Lab Report : The purpose of this experiment is to learn about uncertainty in measurement, and how to perform calculations with those uncertainties. The calculation of the density of a wooden block was used as an example.

Lab Report : The purpose of this experiment is to learn about uncertainty in measurement, and how to perform calculations with those uncertainties. The calculation of the density of a wooden block was used as an example.

info@checkyourstudy.com Lab Report : The purpose of this experiment is … Read More...
Assignment 5 Due: 11:59pm on Wednesday, March 5, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 6.13 A hand presses down on the book in the figure. Part A Is the normal force of the table on the book larger than, smaller than, or equal to ? ANSWER: Correct mg Equal to Larger than Smaller than mg mg mg Problem 6.2 The three ropes in the figure are tied to a small, very light ring. Two of these ropes are anchored to walls at right angles with the tensions shown in the figure. Part A What is the magnitude of the tension in the third rope? Express your answer using two significant figures. ANSWER: Correct Part B What is the direction of the tension in the third rope? Express your answer using two significant figures. T  3 T3 = 94 N T  3 Typesetting math: 100% ANSWER: Correct The Normal Force When an object rests on a surface, there is always a force perpendicular to the surface; we call this the normal force, denoted by . The two questions to the right will explore the normal force. Part A A man attempts to pick up his suitcase of weight by pulling straight up on the handle. However, he is unable to lift the suitcase from the floor. Which statement about the magnitude of the normal force acting on the suitcase is true during the time that the man pulls upward on the suitcase? Hint 1. How to approach this problem First, identify the forces that act on the suitcase and draw a free-body diagram. Then use the fact that the suitcase is in equilibrium, , to examine how the forces acting on the suitcase relate to each other. Hint 2. Identify the correct free-body diagram Which of the figures represents the free-body diagram of the suitcase while the man is pulling on the handle with a force of magnitude ? = 58   below horizontal n ws n F = 0 fpull Typesetting math: 100% ANSWER: ANSWER: Correct Part B A B C D The magnitude of the normal force is equal to the magnitude of the weight of the suitcase. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull. The magnitude of the normal force is equal to the sum of the magnitude of the force of the pull and the magnitude of the suitcase’s weight. The magnitude of the normal force is greater than the magnitude of the weight of the suitcase. Typesetting math: 100% Now assume that the man of weight is tired and decides to sit on his suitcase. Which statement about the magnitude of the normal force acting on the suitcase is true during the time that the man is sitting on the suitcase? Hint 1. Identify the correct free-body diagram. Which of the figures represents the free-body diagram while the man is sitting atop the suitcase? Here the vector labeled is a force that has the same magnitude as the man’s weight. ANSWER: wm n wm Typesetting math: 100% ANSWER: Correct Recognize that the normal force acting on an object is not always equal to the weight of that object. This is an important point to understand. Problem 6.5 A construction worker with a weight of 880 stands on a roof that is sloped at 18 . Part A What is the magnitude of the normal force of the roof on the worker? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct A B C D The magnitude of the normal force is equal to the magnitude of the suitcase’s weight. The magnitude of the normal force is equal to the magnitude of the suitcase’s weight minus the magnitude of the man’s weight. The magnitude of the normal force is equal to the sum of the magnitude of the man’s weight and the magnitude of the suitcase’s weight. The magnitude of the normal force is less than the magnitude of the suitcase’s weight. N  n = 840 N Typesetting math: 100% Problem 6.6 In each of the two free-body diagrams, the forces are acting on a 3.0 object. Part A For diagram , find the value of , the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B For diagram the part A, find the value of the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: kg ax x ax = -0.67 m s2 ay, y Typesetting math: 100% Correct Part C For diagram , find the value of , the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D For diagram the part C, find the value of , the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: ay = 0 m s2 ax x ax = 0.67 m s2 ay y Typesetting math: 100% Correct Problem 6.7 In each of the two free-body diagrams, the forces are acting on a 3.0 object. Part A Find the value of , the component of the acceleration in diagram (a). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct ay = 0 m s2 kg ax x ax = 0.99 m s2 Typesetting math: 100% Part B Find the value of , the component of the acceleration in diagram (a). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C Find the value of , the component of the acceleration in diagram (b). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D Find the value of , the component of the acceleration in diagram (b). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct ay y ay = 0 m s2 ax x ax = -0.18 m s2 ay y ay = 0 m s2 Typesetting math: 100% Problem 6.10 A horizontal rope is tied to a 53.0 box on frictionless ice. What is the tension in the rope if: Part A The box is at rest? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part B The box moves at a steady = 4.80 ? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part C The box = 4.80 and = 4.60 ? Express your answer to three significant figures and include the appropriate units. ANSWER: kg T = 0 N vx m/s T = 0 N vx m/s ax m/s2 Typesetting math: 100% Correct Problem 6.14 It takes the elevator in a skyscraper 4.5 to reach its cruising speed of 11 . A 60 passenger gets aboard on the ground floor. Part A What is the passenger’s weight before the elevator starts moving? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the passenger’s weight while the elevator is speeding up? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the passenger’s weight after the elevator reaches its cruising speed? T = 244 N s m/s kg w = 590 N w = 730 N Typesetting math: 100% Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Block on an Incline A block lies on a plane raised an angle from the horizontal. Three forces act upon the block: , the force of gravity; , the normal force; and , the force of friction. The coefficient of friction is large enough to prevent the block from sliding . Part A Consider coordinate system a, with the x axis along the plane. Which forces lie along the axes? ANSWER: w = 590 N  F  w F n F  f Typesetting math: 100% Correct Part B Which forces lie along the axes of the coordinate system b, in which the y axis is vertical? ANSWER: Correct only only only and and and and and F  f F  n F  w F  f F  n F  f F  w F  n F w F  f F  n F w only only only and and and and and F  f F  n F  w F  f F  n F  f F  w F  n F w F  f F  n F w Typesetting math: 100% Usually the best advice is to choose coordinate system so that the acceleration of the system is directly along one of the coordinate axes. If the system isn’t accelerating, then you are better off choosing the coordinate system with the most vectors along the coordinate axes. But now you are going to ignore that advice. You will find the normal force, , using vertical coordinate system b. In these coordinates you will find the magnitude appearing in both the x and y equations, each multiplied by a trigonometric function. Part C Because the block is not moving, the sum of the y components of the forces acting on the block must be zero. Find an expression for the sum of the y components of the forces acting on the block, using coordinate system b. Express your answer in terms of some or all of the variables , , , and . Hint 1. Find the y component of Write an expression for , the y component of the force , using coordinate system b. Express your answer in terms of and . Hint 1. Some geometry help – a useful angle The smaller angle between and the y-axis is also , as shown in the figure. ANSWER: F  n Fn Fn Ff Fw  F n Fny F  n Fn  F  n  Typesetting math: 100% Hint 2. Find the y component of Write an expression for , the y component of the force , using coordinate system b. Express your answer in terms of and . Hint 1. Some geometry help – a useful angle The smaller angle between and the x-axis is also , as shown in the figure. ANSWER: ANSWER: Fny = Fncos() F f Ffy F f Ff  F  f  Ffy = Ffsin() Fy = 0 = Fncos() + Ffsin() − Fw Typesetting math: 100% Correct Part D Because the block is not moving, the sum of the x components of the forces acting on the block must be zero. Find an expression for the sum of the x components of the forces acting on the block, using coordinate system b. Express your answer in terms of some or all of the variables , , , and . Hint 1. Find the x component of Write an expression for , the x component of the force , using coordinate system b. Express your answer in terms of and . ANSWER: ANSWER: Correct Part E To find the magnitude of the normal force, you must express in terms of since is an unknown. Using the equations you found in the two previous parts, find an expression for involving and but not . Hint 1. How to approach the problem From your answers to the previous two parts you should have two force equations ( and ). Combine these equations to eliminate . The key is to multiply the Fn Ff Fw  F n Fnx F  n Fn  Fnx = −Fnsin() Fx = 0 = −Fnsin() + Ffcos() Fn Fw Ff Fn Fw  Ff Typesetting math: 100% Fy = 0 Fx = 0 Ff equation for the y components by and the equation for the x components by , then add or subtract the two equations to eliminate the term . An alternative motivation for the algebra is to eliminate the trig functions in front of by using the trig identity . At the very least this would result in an equation that is simple to solve for . ANSWER: Correct Congratulations on working this through. Now realize that in coordinate system a, which is aligned with the plane, the y-coordinate equation is , which leads immediately to the result obtained here for . CONCLUSION: A thoughtful examination of which coordinate system to choose can save a lot of algebra. Contact Forces Introduced Learning Goal: To introduce contact forces (normal and friction forces) and to understand that, except for friction forces under certain circumstances, these forces must be determined from: net Force = ma. Two solid objects cannot occupy the same space at the same time. Indeed, when the objects touch, they exert repulsive normal forces on each other, as well as frictional forces that resist their slipping relative to each other. These contact forces arise from a complex interplay between the electrostatic forces between the electrons and ions in the objects and the laws of quantum mechanics. As two surfaces are pushed together these forces increase exponentially over an atomic distance scale, easily becoming strong enough to distort the bulk material in the objects if they approach too close. In everyday experience, contact forces are limited by the deformation or acceleration of the objects, rather than by the fundamental interatomic forces. Hence, we can conclude the following: The magnitude of contact forces is determined by , that is, by the other forces on, and acceleration of, the contacting bodies. The only exception is that the frictional forces cannot exceed (although they can be smaller than this or even zero). Normal and friction forces Two types of contact forces operate in typical mechanics problems, the normal and frictional forces, usually designated by and (or , or something similar) respectively. These are the components of the overall contact force: perpendicular to and parallel to the plane of contact. Kinetic friction when surfaces slide cos  sin  Ff cos() sin() Fn sin2() + cos2 () = 1 Fn Fn = Fwcos() Fy = Fn − FW cos() = 0 Fn F = ma μn n f Ffric n f Typesetting math: 100% When one surface is sliding past the other, experiments show three things about the friction force (denoted ): The frictional force opposes the relative motion at the 1. point of contact, 2. is proportional to the normal force, and 3. the ratio of the magnitude of the frictional force to that of the normal force is fairly constant over a wide range of speeds. The constant of proportionality is called the coefficient of kinetic friction, often designated . As long as the sliding continues, the frictional force is then (valid when the surfaces slide by each other). Static friction when surfaces don’t slide When there is no relative motion of the surfaces, the frictional force can assume any value from zero up to a maximum , where is the coefficient of static friction. Invariably, is larger than , in agreement with the observation that when a force is large enough that something breaks loose and starts to slide, it often accelerates. The frictional force for surfaces with no relative motion is therefore (valid when the contacting surfaces have no relative motion). The actual magnitude and direction of the static friction force are such that it (together with other forces on the object) causes the object to remain motionless with respect to the contacting surface as long as the static friction force required does not exceed . The equation is valid only when the surfaces are on the verge of sliding. Part A When two objects slide by one another, which of the following statements about the force of friction between them, is true? ANSWER: Correct Part B fk fk μk fk = μkn μsn μs μs μk fs ! μsn μsn fs = μsn The frictional force is always equal to . The frictional force is always less than . The frictional force is determined by other forces on the objects so it can be either equal to or less than . μkn μkn μkn Typesetting math: 100% When two objects are in contact with no relative motion, which of the following statements about the frictional force between them, is true? ANSWER: Correct For static friction, the actual magnitude and direction of the friction force are such that it, together with any other forces present, will cause the object to have the observed acceleration. The magnitude of the force cannot exceed . If the magnitude of static friction needed to keep acceleration equal to zero exceeds , then the object will slide subject to the resistance of kinetic friction. Do not automatically assume that unless you are considering a situation in which the magnitude of the static friction force is as large as possible (i.e., when determining at what point an object will just begin to slip). Whether the actual magnitude of the friction force is 0, less than , or equal to depends on the magnitude of the other forces (if any) as well as the acceleration of the object through . Part C When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some point “break loose” and start to accelerate down the board. The box begins to slide once the component of gravity acting parallel to the board just begins to exceeds the maximum force of static friction. Which of the following is the most general explanation for why the box accelerates down the board? ANSWER: The frictional force is always equal to . The frictional force is always less than . The frictional force is determined by other forces on the objects so it can be either equal to or less than . μsn μsn μsn μsn μsn fs = μsn μsn μsn F = ma Fg The force of kinetic friction is smaller than that of maximum static friction, but remains the same. Once the box is moving, is smaller than the force of maximum static friction but larger than the force of kinetic friction. Once the box is moving, is larger than the force of maximum static friction. When the box is stationary, equals the force of static friction, but once the box starts moving, the sliding reduces the normal force, which in turn reduces the friction. Fg Fg Fg Fg Typesetting math: 100% Correct At the point when the box finally does “break loose,” you know that the component of the box’s weight that is parallel to the board just exceeds (i.e., this component of gravitational force on the box has just reached a magnitude such that the force of static friction, which has a maximum value of , can no longer oppose it.) For the box to then accelerate, there must be a net force on the box along the board. Thus, the component of the box’s weight parallel to the board must be greater than the force of kinetic friction. Therefore the force of kinetic friction must be less than the force of static friction which implies , as expected. Part D Consider a problem in which a car of mass is on a road tilted at an angle . The normal force Select the best answer. ANSWER: Correct The key point is that contact forces must be determined from Newton’s equation. In the problem described above, there is not enough information given to determine the normal force (e.g., the acceleration is unknown). Each of the answer options is valid under some conditions ( , the car is sliding down an icy incline, or the car is going around a banked turn), but in fact none is likely to be correct if there are other forces on the car or if the car is accelerating. Do not memorize values for the normal force valid in different problems–you must determine from . Problem 6.17 Bonnie and Clyde are sliding a 323 bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 375 of force while Bonnie pulls forward on a rope with 335 of force. μsn μsn μkn μsn μk < μs M  is found using n = Mg n = Mg cos() n = Mg cos() F  = Ma  = 0 n F = ma kg N N Typesetting math: 100% Part A What is the safe's coefficient of kinetic friction on the bank floor? ANSWER: Correct Problem 6.19 A crate is placed on a horizontal conveyor belt. The materials are such that and . Part A Draw a free-body diagram showing all the forces on the crate if the conveyer belt runs at constant speed. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: 0.224 10 kg μs = 0.5 μk = 0.3 Typesetting math: 100% Correct Part B Draw a free-body diagram showing all the forces on the crate if the conveyer belt is speeding up. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Typesetting math: 100% Correct Part C What is the maximum acceleration the belt can have without the crate slipping? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct amax = 4.9 m s2 Typesetting math: 100% Problem 6.28 A 1100 steel beam is supported by two ropes. Part A What is the tension in rope 1? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the tension in rope 2? Express your answer to two significant figures and include the appropriate units. ANSWER: kg T1 = 7000 N Typesetting math: 100% Correct Problem 6.35 The position of a 1.4 mass is given by , where is in seconds. Part A What is the net horizontal force on the mass at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the net horizontal force on the mass at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 6.39 T2 = 4800 N kg x = (2t3 − 3t2 )m t t = 0 s F = -8.4 N t = 1 s F = 8.4 N Typesetting math: 100% A rifle with a barrel length of 61 fires a 8 bullet with a horizontal speed of 400 . The bullet strikes a block of wood and penetrates to a depth of 11 . Part A What resistive force (assumed to be constant) does the wood exert on the bullet? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How long does it take the bullet to come to rest after entering the wood? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 6.45 You and your friend Peter are putting new shingles on a roof pitched at 21 . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.0 away, asks you for the box of nails. Rather than carry the 2.0 box of nails down to Peter, you decide to give the box a push and have it slide down to him. Part A If the coefficient of kinetic friction between the box and the roof is 0.55, with what speed should you push the box to have it gently come to rest right at the edge of the roof? Express your answer to two significant figures and include the appropriate units. cm g m/s cm fk = 5800 N = 5.5×10−4 t s  m kg Typesetting math: 100% ANSWER: Correct Problem 6.54 The 2.0 wood box in the figure slides down a vertical wood wall while you push on it at a 45 angle. Part A What magnitude of force should you apply to cause the box to slide down at a constant speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct v = 3.9 ms kg  F = 23 N Typesetting math: 100% Score Summary: Your score on this assignment is 98.8%. You received 114.57 out of a possible total of 116 points. Typesetting math: 100%

Assignment 5 Due: 11:59pm on Wednesday, March 5, 2014 You will receive no credit for items you complete after the assignment is due. Grading Policy Conceptual Question 6.13 A hand presses down on the book in the figure. Part A Is the normal force of the table on the book larger than, smaller than, or equal to ? ANSWER: Correct mg Equal to Larger than Smaller than mg mg mg Problem 6.2 The three ropes in the figure are tied to a small, very light ring. Two of these ropes are anchored to walls at right angles with the tensions shown in the figure. Part A What is the magnitude of the tension in the third rope? Express your answer using two significant figures. ANSWER: Correct Part B What is the direction of the tension in the third rope? Express your answer using two significant figures. T  3 T3 = 94 N T  3 Typesetting math: 100% ANSWER: Correct The Normal Force When an object rests on a surface, there is always a force perpendicular to the surface; we call this the normal force, denoted by . The two questions to the right will explore the normal force. Part A A man attempts to pick up his suitcase of weight by pulling straight up on the handle. However, he is unable to lift the suitcase from the floor. Which statement about the magnitude of the normal force acting on the suitcase is true during the time that the man pulls upward on the suitcase? Hint 1. How to approach this problem First, identify the forces that act on the suitcase and draw a free-body diagram. Then use the fact that the suitcase is in equilibrium, , to examine how the forces acting on the suitcase relate to each other. Hint 2. Identify the correct free-body diagram Which of the figures represents the free-body diagram of the suitcase while the man is pulling on the handle with a force of magnitude ? = 58   below horizontal n ws n F = 0 fpull Typesetting math: 100% ANSWER: ANSWER: Correct Part B A B C D The magnitude of the normal force is equal to the magnitude of the weight of the suitcase. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull. The magnitude of the normal force is equal to the sum of the magnitude of the force of the pull and the magnitude of the suitcase’s weight. The magnitude of the normal force is greater than the magnitude of the weight of the suitcase. Typesetting math: 100% Now assume that the man of weight is tired and decides to sit on his suitcase. Which statement about the magnitude of the normal force acting on the suitcase is true during the time that the man is sitting on the suitcase? Hint 1. Identify the correct free-body diagram. Which of the figures represents the free-body diagram while the man is sitting atop the suitcase? Here the vector labeled is a force that has the same magnitude as the man’s weight. ANSWER: wm n wm Typesetting math: 100% ANSWER: Correct Recognize that the normal force acting on an object is not always equal to the weight of that object. This is an important point to understand. Problem 6.5 A construction worker with a weight of 880 stands on a roof that is sloped at 18 . Part A What is the magnitude of the normal force of the roof on the worker? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct A B C D The magnitude of the normal force is equal to the magnitude of the suitcase’s weight. The magnitude of the normal force is equal to the magnitude of the suitcase’s weight minus the magnitude of the man’s weight. The magnitude of the normal force is equal to the sum of the magnitude of the man’s weight and the magnitude of the suitcase’s weight. The magnitude of the normal force is less than the magnitude of the suitcase’s weight. N  n = 840 N Typesetting math: 100% Problem 6.6 In each of the two free-body diagrams, the forces are acting on a 3.0 object. Part A For diagram , find the value of , the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B For diagram the part A, find the value of the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: kg ax x ax = -0.67 m s2 ay, y Typesetting math: 100% Correct Part C For diagram , find the value of , the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D For diagram the part C, find the value of , the -component of the acceleration. Express your answer to two significant figures and include the appropriate units. ANSWER: ay = 0 m s2 ax x ax = 0.67 m s2 ay y Typesetting math: 100% Correct Problem 6.7 In each of the two free-body diagrams, the forces are acting on a 3.0 object. Part A Find the value of , the component of the acceleration in diagram (a). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct ay = 0 m s2 kg ax x ax = 0.99 m s2 Typesetting math: 100% Part B Find the value of , the component of the acceleration in diagram (a). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C Find the value of , the component of the acceleration in diagram (b). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part D Find the value of , the component of the acceleration in diagram (b). Express your answer to two significant figures and include the appropriate units. ANSWER: Correct ay y ay = 0 m s2 ax x ax = -0.18 m s2 ay y ay = 0 m s2 Typesetting math: 100% Problem 6.10 A horizontal rope is tied to a 53.0 box on frictionless ice. What is the tension in the rope if: Part A The box is at rest? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part B The box moves at a steady = 4.80 ? Express your answer as an integer and include the appropriate units. ANSWER: Correct Part C The box = 4.80 and = 4.60 ? Express your answer to three significant figures and include the appropriate units. ANSWER: kg T = 0 N vx m/s T = 0 N vx m/s ax m/s2 Typesetting math: 100% Correct Problem 6.14 It takes the elevator in a skyscraper 4.5 to reach its cruising speed of 11 . A 60 passenger gets aboard on the ground floor. Part A What is the passenger’s weight before the elevator starts moving? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the passenger’s weight while the elevator is speeding up? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part C What is the passenger’s weight after the elevator reaches its cruising speed? T = 244 N s m/s kg w = 590 N w = 730 N Typesetting math: 100% Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Block on an Incline A block lies on a plane raised an angle from the horizontal. Three forces act upon the block: , the force of gravity; , the normal force; and , the force of friction. The coefficient of friction is large enough to prevent the block from sliding . Part A Consider coordinate system a, with the x axis along the plane. Which forces lie along the axes? ANSWER: w = 590 N  F  w F n F  f Typesetting math: 100% Correct Part B Which forces lie along the axes of the coordinate system b, in which the y axis is vertical? ANSWER: Correct only only only and and and and and F  f F  n F  w F  f F  n F  f F  w F  n F w F  f F  n F w only only only and and and and and F  f F  n F  w F  f F  n F  f F  w F  n F w F  f F  n F w Typesetting math: 100% Usually the best advice is to choose coordinate system so that the acceleration of the system is directly along one of the coordinate axes. If the system isn’t accelerating, then you are better off choosing the coordinate system with the most vectors along the coordinate axes. But now you are going to ignore that advice. You will find the normal force, , using vertical coordinate system b. In these coordinates you will find the magnitude appearing in both the x and y equations, each multiplied by a trigonometric function. Part C Because the block is not moving, the sum of the y components of the forces acting on the block must be zero. Find an expression for the sum of the y components of the forces acting on the block, using coordinate system b. Express your answer in terms of some or all of the variables , , , and . Hint 1. Find the y component of Write an expression for , the y component of the force , using coordinate system b. Express your answer in terms of and . Hint 1. Some geometry help – a useful angle The smaller angle between and the y-axis is also , as shown in the figure. ANSWER: F  n Fn Fn Ff Fw  F n Fny F  n Fn  F  n  Typesetting math: 100% Hint 2. Find the y component of Write an expression for , the y component of the force , using coordinate system b. Express your answer in terms of and . Hint 1. Some geometry help – a useful angle The smaller angle between and the x-axis is also , as shown in the figure. ANSWER: ANSWER: Fny = Fncos() F f Ffy F f Ff  F  f  Ffy = Ffsin() Fy = 0 = Fncos() + Ffsin() − Fw Typesetting math: 100% Correct Part D Because the block is not moving, the sum of the x components of the forces acting on the block must be zero. Find an expression for the sum of the x components of the forces acting on the block, using coordinate system b. Express your answer in terms of some or all of the variables , , , and . Hint 1. Find the x component of Write an expression for , the x component of the force , using coordinate system b. Express your answer in terms of and . ANSWER: ANSWER: Correct Part E To find the magnitude of the normal force, you must express in terms of since is an unknown. Using the equations you found in the two previous parts, find an expression for involving and but not . Hint 1. How to approach the problem From your answers to the previous two parts you should have two force equations ( and ). Combine these equations to eliminate . The key is to multiply the Fn Ff Fw  F n Fnx F  n Fn  Fnx = −Fnsin() Fx = 0 = −Fnsin() + Ffcos() Fn Fw Ff Fn Fw  Ff Typesetting math: 100% Fy = 0 Fx = 0 Ff equation for the y components by and the equation for the x components by , then add or subtract the two equations to eliminate the term . An alternative motivation for the algebra is to eliminate the trig functions in front of by using the trig identity . At the very least this would result in an equation that is simple to solve for . ANSWER: Correct Congratulations on working this through. Now realize that in coordinate system a, which is aligned with the plane, the y-coordinate equation is , which leads immediately to the result obtained here for . CONCLUSION: A thoughtful examination of which coordinate system to choose can save a lot of algebra. Contact Forces Introduced Learning Goal: To introduce contact forces (normal and friction forces) and to understand that, except for friction forces under certain circumstances, these forces must be determined from: net Force = ma. Two solid objects cannot occupy the same space at the same time. Indeed, when the objects touch, they exert repulsive normal forces on each other, as well as frictional forces that resist their slipping relative to each other. These contact forces arise from a complex interplay between the electrostatic forces between the electrons and ions in the objects and the laws of quantum mechanics. As two surfaces are pushed together these forces increase exponentially over an atomic distance scale, easily becoming strong enough to distort the bulk material in the objects if they approach too close. In everyday experience, contact forces are limited by the deformation or acceleration of the objects, rather than by the fundamental interatomic forces. Hence, we can conclude the following: The magnitude of contact forces is determined by , that is, by the other forces on, and acceleration of, the contacting bodies. The only exception is that the frictional forces cannot exceed (although they can be smaller than this or even zero). Normal and friction forces Two types of contact forces operate in typical mechanics problems, the normal and frictional forces, usually designated by and (or , or something similar) respectively. These are the components of the overall contact force: perpendicular to and parallel to the plane of contact. Kinetic friction when surfaces slide cos  sin  Ff cos() sin() Fn sin2() + cos2 () = 1 Fn Fn = Fwcos() Fy = Fn − FW cos() = 0 Fn F = ma μn n f Ffric n f Typesetting math: 100% When one surface is sliding past the other, experiments show three things about the friction force (denoted ): The frictional force opposes the relative motion at the 1. point of contact, 2. is proportional to the normal force, and 3. the ratio of the magnitude of the frictional force to that of the normal force is fairly constant over a wide range of speeds. The constant of proportionality is called the coefficient of kinetic friction, often designated . As long as the sliding continues, the frictional force is then (valid when the surfaces slide by each other). Static friction when surfaces don’t slide When there is no relative motion of the surfaces, the frictional force can assume any value from zero up to a maximum , where is the coefficient of static friction. Invariably, is larger than , in agreement with the observation that when a force is large enough that something breaks loose and starts to slide, it often accelerates. The frictional force for surfaces with no relative motion is therefore (valid when the contacting surfaces have no relative motion). The actual magnitude and direction of the static friction force are such that it (together with other forces on the object) causes the object to remain motionless with respect to the contacting surface as long as the static friction force required does not exceed . The equation is valid only when the surfaces are on the verge of sliding. Part A When two objects slide by one another, which of the following statements about the force of friction between them, is true? ANSWER: Correct Part B fk fk μk fk = μkn μsn μs μs μk fs ! μsn μsn fs = μsn The frictional force is always equal to . The frictional force is always less than . The frictional force is determined by other forces on the objects so it can be either equal to or less than . μkn μkn μkn Typesetting math: 100% When two objects are in contact with no relative motion, which of the following statements about the frictional force between them, is true? ANSWER: Correct For static friction, the actual magnitude and direction of the friction force are such that it, together with any other forces present, will cause the object to have the observed acceleration. The magnitude of the force cannot exceed . If the magnitude of static friction needed to keep acceleration equal to zero exceeds , then the object will slide subject to the resistance of kinetic friction. Do not automatically assume that unless you are considering a situation in which the magnitude of the static friction force is as large as possible (i.e., when determining at what point an object will just begin to slip). Whether the actual magnitude of the friction force is 0, less than , or equal to depends on the magnitude of the other forces (if any) as well as the acceleration of the object through . Part C When a board with a box on it is slowly tilted to larger and larger angle, common experience shows that the box will at some point “break loose” and start to accelerate down the board. The box begins to slide once the component of gravity acting parallel to the board just begins to exceeds the maximum force of static friction. Which of the following is the most general explanation for why the box accelerates down the board? ANSWER: The frictional force is always equal to . The frictional force is always less than . The frictional force is determined by other forces on the objects so it can be either equal to or less than . μsn μsn μsn μsn μsn fs = μsn μsn μsn F = ma Fg The force of kinetic friction is smaller than that of maximum static friction, but remains the same. Once the box is moving, is smaller than the force of maximum static friction but larger than the force of kinetic friction. Once the box is moving, is larger than the force of maximum static friction. When the box is stationary, equals the force of static friction, but once the box starts moving, the sliding reduces the normal force, which in turn reduces the friction. Fg Fg Fg Fg Typesetting math: 100% Correct At the point when the box finally does “break loose,” you know that the component of the box’s weight that is parallel to the board just exceeds (i.e., this component of gravitational force on the box has just reached a magnitude such that the force of static friction, which has a maximum value of , can no longer oppose it.) For the box to then accelerate, there must be a net force on the box along the board. Thus, the component of the box’s weight parallel to the board must be greater than the force of kinetic friction. Therefore the force of kinetic friction must be less than the force of static friction which implies , as expected. Part D Consider a problem in which a car of mass is on a road tilted at an angle . The normal force Select the best answer. ANSWER: Correct The key point is that contact forces must be determined from Newton’s equation. In the problem described above, there is not enough information given to determine the normal force (e.g., the acceleration is unknown). Each of the answer options is valid under some conditions ( , the car is sliding down an icy incline, or the car is going around a banked turn), but in fact none is likely to be correct if there are other forces on the car or if the car is accelerating. Do not memorize values for the normal force valid in different problems–you must determine from . Problem 6.17 Bonnie and Clyde are sliding a 323 bank safe across the floor to their getaway car. The safe slides with a constant speed if Clyde pushes from behind with 375 of force while Bonnie pulls forward on a rope with 335 of force. μsn μsn μkn μsn μk < μs M  is found using n = Mg n = Mg cos() n = Mg cos() F  = Ma  = 0 n F = ma kg N N Typesetting math: 100% Part A What is the safe's coefficient of kinetic friction on the bank floor? ANSWER: Correct Problem 6.19 A crate is placed on a horizontal conveyor belt. The materials are such that and . Part A Draw a free-body diagram showing all the forces on the crate if the conveyer belt runs at constant speed. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: 0.224 10 kg μs = 0.5 μk = 0.3 Typesetting math: 100% Correct Part B Draw a free-body diagram showing all the forces on the crate if the conveyer belt is speeding up. Draw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded. ANSWER: Typesetting math: 100% Correct Part C What is the maximum acceleration the belt can have without the crate slipping? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct amax = 4.9 m s2 Typesetting math: 100% Problem 6.28 A 1100 steel beam is supported by two ropes. Part A What is the tension in rope 1? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the tension in rope 2? Express your answer to two significant figures and include the appropriate units. ANSWER: kg T1 = 7000 N Typesetting math: 100% Correct Problem 6.35 The position of a 1.4 mass is given by , where is in seconds. Part A What is the net horizontal force on the mass at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B What is the net horizontal force on the mass at ? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 6.39 T2 = 4800 N kg x = (2t3 − 3t2 )m t t = 0 s F = -8.4 N t = 1 s F = 8.4 N Typesetting math: 100% A rifle with a barrel length of 61 fires a 8 bullet with a horizontal speed of 400 . The bullet strikes a block of wood and penetrates to a depth of 11 . Part A What resistive force (assumed to be constant) does the wood exert on the bullet? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Part B How long does it take the bullet to come to rest after entering the wood? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct Problem 6.45 You and your friend Peter are putting new shingles on a roof pitched at 21 . You're sitting on the very top of the roof when Peter, who is at the edge of the roof directly below you, 5.0 away, asks you for the box of nails. Rather than carry the 2.0 box of nails down to Peter, you decide to give the box a push and have it slide down to him. Part A If the coefficient of kinetic friction between the box and the roof is 0.55, with what speed should you push the box to have it gently come to rest right at the edge of the roof? Express your answer to two significant figures and include the appropriate units. cm g m/s cm fk = 5800 N = 5.5×10−4 t s  m kg Typesetting math: 100% ANSWER: Correct Problem 6.54 The 2.0 wood box in the figure slides down a vertical wood wall while you push on it at a 45 angle. Part A What magnitude of force should you apply to cause the box to slide down at a constant speed? Express your answer to two significant figures and include the appropriate units. ANSWER: Correct v = 3.9 ms kg  F = 23 N Typesetting math: 100% Score Summary: Your score on this assignment is 98.8%. You received 114.57 out of a possible total of 116 points. Typesetting math: 100%

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