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### Presentation CHAPTER5

Course: PHY 2043, Fall 2011
School: FAU
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5 APPLICATIONS CHAPTER OF NEWTONS LAW Friction Drag Forces Motion Along Curved Path The Center of Mass 1 FRICTION Friction is physical phenomenon that appear at contact of two objects, when you move or attempt to move one object relatively another. Polished steel surface Source of friction: Interaction between surfaces on on molecular level Nickel atoms Gold atoms 2 Static Friction Static friction formula f...

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5 APPLICATIONS CHAPTER OF NEWTONS LAW Friction Drag Forces Motion Along Curved Path The Center of Mass 1 FRICTION Friction is physical phenomenon that appear at contact of two objects, when you move or attempt to move one object relatively another. Polished steel surface Source of friction: Interaction between surfaces on on molecular level Nickel atoms Gold atoms 2 Static Friction Static friction formula f s max = s Fn Where constant s is called coefficient static friction. Object begin to slide if magnitude applied force F s Fn Friction forces Free body diagram for the block Fn Direction of friction force always opposite direction of motion In case sliding block on horizontal surface the F fs magnitude of normal force is equal to weight Fg Measuring coefficient static friction using incline plane 3 Kinetic Friction Kinetic friction formula is applied to the mowing object. f k = k Fn Constant k is the coefficient of kinetic friction. Usually for the same couple surfaces s k From your experience, it is difficult to start slide object but when it start it is easy to continue push. The figure demonstrate how friction force depend from applying force. When object moving kinetic friction force remain constant Balanced forces 1st Newtown law at rest In motion 4 5 The frictional force depends only on the type of surfaces and how hard the surfaces are pressed together. Normal force Friction force We can reduce friction by lubricating the surfaces. This means that the surfaces no longer rub directly on each other, but slide past on a layer of liquid. Fgn Fg Acting force 6 Four identical blocks of same mass are moving on a surface for which the coefficient of kinetic friction between each block and the surface is k. The velocity of each block is indicated by the vector on the block. For which block is the force of friction between the surface and the block greatest? A. B C D E 1 2 3 4 The force of friction is the same for all blocks. 7 Rolling Friction Rolling friction formula f r = r Fn Constant r is coefficient of rolling friction. r Depends on nature of contacting surfaces v 8 Conceptual problem: Blocks made of the same material and similar shape but differing in mass lie on the bed of a truck that is moving along a straight horizontal road. When track accelerate all of the blocks will slide. What is relation between minimum accelerations to start sliding of each block with different mass ? Solution: Of these three forces, the only one that acts in the direction of the acceleration is the static friction force. Apply Newtons 2nd law f s = s Fg = s mg = ma x ax = s g Because ax is independent of m and Fg, the critical accelerations are the same. 9 Conceptual problem: Attach a rubber band to the block and pull gently and steadily on the rubber band in the horizontal direction. At some point, the block will start moving, but it will not move smoothly. Instead, it will start moving, stop again, start moving again, stop again, and so on. Explain why the block moves this way. Answer: As the rubber band (spring) is extended, the force exerted by the rubber band on the block increases. Once that force is greater than the maximum value of the force of static friction on the block, the block will begin to move. However, as it accelerates, it will shorten the length of the band, decreasing the force that the rubber band exerts on the block. As this happens, the force of kinetic friction can then slow the block to a stop, which starts the cycle over again. 10 A block of mass m is at rest on an inclined plane that makes an angle of 30 with the horizontal, as shown in the figure. Which of the following statements about the force of static friction is true? A B C D E fs > mg fs > mg cos 30 fs = mg cos 30 fs = mg sin 30 None of the statements is true. 11 Problem: On an icy roadway, the coefficient of friction between the tires of car and roadway is reduced to one-quarter of its value on a dry road. As a result, the maximum speed vmax at which the car can safely turn on curve of radius R is reduced. What is a new value for this speed. Solution: Fx = f s,max f s,max = s mg 2 vmax =m R vmax = s gR F y = Fn mg = 0 v'max = 's gR 's gR v'max ' s = = v max s gR s v'max = ' s v max v' = max s v 1 4 max = 0.5vmax = 50%vmax 12 A block of mass m rests on a plane that is inclined at an angle with the horizontal. What is maxim angle when block start slide if the coefficient of static friction is s Apply Fx = ma x f s Fg sin = 0 or, because Fg = mg, f s mg sin = 0 Apply F y = ma y (1) Fn mg cos = 0 (2) Divide equation (1) by equation (2) to obtain: fs tan = Fn Substitute for fs = sFn and simplify to obtain: tan = s Fn = s Fn = tan 1 s 13 Two blocks of masses m1 and m2 are sliding down an incline as shown in figure. They are connected by a massless rod. The coefficients of kinetic friction between the block and the surface are 1 for block 1 and 2 for block 2. Determine the acceleration of the two blocks. Apply F = m a to block 1 Fx = T1 + m1 g sin f k,1 = m1a x (1) F y = Fn,1 m1 g cos = 0 f k,1 = 1Fn,1 Apply F = m a to block 2 Fx = m2 g sin T2 f k,2 = m2a x (2) F y = Fn,2 m 2 g cos = 0 f k,2 = 2 Fn,2 m 2 g sin 14 Let T1 = T2 = T, Use the definition of the kinetic friction force to eliminate fk and Fn from equations (1) and (2) m1a x = m1 g sin + T 1m1 g cos (3) m 2a x = m 2 g sin T 2 m 2 g cos (4) Add equations (3) and (4) to eliminate T and solve for ax: 1m1 + 2 m 2 a x = g sin cos m1 + m 2 15 In figure m1 = 4.0 kg and the coefficient of static friction between the block and the incline is 0.40. Find the range of possible values for m2 for which the system will be in static equilibrium. Noting that Fg,1 = m1 g , apply Fx = T f s,max m1 g sin 30 = 0 (1) F = m a to the block whose mass is m1: F y = Fn,1 m1 g cos 30 = 0 (2) Using f s, max = s Fn , substitute equation (2) in equation (1) to obtain: T m g cos 30 m g sin 30 = 0 (3) s 1 1 Noting that Fg,2 = m 2 g , m 2 g T = 0 (4) Add equations (3) and (4) to eliminate T m 2 = m1 ( s cos 30 + sin 30) 6 m 2 = ( 4.0 kg ) [ ( 0.40 ) cos 30 + sin 30] m 2, + = 3.4 kg1and m 2,- = 0.61 kg Summary Horizontal surface the maximum static friction force f s,max = s mg f s, max = s Fn if Fapp f s ,max object is rest and f s , frict = Fapp if object have const . velocity Fapp = f k , frict if object move with acceleration Fapp f k , frict = ma x Incline plane same condition we can apply to incline only Fn = mg cos and Fapp = mg sin Fapp if Fapp f s ,max object is rest and f s , frict = Fapp If additional force F applied to same direction as Fapp then: Fapp ,total = F + Fapp object move with acceleration Fapp ,total f k , frict = ma x 17 If force is applied under some angle Fapp the maximum static friction force f s, max = s Fn Fn = Fg Fapp sin x f s ,max = s ( Fg Fapp sin ) if Fapp , x Fg = mg Fapp , x = Fapp cos f s ,max object is rest and f s , frict = Fapp , x if object have const . velocity Fapp , x = f k , frict if object move with acceleration Fapp , x f k , frict = ma x 18 Resistive Force Resistive force act on object that move in media always is in opposite direction of velocity. General formula for this force is: ( ) Fres = b1v + b2 v 2 v Resistive force on spherical object with radius r, mass m, and speed v is ( ) Fres = C1rv + C 2 r 2v 2 v kg m3 kg ms Firs term associate with viscosity and second with pressure Fres At condition when acceleration is 0 and 2 Fg = mg = (C1rvterm + C 2 r 2v term ) Velocity of object is terminal velocity. 2 C1rvcrit = C2r 2vcrit Fg Fres = Fg vcrit = C1 C2r C v Fres = C1rv 1 + 2 rv = C1rv 1 + v C1 crit Regime 1 when v < <vcrit (viscose liquids) For air C1=3.1*10^-4 19 Regime 2 when v >> v crit (gases) and C2=0.85 for liquids Drag Force 2 Drag force formula Fd = b v The parameter b is function of area of object exposed to the flow. If object falling under influence of gravity then acceleration will define by dragging force, mg bv 2 = ma y bv 2 a y= g m Terminal speed define when a y approach zero (drag force balancing gravitational force). 1/ 2 mg vT = b bv 2 Terminal speed mean that object move with ma y constant velocity (1st Newtons Law in action) mg v2 a y = g 1 2 20 vT As an object falls faster and faster, the air resistance will get more and more. Eventually, the force of air resistance will be equal to the object's weight, so the object will stay at that speed, and stop 21 accelerating. A skydiver of mass 60.0 kg can slow himself to a constant speed of 200 km/h by orienting body horizontally in arc position. In this position he presents the maximum cross-sectional area and thus maximize the air-drag (a) force. What is the magnitude of the drag force on the skydiver? (b) If the drag force is given by bv^2, what is the value of b? (a) Apply F y = ma y (1) Fd mg = ma y or, because ay = 0, F = mg d ( ) Fd = ( 60.0 kg ) 9.81 m/s 2 = 588.6 N = 589 N (b) Substitute Fd = bv t2 in equation (1) mg Fd bv t2 = mg b= 2 = 2 vt vt b= 588.6 N km 1h 200 h 3600 s 2 0.42 kg/m 22 (1) MOTION ALONG A CURVED PATH Satellite moving in circular orbit around Earth with constant tangential velocity. ( vt ) 2 + r 2 = ( r + h ) 2 ( vt ) 2 + r 2 = r 2 + 2rh + h 2 h( 2r + h ) = ( vt ) 2 For short int erval time we can neglect h compared with 2r ( v 2 r ) t 2 h 2 a t2 this formula is similar to x = 2 The satellite moving with constant speed and appear falling every time distance h on circular orbit with v2 acceleration r 23 vt MOTION ALONG A CURVED PATH Conical pendulum Newtowns Second Law v2 Fx = ma x T sin = m = Fnet r F y = ma y T cos mg = 0 The net force is centripetal force. v2 m Fnet r T= = sin sin Fnet v 2 ax = = m r v = axr Period rotation = 2r v 24 A block of mass m1 is attached to a cord of length L1, which is fixed at one end. The block moves in a horizontal circle on a frictionless tabletop. A second block of mass m2 is attached to the first by a cord of length L2 and also moves in a circle on the same frictionless tabletop, as shown in figure. If the period of the motion is P, find the tension in each cord. Free body diagram 25 Apply Fx = max 2 v1 T1 T2 = m1 L1 Apply to the block whose mass is m1: (1) Fx = max 2 v2 T2 = m2 L1 + L2 to the block whose mass is m2: (2) Relate the speeds of each block to their common period P v1 = 2L1 2 ( L1 + L2 ) and v2 = P P In the second an first force equations, substitute for v2 and v1 , and simplify to obtain: T2 = [ m2 ( L1 + L2 ) ] 2 P 2 2 T1 = [ m2 ( L1 + L2 ) + m1L1 ] P 2 26 Problem: You swing a bucket containing mass m of water in a vertical circle of radius r. If the speed is vtop at the top of the circle find (a) the force FPW exerted by the bucket on the water at the top of the circle, and (b) the minimum value of vtop for the water to remain in the bucket (c) What is the force exerted by the bucket on the water at the bottom of the circle, where the speed of the bucket is vbot. 27 Solution: For position of bucket on the top. 2 m v top FPW + mg = r 2 v top FPW = m g r For minimum speed on the top FPW = 0 mg = 2 m v top min v top min = rg r For position of bucket on the bottom. FPW 2 2 v bot m vbot mg = FPW = m + g r r 28 An airplane is flying in a horizontal circle at a speed of 480 km/h. The plane is banked for this turn, its wings tilted at an angle of 40 from the horizontal. Assume that a lift force acting perpendicular to the wings acts on the aircraft as it moves through the air. What is the radius of the circle in which the plane is flying? 29 v2 Apply F = m a to the plane: F = F sin = m x lift r and F y = Flift cos mg = 0 Eliminate Flift between these equations to obtain by dividing: v2 v2 r= tan = g tan rg 2 km 1h 480 h 3600 s = 2.2 km r= 9.81 m/s 2 tan40 ( ) 30 A curve of radius 150 m is banked at an angle of 10. An 800-kg car negotiates the curve at maximum 85 km/h without skidding. Neglect the effects of air drag and rolling friction. Find (a) the normal force exerted by the pavement on the tires, (b) the frictional force exerted by the pavement on the tires, (c) the coefficient of static friction between the pavement and the tires. v2 Fx = Fn sin + f s cos = m r F y = Fn cos f s sin mg = 0 Multiply the x equation by sin and y equation by cos v2 2 f s sin cos + Fn sin = m sin r 2 Fn cos f s sin cos mg cos = 0 v2 Add these equations to eliminate fs: Fn mg cos = m sin r 2 2 v v Fn = mg cos + m sin = m g cos + sin r r fs s = Fn cos mg fs = 31 Fn sin You ride a bicycle in a 20-m-radius circle on a horizontal surface. The resultant force exerted by the surface on the bicycle (normal force plus frictional force) makes an angle of 15 with the vertical. What is your speed? Apply F = m a to the bicycle: mv 2 and F y = Fn mg = 0 Fx = f s = r Relate Fn and fs to : mv 2 fs v2 tan = =r= Fn mg rg Solving for v yields: v = rg tan = 7.3 m / s 32 Summary v2 Centripetal acceleration: a c = r v2 Centripetal force:Fc = ma c = m r d vt Tangential acceleration: a t = dt The magnitude of total acceleration at curve path is: a = a = ac + at At constant speed (uniform circular motion) a t = 0 2 ac + at 2 When you make a turn on road with friction at constant maxim speed without skidding define by: mv 2 f s = s Fn = s mg = r 33 THE CENTER OF MASS The center of mass call: The point in a system of bodies at which the mass of the system may be considered to be concentrated and at which external forces may be considered to be applied. Parabola Center mass define by; Mxcm = m1 x1 + m 2 x2 where M = m1 + m 2 Choose origin of coordinate for location of first mass and distance between mass is d. baton thrown into the air Mxcm = m1 0 + m 2 d xcm m2d = m1 + m 2 34 35 Center mass in three dimension for N particles. Mxcm = m1 x1 + m 2 x2 + m3 x3 + .... + m N x N or for x coordinate Mxcmr = m i x i i Where M = m i total mass i for y coordinate Mycm = m i yi i for z coordinate Mzcm = m i z i i in vector notation Mrcm = m i ri Where ri = xi i + yi + z i k j j and rcm = xcm i + ycm + zcm k r d m = r d m Center mass in three rcm d m = r d m rcm = dimension for solid body. M 36 dm Three balls A, B, and C, with masses of 3.0 kg, 1.0 kg, and 1.0 kg, respectively, are connected by massless rods, as shown in figure. What are the coordinates of the center of mass of this system? The x coordinate of the center of mass is given by: x cm m A x A + m B xB + mC xC = = 2m m A + mB + mC The y coordinate of the center of mass is given by: y cm 1.4 c.m. m A y A + m B y B + m C yC = = 1.4m mA + mB + mC 37 Alternate Solution Using Vectors The vector expression for the center of mass is: m i ri M rcm = m i ri rcm = i i M i j where rcm = xcm + ycm + zcm k 1.4 c.m. The position vectors for the objects located at A, B, and C are: i j rA = 2.0 + 2.0 + 0k m rB = ( + + 0k )m rC = ( 3.0 + 0 + 0k ) m i j ij m A rA + m B rB + mC rC rcm = m A + m B + mC ( [ ) ( ) ( ) ( )] ( 3.0 kg ) 2.0 + 2.0j + 0k m + (1.0 kg ) + j + 0k m + (1.0 kg ) 3.0 + 0j + 0k m i i i rcm = ( 3.0 kg + 1.0 kg + 1.0 kg ) = ( 2.0 m ) + (1.4 m ) + 0k i j 38 Two identical uniform rods each of length L are glued together so that the angle at the joint is 90. Determine the location of the center of mass (in terms of L) of this configuration relative to the origin taken to be at the joint. The x coordinate of the center of mass is given by: ( 0) m1 + ( 1 L) m2 x1,cm m1 + x2,cm m 2 2 xcm = xcm = m1 + m 2 m1 + m 2 or, because m1 = m2 = m, xcm = ycm = ( 0) m + ( 1 L) m 2 m+m ( 1 L) m + ( 0) m 2 m+m =1L 4 =1L 4 The center of mass of this system is located at ( 1 L, 1 L) 4 4 39 Problem: Find the center of mass of the uniform sheet of plywood. Solution: Divide sheet of plywood on two rectangle 1 and 2. m1 xcm1 + m 2 xcm 2 xcm = m1 + m 2 = M m1 + m 2 m1 ycm1 + m 2 ycm 2 M Area of sheet1 = A1 sheet 2 = A2 m i M = Ai A ycm = A1 A2 A1 A2 xcm = xcm1 + xcm 2 ycm = ycm1 + ycm 2 A A A A A1 = 0.32m 2 A2 = 0.04m 2 A = A1 + A2 xcm1 = 0.4m ycm1 = 0.2m xcm 2 = 0.7 m ycm 2 = 0.5m xcm = 0.43m ycm = 0.23m 40 Find the location of the center of mass of a nonuniform rod 0.40 m in length if its linear mass density ( x ) varies linearly from 1.00 g/cm at one end to 5.00 g/cm at the other end. Specify the center-of-mass location relative to the less-massive end of the rod. The x coordinate of the center of mass xdm xcm = dm = ( x ) dx , xcm = x ( x ) dx ( x ) = b + ax dm ( x ) dx ( 0 ) = b = 1.00 g / cm ( 40 ) = 1.0 g / cm + a 40 cm a = 0.1 g / cm 2 xcm ( x ) = 1.00 g/cm + ( 0.10 g/cm 2 ) x 40 cm 2 x[1.00 g/cm + ( 0.10 g/cm ) x ]dx = 40 cm = 24 cm 0 [1.00 g/cm + (0.10 g/cm 2 ) x ]dx 41 0 Find the center of mass of a uniform semicircular disk of radius R. y dA ycm = y = r sin (1) M Express dA in terms of r and : dA = r d dr M = Ahalf disk = 1 R 2 2 Substitute in equation (1) and evaluate ycm: R y cm = r 2 sin d dr 00 M 2 R 2 2 3 4 = r dr = 3M R = 3 R M0 42 Motion of the Center Mass drcm dri dr1 dr2 M = m1 + m2 + ... = m i dt dt dt dt i Mv cm = m1v1 + m 2 v 2 + ... = m i v i i Macm = m1a1 + m 2a2 + ... = m i ai i mi ai = Fi i i Fi = Fint i + Fexti i i i Fint i = 0 Third Newtowns law i for stable system of particles Fnet = Fexti = Macm i The center mass of system move like single mass M = m i under the influence of external force. 43
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Editing Test 6Dear Ms. Allison:During your remodeling planning, think about the amount of natural light in eachroom, the use you make of the room, and the layout of its furniture. Then consider whatquality of light you need and where you need itbright
Rutgers - ENGLISH: C - 203
Editing Test 2Urban Systems has have just introduced WiteLite, an ultraviolet-filtered, fullspectrum light source that combats Seasonal Affective Disorder (SAD), a type ofdepression caused by the decrease of daylight during the fall and winter months.M
FAU - PHYSICS - 2043
Robert NelsonGroup: Michael Schultz, Eloise CaveGeneral Physics IPHY2048L9/8/11
FAU - PHYSICS - 2043
Robert NelsonGroup: Michael Schultz, Eloise CaveGeneral Physics IPHY2048L9/8/11
Rutgers - ENGLISH: C - 203
Editing Test 1McCormick Place in Chicago, Illinois has been selected to host thelargest, most comprehensive lighting exhibit and conference ever held in theUnited States. This conference, which is known as the International LightingExpo, will be held
FAU - PHYSICS - 2043
Crescent Nebula - Lusarn Cluster - Planet TarithCrescent Nebula - Zelene Cluster - Planet HelymeHourglass Nebula - Faryar Cluster - Planet DaratarHourglass Nebula - Ploitari Cluster - Planet TheganRosetta Nebula - Alpha Draconis Cluster - Planet 2175
FAU - PHYSICS - 2043
Crescent Nebula - Lusarn Cluster - Planet TarithCrescent Nebula - Zelene Cluster - Planet HelymeHourglass Nebula - Faryar Cluster - Planet DaratarHourglass Nebula - Ploitari Cluster - Planet TheganRosetta Nebula - Alpha Draconis Cluster - Planet 2175
Rutgers - ENVIRONMEN - 101
DEAL IS REACHED TO SAVE CALIFORNIA REDWOOD FOREST (Clifford) Pacific Lumber Co. makes deal to save Californias Forest. Transfers about 10,000 acres into public ownership and set strict newguidelines for protecting water quality and wildlife on over 200
FAU - PHYSICS - 2043
Robert NelsonGroup: Michael SchultzGeneral Physics I LabPHY2048LExperiment11/10/11
FAU - PHYSICS - 2043
Robert NelsonGroup: Michael SchultzGeneral Physics I LabPHY2048LExperiment11/10/11
Rutgers - ENVIRONMEN - 101
The price of everything-Talks about free market environmentalism on our natural resources-Dont take into account externalities-Automobiles in urban settings-They proposed something that will protect the environment against externalities (what isit?)
Rutgers - ENVIRONMEN - 101
What is Human Ecology?-Media plays a role in the growth of human ecology-People made up from different disciplines. All interested in how people act with theenvironment-Each has different theories, methods, questions, and approaches-Sometimes causes
Rutgers - SOC. PSYCH - 1:830:321
November 8, 2010A-Social Fascilitation1- Others presencea. Individual efforts evaluated evaluation apprehension and distraction arousalB-Social Loafing1- Others presencea. Individual efforts pooled and NOT evaluated no evaluation apprehension relax
Rutgers - SOC. PSYCH - 1:830:321
A-Cognitive Sources1- Categorization &amp; Sterotyping2- Out-group homogeneity effecta. Tendency that once split into categories or groups, to see them as more similarthan they actually areb. Due to a lack of experience with members of out-groupsb.i. We
Rutgers - SOC. PSYCH - 1:830:321
December 6, 2010A-Measuring Implicit Prejudice/Stereotypes: The IAT1- Implicit and Explicit attitudes are not correlated2- bonafide pipeline study and Video studya. Supraliminal priming with black and white facesb. Classify words as good or bad3- Re
Rutgers - SOC. PSYCH - 1:830:321
December 8, 2010Notes from the Movie-students predicted that their mitochondria would be more similar to their own race-There is more variation within racial groups than between racial groups-are the peoplewho we call black more like blacks geneticall
Rutgers - SOC. PSYCH - 1:830:321
December 13, 2010A-Defendant Characteristics1- Physical attractivenessa. Lighter sentences to physically attractive defendants2- Similarity to jurors (similarity-attraction and False Consensus effects)a. People tend to feel more comfortable with thos
Rutgers - SOC. PSYCH - 1:830:321
November 1, 2010A-Compliance with Requests1- Mindless compliance- the role of automaticity in compliancea. Peoples tendency to agree than disagreeb. We would rather give people what they want than prevent them for getting whatthey want2- Two techniq
Rutgers - SOC. PSYCH - 1:830:321
November 3, 2010A-Varying the Conditions-Increasing or Decreasing Compliance1- Distance of Authoritya. Further away, the less the compliance2- Distance of Victima. The closer the victim, the less the compliance3- Status of Authoritya. The higher th
Rutgers - SOC. PSYCH - 1:830:321
Social Psychology Lecture NotesSeptember 8, 2010What is Social Psychology?Scientific study of the way in which peoples thoughts, feelings, and actions areinfluenced by the real or imagined presence of other people.It is what people believe that is im
Rutgers - SOC. PSYCH - 1:830:321
November 10, 2010A-Majority and Minority Influence in Groups1- Majority influence:a. Normative conformity (public compliance)2- Minority influence:a. Persuasion through informational influence (private acceptance)b. Consistency, confidence, and reas
Rutgers - SOC. PSYCH - 1:830:321
November 15, 2010A-Why does watching pro-social behavior increase helping?1- Cognitive factors:a. Observational learningb. Priming pro-social schemas2- Affective factorsa. elevation of mood3- Neurobiological factorsa. Activation of mirror neurons
Rutgers - SOC. PSYCH - 1:830:321
November 17, 2010A-Social-situational causes of Aggression1- Objects as Aggressive Priming Cues Increase Aggression (Berkowitz &amp; LePage)a. One half sat at desk with a gun, and other half sat at dest with racketb. Operationalizing aggressionduration of
Rutgers - SOC. PSYCH - 1:830:321
November 29, 2010FINAL EXAM: DECEMBER 20 -Monday night8-11College aveCh.13 stereotypingA-Definitions of Prejudice1- *A negative attitude toward a socially defined group, and any person perceived to be amember of that groupa. Mrs. Elliot classroom e
Rutgers - SOC. PSYCH - 1:830:321
October 6, 2010A-How we come to know ourselves1- Self-Perception Theorya. We infer our attitudes and feelings by observing our own behavior (and itscontext)b. If we know what our attitudes and beliefs are we dont need to look at ourbehaviorsc. It i