resulting wire? Hint 1 The spring constant Use the expression for the spring constant determined in Part B From the expression derived in Part B, you can determine what happens to the spring constant when the area of the spring increases 01k k 10k 100k Part E Ten identical steel wires have equal lengths L and equal "spring constants" k The Young's modulus of each wire is Y The wires are connected end to end, so that the resultant wire has length 10L What is the Young's modulus of the resulting wire? 01Y Y 10Y 100Y
6 Part F Ten identical steel wires have equal lengths L and equal "spring constants" k The Young's modulus of each wire is Y The wires are slightly twisted together, so that the resultant wire has length L and is ten times as thick as the individual wire What is the Young's modulus of the resulting wire? 01Y Y 10Y 100Y 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 L = 100 A = 0500 Consider a steel guitar string of initial length meter and cross sectional area square millimeters The Young's modulus of the steel is Y = pascals How far ( ΔL) would such a string stretch under a tension of 1500 newtons? Use two significant figures in your answer Express your answer in millimeters ΔL = 15 mm 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 98 meters per second per second Use two significant figures in your answer, expressed in kilometers Hint 1 Why does the cable break? The cable breaks because of the stress exerted on it by its own weight At the moment that the breaking stress is reached, the stress at the top of the cable reaches its maximum, and the material begins to deteriorate
7 Introduce an arbitrary cross sectional area of the cable (which will cancel out of the final answer) The mass of the cable below the top point can be found as the product of its volume and its density Use this to find the force at the top that will lead to the breaking stress Hint Find the stress in the cable Assume that the cable has cross sectional area A and length L The density is ρ The maximum stress in the cable is at the very top, where it has to support its own weight What is this maximum stress? Express your answer in terms of ρ, L, and g, the magnitude of the acceleration due to gravity Recall that the stress is the force per unit area, so the area will not appear in your expression maximum stress = ρlg 6 km 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 Problem 1176 N Two identical, uniform beams weighing 60 each are connected at one end by a frictionless hinge A light horizontal crossbar attached at the midpoints of the beams maintains an angle of 530 between the beams The beams are suspended from the ceiling by vertical wires such that they form a " ", as shown in the figure V Part A What force does the crossbar exert on each beam? F = 130 N
8 Part B Is the crossbar under tension or compression? tension compression Part C What is the magnitude of the force that the hinge at point A exerts on each beam? F = 130 N Part D What is the direction of the force that the hinge at point A exerts on the right hand beam? ϕ = 180 with the direction to the right Part E What is the direction of the force that the hinge at point A exerts on the left hand beam? ϕ = 0 with the direction to the right Understanding Bernoulli's Equation Bernoulli's equation is a simple relation that can give useful insight into the balance among fluid pressure, flow speed, and elevation It applies exclusively to ideal fluids with steady flow, that is, fluids with a constant density and no internal friction
9 forces, whose flow patterns do not change with time Despite its limitations, however, Bernoulli's equation is an essential tool in understanding the behavior of fluids in many practical applications, from plumbing systems to the flight of airplanes For a fluid element of density ρ that flows along a streamline, Bernoulli's equation states that p 1 +ρg + ρ = +ρg + ρ h1 1 v 1 p h where p is the pressure, v is the flow speed, h is the height, g is the acceleration due to gravity, and subscripts 1 and refer to any two points along the streamline The physical interpretation of Bernoulli's equation becomes clearer if we rearrange the terms of the equation as follows: The term p 1 p on the left hand side represents the total work done on a unit volume of fluid by the pressure forces of the surrounding fluid to move that volume of fluid from point 1 to point The two terms on the right hand side represent, 1 respectively, the change in potential energy,, and the change in kinetic energy,, of the unit volume during its flow from point 1 to point In other words, Bernoulli's equation states that the work done on a unit volume of fluid by the surrounding fluid is equal to the sum of the change in potential and kinetic energy per unit volume that occurs during the flow This is nothing more than the statement of conservation of mechanical energy for an ideal fluid flowing along a streamline 1 v = ρg( )+ 1 ρ( ) p 1 p h h1 ρg( ) v v 1 h h1 ρ( ), v v 1 Part A Consider the portion of a flow tube shown in the figure Point 1 and point are at the same height An ideal fluid enters the flow tube at point 1 and moves steadily toward point If the cross section of the flow tube at point 1 is greater than that at point, what can you say about the pressure at point? Hint 1 How to approach the problem Apply Bernoulli's equation to point 1 and to point Since the points are both at the same height, their elevations cancel out in the equation and you are left with a relation between pressure and flow speeds Even though the problem does not give direct information on the flow speed along the flow tube, it does tell you that the cross section of the flow tube decreases as the fluid flows toward point Apply the continuity equation to points 1 and and determine whether the flow speed at point is greater than or smaller than the flow speed at point 1 With that information and Bernoulli's equation, you will be able to determine the pressure at point with respect to the pressure at point 1 Hint Apply Bernoulli's equation Apply Bernoulli's equation to point 1 and to point to complete the expression below Here p and v are the pressure and flow speed, respectively, and subscripts 1 and refer to point 1 and point Also, use h for elevation with the appropriate subscript, and use ρ for the density of the fluid Express your answer in terms of some or all of the variables p1, v1, h1, p, v, h, and ρ Hint 1 Flow along a horizontal streamline Along a horizontal streamline, the change in potential energy of the flowing fluid is zero In other words, when applying Bernoulli's equation to any two points of the streamline, h1 = h and they cancel out
10 + ρ p 1 1 v 1 = + ρv p Hint 3 Determine v with respect to v1 By applying the continuity equation, determine which of the following is true Hint 1 The continuity equation The continuity equation expresses conservation of mass for incompressible fluids flowing in a tube It says that the amount of fluid ΔV flowing through a cross section A of the tube in a time interval Δt must be the same for all cross sections, or ΔV Δt = = A1v1 Av Therefore, the flow speed must increase when the cross section of the flow tube decreases, and vice versa v v v > = < v1 v1 v1 The pressure at point is lower than the pressure at point 1 equal to the pressure at point 1 higher than the pressure at point 1 Thus, by combining the continuity equation and Bernoulli's equation, one can characterize the flow of an ideal fluidwhen the cross section of the flow tube decreases, the flow speed increases, and therefore the pressure decreases In other words, if A < A1, then v > v1 and p < p1 Part B As you found out in the previous part, Bernoulli's equation tells us that a fluid element that flows through a flow tube with decreasing cross section moves toward a region of lower pressure Physically, the pressure drop experienced by the fluid element between points 1 and acts on the fluid element as a net force that causes the fluid to Hint 1 Effects from conservation of mass Recall that, if the cross section A of the flow tube varies, the flow speed v must change to conserve mass This means that there is a nonzero net force acting on the fluid that causes the fluid to increase or decrease speed depending on whether the fluid is flowing through a portion of the tube with a smaller or larger cross section
11 decrease in speed increase in speed remain in equilibrium Part C Now assume that point is at height h with respect to point 1, as shown in the figure The ends of the flow tube have the same areas as the ends of the horizontal flow tube shown in Part A Since the cross section of the flow tube is decreasing, Bernoulli's equation tells us that a fluid element flowing toward point from point 1 moves toward a region of lower pressure In this case, what is the pressure drop experienced by the fluid element? Hint 1 How to approach the problem Apply Bernoulli's equation to point 1 and to point, as you did in Part A Note that this time you must take into account the difference in elevation between points 1 and Do you need to add this additional term to the other term representing the pressure drop between the two ends of the flow tube or do you subtract it? The pressure drop is smaller than the pressure drop occurring in a purely horizontal flow equal to the pressure drop occurring in a purely horizontal flow larger than the pressure drop occurring in a purely horizontal flow Part D From a physical point of view, how do you explain the fact that the pressure drop at the ends of the elevated flow tube from Part C is larger than the pressure drop occurring in the similar but purely horizontal flow from Part A? Hint 1 Physical meaning of the pressure drop in a tube As explained in the introduction, the difference in pressure p 1 p between the ends of a flow tube represents the total work done on a unit volume of fluid by the pressure forces of the surrounding fluid to move that volume of fluid from one end to the other end of the flow tube
12 A greater amount of work is needed to balance the increase in potential energy from the elevation change decrease in potential energy from the elevation change larger increase in kinetic energy larger decrease in kinetic energy In the case of purely horizontal flow, the difference in pressure between the two ends of the flow tube had to balance only the increase in kinetic energy resulting from the acceleration of the fluid In an elevated flow tube, the difference in pressure must also balance the increase in potential energy of the fluid; therefore a higher pressure is needed for the flow to occur Water Flowing from a Tank Water flows steadily from an open tank as shown in the figure The elevation of point 1 is 100 meters, and the elevation of points and 3 is 00 meters The cross sectional area at point is square meters; at point 3, where the water is discharged, it is square meters The cross sectional area of the tank is very large compared with the cross sectional area of the pipe Part A dv Assuming that Bernoulli's equation applies, compute the discharge rate dt Express your answer in cubic meters per second Hint 1 How to approach the problem The discharge rate is the rate at which a given volume of water flows across the exit of the pipe per unit time It is also defined as volume flow rate, and it depends on both the cross sectional area of the pipe at the exit and the fluid speed at that point Hint The volume flow rate Consider a steadily moving incompressible fluid, and let A denote the cross sectional area of a flow tube The volume ΔV of fluid flowing across the cross section of area A at speed v during a small interval of time Δt is given by AvΔt Therefore, the rate at which fluid volumes cross a portion of the flow tube is dv dt Hint 3 Find the fluid speed at the end of the pipe = Av
13 Assuming that Bernoulli's equation applies, find the speed v3 of the water at point 3 Recall that the area of the tank is very large compared to the cross sectional area of the pipe, and consequently, the velocity of water at a point on the surface of the water in the tank may be considered to be zero Express your answer in meters per second to three significant figures Hint 1 Apply Bernoulli's principle Let pa be the atmospheric pressure and ρ the density of water Consider the entire volume of water as a single flow tube and apply Bernoulli's principle to point 3 and to point 1 Complete the expression below, where v3 is the fluid speed at point 3 Express your answer in terms of p a, ρ, and g the free fall acceleration due to gravity Hint 1 Bernoulli's principle For the steady flow of an incompressible fluid with no internal friction, the pressure p and the flow speed v at depth H below the surface are linked by an important relationship, known as Bernoulli's principle In particular, at any point at depth H along a flow tube, the following relation holds: p +ρgh + 1 ρ v = constant, where ρ is the density of the fluid and g is the accereleration due to gravity Since Bernoulli's principle is valid at any point along a flow tube, it takes the form p 1 +ρg H ρ v 1 = p +ρg H + 1 ρv when applied to two distinct points along a flow tube The subscripts 1 and refer to such points p +ρg + 1 a ρv 3 = + 10ρg p a v3 = 15 m/s dv dt = 000 m 3 /s Part B What is the gauge pressure at point? Express your answer in pascals Hint 1 Definition of gauge pressure
14 Gauge pressure is defined as the excess pressure above atmospheric pressure Let p a pressure and p the total pressure of a fluid Then the gauge pressure is p p a be the atmospheric Hint How to approach the problem You can relate the fluid pressure at point with the atmospheric pressure by applying Bernoulli's principle to point and point 1, or alternatively to point and point 3 To determine the fluid speed at point you can use the continuity equation, using the fluid speed at the exit of the pipe found in Part A Hint 3 Apply Bernoulli's principle Consider the entire volume of water as a single flow tube Let p and v be respectively the pressure and the fluid speed at point Let the atmospheric pressure be p a and the density of water ρ Apply Bernoulli's principle to point 1 and point and complete the expression below Express your answer in terms of v, ρ, and g, the free fall acceleration Hint 1 Bernoulli's principle For the steady flow of an incompressible fluid with no internal friction, the pressure p and the flow speed v at depth H below the surface are linked by an important relationship, known as Bernoulli's principle In particular, at any point at depth H along a flow tube, the following relation holds: p +ρgh + 1 ρ v = constant, where ρ is the density of the fluid and g is the accereleration due to gravity Since Bernoulli's principle is valid at any point along a flow tube, it takes the form p 1 +ρg + ρ = +ρg + ρ H1 1 v 1 p H when applied to two distinct points along a flow tube The subscripts 1 and refer to such points 1 v p p a = 05ρ v + 8gρ Hint 4 Find the fluid speed at point Find v, the speed of the water at point Express your answer in meters per second to three significant figures Hint 1 The continuity equation In a steadily moving incompressible fluid, the mass of fluid flowing along a flow tube is constant In particular, consider a flow tube between two stationary cross sections with areas A1 and A Let v1 and v be the fluid speeds at these sections, respectively Then conservation of mass takes the form A1v1 which is known as the continuity equation = Av, v = 417 m/s Hint 5 Density of Water Recall that the density of water is 3
15 Recall that the density of water is 1000 kg/m Pa Score Summary: Your score on this assignment is 101% You received 707 out of a possible total of 7 points
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