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# GENERAL PHYSICS PHYS 2001

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This 80 page Class Notes was uploaded by Elva Fahey on Tuesday October 13, 2015. The Class Notes belongs to PHYS 2001 at Louisiana State University taught by Staff in Fall. Since its upload, it has received 26 views. For similar materials see /class/223006/phys-2001-louisiana-state-university in Physics 2 at Louisiana State University.

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Date Created: 10/13/15

Review chapters 789 v1 Exam on Wednesday night 15 April 600 pm sharp Students with serious con icts should send me an email by tomorrow night not wed afternoon stating reason Con ict exam will start at 500 pm 0 Will cover chapters 789 not 10 Chapter 7 Impulse and momentum Impulse de ned f Fm Linear momentum of one particle I m5 Impulse momentum theorem I A Newton s 2nd again 7 M 7 At total momentum when 2 Fm 07 momentum conserved7 so AP07 or 3013 center of mass 2mm 0M 7 CM mtot Ptot UCM 1 dim 2 my mtot Chapter 8 Rotational Kinematics angular displacement average angular velocity average angular acceleration A66f 00 im At aamp At equations When angular acceleration is constant tangential variables centripetal acceleration wfw0at 1 2 A0w0t at w tug ZQAO sr0 UTrw aTroz 2 v aC irwg T Chapter 9 Rigid objects in equilibrium torque r distance from axis to F 739 Frsin6 Fl rotational equilbrium the 3rd equation 2 T 0 center of gravity aka center of mass 2 VV z 956M VVtot moment of inertia I mrg point I Imb rigid see table I Ital mrg rigid point etc Newton7s 2nd law for rotation 7I0z Rotational work WR 7396 Rotational kinetic energy 1 KER Iw2 2 total mechanical energy 1 1 Em Emu2 EIQJ2 mgh if WNC 0 and 76 0 then tot energy conserved Etot 0 Etot f angular momentum L I w if E Tm 0 angular momentum is conserved Iowo Iftdf 4 Question DetailsCJ7 7P010 568708 A student m 63 kg falls freely from rest and strikes the ground During the collision with the ground he comes to rest in a time of00140 s The average force exerted on him by the ground is 16000 N where the upward direction is taken to be the positive direction From what height did the student fall Assume that the only force acting on him during the collision is that due to the ground 0645 m lull g l39 i 6 lO l 4 COMECk finalquot 1 1quot i7 l val 1quot la quotf39 M A 1r T gim N V1 i l if K sf 39 20 h 33 H quot infant A e 4 lull 39v I 68er Ms l f 39 a 7 I N 5 U m up 3 fDE YY r Ql u law 1 i NM MU 7 it i E ll 5 Ml l l W 1 l J L V l 4 Question Details CJ7 7P029 568683 A 250 g bullet traveling at a speed of420 ms strikes the wooden block ofa ballistic pendulum as shown The block has a mass of 270 g m 1 m w 7 9 quLe seed of me bullelblock comblnatlon lmmedlalely after me colllslon M b How mg comblnatlon rlse above lls lmtlal posmonv m MM new a Win vmWL A MUM gt gt A 7 r 070 Vic 95quot W 7 T 39 m 238m Q xz0 3 ml Q653m 2 651 4 zo 1 WM 0 f wmmm mmk W1Qwa 1 3 miquot 2511 39 1990 2900 t 3339 b F 39E L G165 W 7 Question DetailsCJ7 9P020 568768 end A W 1860 N A Crate hangs from the far end of the beam Iw liailn 5 a l n a Catcutate the maqmtude of the tehston m the we 222D N at the beam Fx Fy FEM mgmf amm 7 4924332544031 17 my l V f g fi f i mm 5 quotT WE TME S E ITZBGTZOW M v quot 77quot 70 443 wbuw Hugo 3 ng i A s 5 a f rV 7A r 7 Li li Wig 7 if x f Ch 4 Forces and Newton s Laws If I have seen far it is because I have stood on the shoulders of giants Isaac Newton In the previous chapters we used the concepts of displacement velocity and acceleration to study the motion of objects Now we want to know how the object moves and why it moves or just as importantly why an object doesn t move To do this we need to develop the concept of a Force Simply put a Force in physics is a push or a pull on an object We will discover that there are several types of forces that can act on an object but they all fall into two broad categories Contact forces and Non contact forces But there is one commonality among all forces They each have the ability to change the motion of an object When an object falls and hits the ground it remains at rest Aristotle 350 BC It wasn t until the 1600 s that these ideas were expanded upon Objects continue with uniform motion if no external forces act on them Galileo 1609 1687 Newton publishes his Principia or The Mathematical Principles of Natural Philosophy Newton was born on Christmas in 1642 the same year Galileo died He was born premature and not expected to survive He enrolled in Trinity College at Cambridge when he was 18 There he studied math and optics After graduation he lived in seclusion in the countryside to avoid the second Plague in Europe Over the next year and a half he did the following Developed the binomial theorem of mathematics Began his study of mechanics Developed the Universal Law of Gravitation Analyzed the decomposition of light into its spectrum Developed both forms of calculus integral and differential When he finished he was 25 In this course we will be interested in his studies of mechanics and gravitation Newton expanded on Galileo s ideas about the motion of an object in the absences of forces An object at rest tends to stay at rest and an object in motion tends to stay in motion at a constant speed moving in a straight line unless acted upon by some external force Newton s First Law The property of objects that make them obey Newton s First Law is something called lnertia Newton s First Law is often referred to as the Law of Inertia Inertia is a measure of how much mass an object has In other words how much matter it contains For example a lead brick has a lot more mass and thus inertia than does a penny The quantitative measure of inertia is Mass ln SI units mass comes in kilograms kg Key point If the net force on an object is zero then the object will remain at rest or continue on at constant velocity Thus if there is a net force acting on an object then the object s vzlocity WILL v change Thus the object must be accelerating a T So forces are synonymous with accelerations In other words a force is a push or a pull on an object that can overcome the object s inertia and produce an acceleration Now for a given force what determines how much acceleration I m going to get It s the object s INERTIA The more inertia and thus mass an object has the more it resists a change in its motion 2 F I ma Newton s 2nd Law This is Newton s 2nd Law In words it says The sum of the forces acting on an object is equal to its mass times its acceleration Notice that forces are vectors They require both magnitude and direction to completely describe them Units kgm MassAcceleration 41 82 N Newton It takes about 1 N of force to push a stamp on an envelope Frames of reference Newton s First Law and 2nd Law may be invalid depending on an observer s frame of reference Your frame of reference or reference frame is your perspective from which you observe a physical system In physics we attach a set of coordinate axes to our frame of reference so that we can measure things like position velocity and acceleration For example let s say you are standing at rest watching a passenger train go by you at constant speed I m sitting in my seat on the train with my lap tray down as I enjoy my favorite cold beverage An ice cube from my drink is sitting on the tray in front of me As the train goes by you we both agree that the ice cube is obeying Newton s First Law You see the ice cube continue to move in a straight line at constant speed Relative to me see the ice cube remain at rest But now the train slams on its brakes What do we see The train comes to a halt but the ice cube continues to move at the same speed before the train s deceleration now sliding forward on my tray You standing next to the train would say that the ice cube continues to obey Newton s First Law it continues on in straight line motion at constant speed I however would say that some force has been applied to the ice cube causing it to accelerate and move forward on my tray When actually there is no net force on the ice cube So who is right How can the ice cube obey Newton s First Law in one reference frame but not the other It turns out that Newton s Laws are only valid in Inertial Reference Frames An Inertial Reference Frame is one in which the acceleration of the frame is zero You are standing on the Earth which is a good approximation of an inertial reference frame so when the trains stops you see the ice cube keep moving at constant speed thereby obeying Newton s First Law I am attached to the reference frame of the train which has a nonzero acceleration when the brakes are applied From my point of view it looks like a force has been applied to the ice cube when actually there is no force I am in a noninertial reference fr Concept Question True or False While driving in a circle at constant speed I am in an inertial reference frame 1 True 2 In practice we often refer to Newton s 2quot Law as just We do this however with the following F ma understanding 1 That this is a vector equation 2 That F means the net total vector sum of all the forces that act on the object Cl m Now it is easy to see that the more force I apply to an object the greater its acceleration We can rewrite Newton s 2ndl Law this way But the more massive an object is the more it resists acceleration Example A 71 0kg crate is pulled along a horizontal frictionless surface The puller applies a force of 300 N at an angle of 40 wrt the horizontal What is the acceleration of the crate Example A 71 0kg crate is pulled along a horizontal frictionless surface The puller applies a force of 300 N at an angle of 400 wrt the horizontal What is the acceleration of the crate Draw a figure Since there is no friction the crate will slide in the positive x direction when the force is applied Soto calculate the acceleration in the xdirection ax we need the component of the force in the x dll eCthn Let s breakF down into its 96 and y components FX Fcos400 300 Ncos 400 230 N 5 7 23W 7 m 7710kg7 Example Two forces F1 and F2 act on the 700kg block shown in the drawing The magnitudes of the forces are F1 450 N and F2 250 N What is the horizontal acceleration magnitude and direction of the block Again assume the block is sitting on a horizontal frictionless surface We have the figure let s choose axes directions The acceleration in the horizontal direction is ax To calculate this we need the net force in the xdirection Then we can use Newton39s 2quotd Law x Break F1 down into its A L V L F1 E00570 154N iFlsin70 423N F2 already lIes along an aXIs F2 in 7250N 0N ZFX 154 N 250 N 2 The horizontal acceleration is ax 137 ms m 700 kg 137 ms2 in the negative x direction Newton s 3ml Law Law of Action and Reaction When an object exerts a force on a second object the second object exerts an equal but opposite force on the first Example Two ice skaters are standing facetoface on frictionless ice The skater on the left has a mass of 500 kg and pushes the skater on the right with a force of 400 N The skater on the right has a mass of 700 kg What is the acceleration of both skaters m 50Woo kg F40N fV F4ON g e By Newton s 3rel Law the skater on the right pushes back against the skater on the left with an equal but opposite force Fm 400 N Skater on the left artgm gm 2 70 0 k 057 ms2 has a larger mrl ghl g acceleration since F her mass is ale W 2 40 0 N 080 ms2 smaller m 500 kg left 46 Fundamental Forces Currently there are 3 fundamental forces 1 Gravitational Force 2 Strong Nuclear Force 3 Electroweak Force With the exception of the gravitational force all the forces we will study this semester are related to the third fundamental force the electroweak force A few years ago there were considered to be 4 fundamental forces gravity electromagnetc strong nuclear and weak nuclear Many theoretical scientists have spent much of their careers trying to reduce the of fundamental forces down to one Now let s study the gravitational force 47 The Gravitational Force It s May 16 1666 and Sir Isaac Newton is outside contemplating how the moon revolves around the Earth While thinking he notices an apple fall from a nearby tree and strike the ground At that moment he realized that the same pull that made the apple fall to the ground must also apply to the moon The moon moves in a nearly perfect circle around the earth Since it s moving in a circle its velocity is changing Since its velocity is changing there must be an acceleration Since there is an acceleration there must be a force The same force that pulls the apple toward the earth must also pull the moon toward the earth This is the Gravitational Force If gravity didn t pull on the moon it would continue to move in a straight line The gravitational force from the earth pulls the moon away from its straightline motion Newton s Universal Law of Gravitation So How do you calculate the gravitational force between two objects Let s consider two objects with masses m and m2 and separate them by some distance r m2 m1 F r F mlmz O t lt 0 Then F G r2 Notice that the force is directly proportional to the product of the masses of the two objects and inversely proportional to the square of the distance between them r is the distance between the centers of each mass The gravitational force equation is called an inverse square law Nature loves inverse square laws and we will encounter them again 2 11 G is the universal gravitational constant G 667259x10 Properties of Newton s Law of Gravitation Note The forces on the It grows weaker with distance two masses are equal but It gets stronger for increasing masses opposite as dictated by Newton s 3lrd Law 1 2 3 It is always attractive 4 It s directed along the line containing the two masses Concept Question If the sun suddenly collapsed to become a black hole the earth would 1 Leave the solar system in a straightline path 2 Spiral into the black hole 3 Undergo a major increase in tidal forces Continue to move in it usual orbit 5 What is the acceleration of a tennis ball falling near the surface of the earth It s mlsz but Mass of the ball Mass of the earth mBmE We can calculate the force on the ball F G z rE Radius of the earth But we can also write the force on the ball using Newton s 2nd Law m These two forces have to be equal a G FE mE a G 2 This must be the acceleration of the ball Let s calculate it I E 2 24 a667X1011 N m 598gtlt10 kg kg 638x1o6m2 ms g COOL Thus the acceleration due to gravity depends on the mass of the earth and the distance I am from the earth s center If I m standing on the earth s surface then this distance is just the radius of the earth Now we can calculate the acceleration of gravity on any planet or body m 735gtlt1022 kg Let s take the moon for example 61mm GM 2 G 16 ms2 r2 174x106 m2 moon A amoon N 6 aearth Notice also that the acceleration decreases with the square of the distance On the surface of the earth the acceleration due to gravity is 98 msz but if I move farther away from the surface the acceleration decreases So the acceleration due to gravity on top of Mount Everest is less than it is on the surface at sea level Remember 9 is an acceleration The force due to gravity is called weight You weigh less on the moon since the acceleration due to the moon s gravity is less than that of the earth You weigh less on the top of Mount Everest too Near the surface of the earth where g 98 msz an apple falls 49 m in 1 second However where the moon is the acceleration due to earth s gravity is only about 00027 msz so the moon only falls about 1 mm toward the earth each second Concept Question A 1kg projectile is launched into the air with an initial speed of 25 ms at a launch angle of 55 What is the magnitude of the force on the projectile when it s halfway to the peak in its trajectory 1 O N 2 456 N 3 242 N 4 Mass and Weight Remember your weight W mg is the force on you due to gravity It can change depending on the value ofg ie whether you re on earth or the moon etc Your mass m is a measure of how much matter your body contains how many atoms you are made of Your mass never changes It s the same on the moon on the earth on mars etc Example The drawing below shows one possible alignment forthe sun earth and moon The gravitational force Fm that the sun exerts on the moon is perpendicular to the gravitational force FEM that the earth exerts on the moon Determine the magnitude and direction of the net gravitational force on the moon rSM150gtlt1011m Moon Q A FSM zM 735 X 1022 kg FEMl 7 mS199gtlt1030kg rEM385X108m i 5 5 98 x 1024 k Earth mE I g rSM 150gtlt1011m Moon mM 735 x1022 kg ms 199 x 1030 kg rEM 385 x108 m quot6 m5 598 X 1024 kg Earth First let s draw in the resultant vector FNet To calculate FNet we need to find FSM and FEM and then use Pythagorean Thm FSM w 434x1020 N VSM Fm JFSjl FM 477x1020 N mEmM 20 FEM G rEzM 13998X10 N Fm is 477 x 1020 N directed at 2450 below the horizontal line Direction tan FE M 2 6 tan1FE M between the sun and SM FSM moon Before we study more forces let s discuss Free Body Diagrams FBD A FBD is a drawing or sketch that shows the fundamental forces that act on an object The object or body is drawn free from the rest of its surroundings Often times the body is represented as a single point located at its center of mass In dynamics problems with multiple forces acting on a body the FBD is extremely important So we should really amend our list of problem solving steps as follows 1 Relax breathe It s just a word problem 2 Draw a gure for the problem 3 Draw a FBD for the problem and label the forces 0 O 48 The Normal Force So we ve been saying that it s important to understand how and why an object moves under the influence of forces Just as important is why an object doesn t move Let s take the example of a concrete block sitting at rest on the top of a table Since the block is not moving and thus not accelerating there must be no forces acting on the block right Wrong There are forces acting on the block Newton s 2nd Law tells us that if the object is not accelerating then the sum of the forces 2F must be zero Therefore whatever forces there are acting on the block must sum to zero ie they cancel each other out So what forces do we have We must have the weight of the block There must be another force one to cancel the weight What is it It s the table pushing back up on the block This is the Normal Force FE The Normal Force cancels the weight so the net force on the block is zero Normal Force The component of a force that a surface exerts on an object which is perpendicular to the surface Whenever there is an object in contact with a surface that surface will apply a normal force to the object Let s look at the block on the table some more What would the FBD look like FN y Define y upward We only have forces in the ydirection Newton s 2nd Law says ZFy 2 may 2 0 since there is no acceleration ZFy 2 FN W 0 gt FN W Does the normal force always equal the weight No The normal force equals the weight if the object is sitting on a horizontal surface and no other vertical forces act except gravity What if the surface is not horizontal Example A 60kg box is placed on a frictionless ramp that is inclined at an angle of 30 wrt the horizontal a What is the normal force acting on the box b What force must be applied to the box along the inclined plane to hold the box in place v3 46 gt What forces do we have acting on the box Weight which is always downward And the normal force which is always perpendicular to the surface of contact We can orient our axes however we wish but a real convenient way is to set x up the ramp and then y will be perpendicular to the ramp Draw FBD x Now break the vectors down in the FBD into their x and ycomponents FN is done since it lies along the yaxis but Wcan be broken down xcomponent ycomponent FN 0 FN Wsin3 O0 Wcos300 Now we can apply Newton s 2nd Law for each direction xcomponent ycomponent FN 0 FN W Wsin3 0 Wcos3 0 Since there is no acceleration in the ydirection the sum of the forces in that direction will be zero ZFy 2 FN Wcos300 0 gt FN Woos300 mg cos 300 60 kg98ms2 cos300 b In order to keep the box in place I must apply a force F that is equal but opposite to the component of the weight directed down the ramp in this case Wx F sz W sin 300 mg sin 39060 kg98 ms2 sin 300 294 N Physical insight from previous example In the previous example we found out that when you have an object on a frictionless inclined plane the normal force on the object is equal to the y component of the weight and the force required to keep the box from sliding down the ramp is equal to the xcomponent of the weight F 2 W COS 9 Let s see if our answers make sense in the limiting N cases where 6 0 and 6 90 F W sin 6 6 90 6 0 FNWcos6WcosOW FNWcos6Wcos9OOO FWsinlt9WSin00 FWsin6Wsin90oW The normal force goes to zero It s no The normal force equa5 the welghfl longer in contact with the ramp The object quot 5 herlzehfa The Obleet WOh t Shde is in freefall so the force required to down a herlzehtal ramp maintain its position is its entire weight Apparent Weight We have all had the experience of feeling heavier or lighter when traveling in an elevator If the elevator is at rest and then accelerates upward we feel heavier If the elevator is at rest and then accelerates downward we feel lighter The motion of the elevator can give rise to an apparent weight which is different from our actual weight Why The sensation of weight that we feel comes from the force that the floor of the elevator applies to our feet That is it s the normal force from the elevator floor a No acceleration b Upward acceleration c Downward acceleration d Freefall How do we calculate apparent weight We use Newton s 2nd Law Let s look at the FBD for the guy in the elevator J We have his weight W downward and the normal gquot force FN from the floor upward FN Newton s 2nd Law tells us that Z Fy FN W may 1 Here 139 is the acceleration of the man and elevator FN is the apparent weight that we feel It s the normal force against our feet F N Wapparent 739 3 I W mg WaPParem W may mg may 3 Wapparent ay I Notice if ay 0 Le the elevator is not accelerating then WWWquot mg your true weight Remember g is always a positive number but ay could be positive or negative If I cut the rope holding the elevator then it accelerates downward at ay g 39 l and your apparent weIght goes to zero Youre in freefaquot Concept Question An elevator is moving upward at constant speed If you weigh 700 N on the scale when the elevator is at rest then the scale now reads 1 700 N 2 Less than 700 N 3 More than 700 N 4 O N If the elevator is moving straight up at constant speed then its acceleration is zero Einstein s Theory of General Relativity The Principle of Equivalence lfl drop an apple on the I measur the same surface of the earth I acceleration on the apple measure its acceleration Opposite the direction of the to be equal to g 98 msz elevator s acceleration Repeat the same experiment in an elevatorthat is moving upward with an acceleration g 98 mls2 and the apple is held fixed You can t tell the difference between the acceleration of gravity and the acceleration in some other reference frame like an elevator The Principle of Equivalence A uniformly accelerated reference frame is exactly equivalent to a gravitational field in the opposite direction W A helium filled balloon is tied by a string to the floor of a minivan The minivan is moving to the left at constant speed What happens when the driver slams on the brakes Gravitational Field 49 Frictional Forces We ve been discussing forces on objects who have surfaces in contact The Normal Force is perpendicular to the surface There is another force due to surfaces in contact in which the force is parallel to the surface This is call the Frictional Force We will often just refer to this force as Friction Static Friction Friction comes In two types Kinetic Friction The Frictional Force tends to oppose the direction of motion ie it tries to slow the object down Consider the situation of a block sitting on a table If I want to slide the block across the table then I have to overcome a frictional force just to get it to move static friction And then I have to overcome another frictional force to keep it moving That is the kinetic friction or friction with motion Let s say I want to try and pull the block to the right As start to pull with some force F the block doesn t move because of the static frictional force fs Notice that fs is parallel to the surface of contact and it opposes the direction that I m trying to pull the block So I keep making my pulling force F bigger and bigger fs keeps getting bigger and bigger too canceling out my pulling force untilfs reaches its maximum value Once I go past this point so that F is now greater thanfs the block breaks free and starts to slide Now that the block is sliding no longer have to overcome the static frictional force fs but I have to overcome the kinetic frictional force fk to keep the block moving Static Frictional Force Force you must overcome to get an object at rest to move maximum static frictional force Kinetic Frictional Force Force you must overcome to keep an object moving f k LlCF N us and gk are the coefficients of static and kinetic friction respectively us and uk are material dependent ie it depends on what the two surfaces are that are in contact F l p r exam e Rubber on concrete pk 080 Rubber on wet concrete pk 025 It requires more force to get an object at rest to move then it does to keep it moving Concept Question A speeding truck slams on its brakes and skids to a stop If the truck was heavily loaded so that its mass was doubled what can be said about the stopping distance 1 It s the samd 2 1 12times as far 3 Twice as far 4 4times as far Now let s revisit the block on the inclined plane again but this time let s include friction Example A 60kg block is at rest on an inclined plane that makes an angle of 30 wrt the horizontal What is the coefficient of static friction between the block and the plane if the static frictional force is at its maximum value Let s draw the FBD The block is not accelerating so I know the following must be true by Newton s 2nd Law Zszmaxzo Zszfs Wx20 yx ZFymay0 ZFyzFN Wy20 gt ysF Wsin gtFN m0 Combine the two equations lady COS 6 ySine gtys tan6 gtys 2058 43 423 410 The Tension Force Forces are often applied to objects by ropes or cables Pulling a box for example with a rope This gives rise to a force T called tension We will assume that the ropes are massless unless otherwise stated That way the force we apply to the rope is equal to the force transmitted to the object In other words no force is needed to accelerate the rope Massless ropes transfer the pulling force to the object Example A block with a weight of 422 N sits on a horizontal table connected to a hanging mass M The blocks are connected by a massless rope over a frictionless and massless pulley lfthe coefficient of static friction between the block and the table is 075 what is the maximum mass for M2 such that the 422N block remains stationary Set up a coordinate system Now we need to draw the FBD Whenever you have a mechanical system involving multiple masses objects you should draw a FBD for each ob39ect Let s call the mass on the table M1 Notice the tension on each mass is the same since they are connected by the same rope and our pulleys are Mass on the table Hanging mass idea39 Now we can apply Newton s 2nd law to each mass and in each direction Mass on the table X Hanging mass T W2 ZFy FNl WI 0 The sum of the forces in the y direction for mass 1 is zero since the mass does not accelerate in that direction ZszT fszOgtTfs DVZIUSFM lusIVl The sum of the forces in the xdirection for mass 1 is also zero since we don t want the mass to move so T f sets that condition The sum of the forces in the y direction for mass 2 is zero since the mass does not accelerate in that direction From the two boxed equations on the right they both equal the tension so I can set them equal to each other usz 075422 1 PL 2 I k us 1 2g 2 g 98ms2 DOES THE SCALE READ 1OON ZOON OR ZERO lN BOTH CASES AN APPLIED FORCE o 100 N ACCELERATES THE 100 N BLOCK 411 Equilibrium Conditions An object is in eguilibrium when its acceleration is zero Thismeansthat 21321716720 In 2 dimensions then 2 Fx 2 0 Z Fy 0 The forces acting on an object must balance in each direction Remember zero acceleration does not necessarily mean no motion As long as the velocity does not change in time which means that a system moving at constant speed could be in equilibrium 412 Nonequilibrium Conditions If a system is in nonequilibrium then it must be accelerating By Newton s 2nd Law then 2 F 2 me In 2 dimensions then 2 Fx 2 max ZFy 2 may All the problems we will do can be classified as either an equilibrium or a nonequilibrium problem If it s an equilibrium problem then the sum of the forces in both the x and y direction will be zero If it s a nonequilibrium problem then the sum of the forces in the xdirection will be max and the sum of the forces in the ydirection will be may Example The steel lbeam in the drawing has a weight of 80 kN and is being lifted upward at a constant velocity What is the tension in each cable attached to its ends Is this an equilibrium or nonequilibrium problem It is an equilibrium problem since the beam is not accelerating Therefore ZFX 0 ZFy 0 Choose a coordinate system Draw the FBD Now break the vectors down into their components W is along y so it s done but T needs to be broken down 2 Tx Tx 0 ZFyTyTy W0 32TyW 70 From the figure Ty T sin 70quot W 8000 N 260N 25in 70quot 25in 70 2Tsin70 W 3T Example Ignore all frictional effects and assume the rope and pulley are massless What is the acceleration of the two blocks and the tension in the rope When the 185N block is hung on the rope this system will begin to accelerate since there is no friction Therefore this is a noneguilibrium problem W2 Thus the nonequilibrium conditions apply ZFX max ZFJ may Choose coordinate axes Draw the FBD for each mass W accelerates in the xdirection and VPquot W accelerates in the ydirection Thus the following must be true FNI T 11 T ZFX max Z x O W W2 2F 0 2F may 1 E LE 2Fxmlax22a ZFy OgtZFy FN1 W1Ogt Must include a minus sign here since W2 is accelerating in ydirection ZFy m2aygtZFy T W2 imzaygt Since the two masses are connected by the rope we know that their accelerations will be the same Thus ax y cquot T mla Since 711 Wg we can write them Now our 3 equations look like this FN1 W1 ke thls W1a T 2 W2 mza g FNI W1 I can set the two equations for tension T equal to each W20 other and solver for a the acceleration T 2 W2 g W1aZWZW2a j szaszjaz Wzg g g g g W1W2 185N98 2 a 298ms 422N185N Do the limits on the expression for the acceleration a make sense If W2 0 then a 0 And if W1 0 then a g Now that we have a we can go back and calculate the tension T T zwzlng g Notice that the tension does not equal the weight of the hanging mass 185 N since the system is accelerating Example Two blocks are connected by a rope as shown The coefficient of static friction between the block and the incline is 045 The incline makes an angle of 42 wrt the horizontal What is the maximum value for m such that the system remains in equilibrium Ch 3 Kinematics in 2D Now let s consider the concepts of displacement velocity and acceleration in 2 dimensions 1D 2D szxf xo Afsz O xf xo 7E 7fquot3 AZ I At Z aZQZH az z AZ I At Z xfx0votat2 Ffz ot t2 2 2 A2A2 Ax vf vquot A vf Vquot 2a 267 In general for 2D displacements velocities and accelerations will have components in both the x and ydirection gt Concepts from Ch 1 Thus 1717 17 2 x y xfx0v0taxt 1717 17 x x y 1 2 A A v t at azaxay yf yo 0y 2 y We get 2 equations of motion under constant acceleration one for the x direction and one for the ydirection Example A car drives 60 N of E at a constant speed of 35 ms How far east has the car traveled after 10 s We can break the velocity down into its x and ycomponents Now I can calculate the values for vx and vy vx vcos 60quot 35 ms 2 175 ms vy vsin 600 35 ms 303 ms For distance in the xdirection east we use vx Ax vxt 175 ms105 33 Projectile Motion Now let s analyze the motion of projectiles launched into the air but this time the motion is not completely vertical 2D projectile motion v0 v0 Vertical motion 2D Projectile motion Launch angle 0 90 Launch angle 0lt 90 For the 2D projectile motion case we can break v0 down into its x and y components vox 2 v0 cos 9 0 v 2120 sint9 y So here s the most important thing to remember about projectile motion The only acceleration we have to deal with is that due to gravity and gravity only acts in the vertical 5 direction I ay 2 g and ax 0 I Remember acceleration is defined as the change in velocity over time So if the acceleration in the xdirection is zero the velocity of a projectile in the x direction does not change Let s look at the velocity components of a projectile as it travels along its trajectory Aw lt vay vx vax v v W s lt Ivayl 5 39 Y V vx vax Level Ground v v y 0 v0 Gravity only acts in the vertical direction The x and ymotion can be treated independently v i 0 n L vax I vx vax Level Ground Projectile Motion Summam vy 39Vay There is no acceleration in the horizontal x direction The xcomponent of the v velocity is the same everywhere along the trajectory The velocity of the projectile at any point along the trajectory is the vector sum of its horizontal and vertical components The velocity of the projectile at the top of its trajectory is completely horizontal ie vy O The acceleration of the projectile is 98 ms2 in the negative ydirection at every point along its trajectory Over level ground and with no air resistance projectile motion is completely symmetrical The projectile hits the ground with a speed equal to its initial velocity Example A projectile is fired with initial velocity v0 at angle 6above level ground Find the following a The time the projectile is in the air time of flight b The maximum height of the projectile c The horizontal distance the projectile covers along the ground called the range vox 2 v0 cos 6 v vosin6 0y a Time of flight Use the ycomponent We know that at max height vy O nyV0y nyvo 0 v0sini9 v0sini9 y This is the time it takes to t ay g g t 2V0 sin 9 Time of get to the top So the time of flight is just twice this Fl g ight b Max height Use vertical component again Max height is achieved at 12 the time of flight or I V0 5m g yf yo v0ytayf2 0v0 sin9fgf2 v0 SmQX v3 sin2 6 gt yMax 2g v0 sm 6 Maximum Height 2 v0 sm 6 g g c Range Now we use the xcomponent As long as the projectile is still in the air it will keep covering horizontal ground The time it s in the air is just the time of flight so the horizontal distance covered is Ax 2 v0 1 2 v0 cos 6Mj x 8 Use trig identity sin 26 2 sin 6 cos 6 g 2 2120 sm 6 cos 6 g 2 Ax v0 Sln 26 Max Notice that the range depends on sin26 so which launch anges will give me the maximum range It will be maximum when sin26is maximum ie1 this occurs for 6 45 Basic Equations of Projectile Motion based on kinematic equations with constant acceleration due to gravity Time of t 2V0 sin 9 Flight g v2 sin2 6 Maxnmum Height gt yMax 0 2g 2 g Ranqe of a projectile at different firinq anqles with the same initia velocitv y v3 sin2 6 yMax Maxnmum Height 2g 0 0 x 1 0 0 no range 2 0 20 2 3950o AxmczM 4 6 45 g 5 0 60 6 9 70 6 6 90 no range Example As an airplane takes off it climbs with a constant speed of 75 ms The plane s velocity vector makes an angle of 15 deg with the horizontal Unfortunately when the plane reaches an altitude of 2500 m during this climb one of its engines falls off How long does it take the engine to hit the ground below Vo 75 ms y yo 2500 m X yf O m 2 ay98 W t 194iJ194 4 492500 2 49 yf 2 Hay aytz 194i222 voy 2 v0 sin150 75sin150 194 ms 2 49 t 207 s or O 2500 194t 49l2 Example A golf ball rolls off a horizontal Cliff with an initial speed of 114 ms The ball falls a vertical distance of 155 m into a lake below a How much time does the ball spend in the air b What is the speed v of the ball just before it strikes the water a yf 2 yo voylayl2 O 155 0 98l2 v0 114 ms 0 yo 2 155m 49l 2 W my 0702 vox 114mS 0 Vo ay l ij v0yayl vox V 0 98178 174mS v VyVry V1142 1742 208mS 34 Relative Velocity Let s say that you are standing on the ground at rest beside a road when a pickup truck drives by you at 25 ms VTG 25 ms Relative to you the truck is moving at 25 ms Velocity of the truck relative to the ground Now suppose that someone riding in the back ofthe truck can throw a baseball at 25 ms First considerthe case when he throws the baseball forward the same direction as the truck is moving Now how fast is the baseball moving relative to you VB 25 ms The ball whizzes by you at 50 ms vBG vBT VTG2525 Now suppose that the person riding in the back ofthe truck throws the baseball at 25 ms off the back ofthe truck ie in the opposite direction that the truck is moving Now how fast is the baseball moving relative to you Relative to you on the ground the vBG VBT vTG 25 25 2 ball just drops straight down The above was a collinear example but in general when calculating relative velocities one must take into account the vector nature of the velocity Example You are in a hotair balloon that relative to the ground has a velocity of 51 ms in a direction due east You see a hawk moving directly away from the balloon in a direction due north The speed of the hawk relative to you is 29 ms What are the magnitude and direction of the hawk39s velocity relative to the ground Express the directional angle relative to due east 34 Relative Velocity F 90 ms TG Groundbased observer P passenger G ground quotPT quotTG 11G z 110 ms T train 34 Relative Velocity Example 11 Crossing a River The engine of a boat drives it across a river that is 1800m wide The velocity of the boat relative to the water is 40ms directed perpendicular to the current The velocity of the water relative to the shore is 20ms a What is the velocity of the boat relative to the shore a b How long does it take for the boat to cross the river b r39 J Current W ll l l Current 1 34 Relative Velocity VBS VBW VWS 39 I VHS 1 39 I VIHV I 40 ms H 4 0 I gt 1 o u v 20 ms 6 tan ij M He Jrrr 77 77 VBS 1ng 125 4Oms2 20ms2 45ms 34 Relative Velocity Example You are in a hotair balloon that relative to the ground has a velocity of 51 ms in a direction due east You see a hawk moving directly away from the balloon in a direction due north The speed of the hawk relative to you is 29 ms What are the magnitude and direction of the hawk39s velocity relative to the ground Express the directional angle relative to due east Let s draw a figure and view this from above TVHB 29 mS vHG VH3 vBG vHGx vHBx vBGx v v v HGy HBy BGy gt vBG 51ms Balloon vHGx 05151ms VH6 vHGy 29 0 29 ms 29 ms 51 ms 29 0 vHG 21512292 tan zijez

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