Physics 121, April 1, Equilibrium. Physics 121. April 1, Physics 121. April 1, Course Information. Discussion of Exam # 2

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1 Pysics 121, April 1, Pysics 121. April 1, Course Information Discussion of Exam # 2 Topics to be discussed today: Requirements for Equilibrium Gravitational Equilibrium Sample problems Pysics 121. April 1, Homework set # 7 is due on Saturday morning, April 5, at 8.30 am. Tis assignment as two components: WeBWorK (75%) Video analysis (25%) Homework set # 8 will be available later tis week. Tis assignment will be due on Saturday morning, April 12, at 8.30 am. Exam # 2 will be returned in worksops tis week. 1

2 Midterm Exam # 2. Results. Midterm Exam # 2. Results. Midterm Exam # 2. Results. 2

3 Midterm Exam # 2. Results. Midterm Exam # 2. Results. Midterm Exam # 2. Wat now? Wat do you learn from your results up to now? Exam 1 and Exam 2 > 60%: everyting is OK. Exam 1 and Exam 2 < 40%: you need elp! Please set up a time to meet wit me. Tings are not going to get easier. 40% < Exam 1 and Exam 2 < 60%: you probably will pass te course, but canging your work abits migt result in a better grade. Look at te exam and its solutions and determine wat you are missing (e.g. do you ave a problem applying te correct approac, do you ave problems working wit variables, etc.). Note: even tose students wit 0% on Exam # 1 and Exam # 2 can still pass te course wit a B+ or A-! But you need to act now!!!!! 3

4 An object is in equilibrium is te following conditions are met: Net force N (first condition for equilibrium) and Net torque Nm (second condition for equilibrium) Note: bot conditions must be satisfied. Even if te net force is 0 N, te system can start to rotate if te net torque is not equal to 0 Nm. Static Wat appens wen te net force is equal to 0 N? P = constant Wat appens wen te net torque is equal to 0 Nm? L = constant We conclude tat an object in equilibrium can still move (wit constant linear velocity) and rotate (wit constant angular velocity). Conditions for static equilibrium: P kg m/s L kg m 2 /s Summary of conditions. Equilibrium in 3D:! F x!" x! F y and!" y! F z!" z Equilibrium in 2D:! F x F y!!" z 4

5 Be sure to include all forces!!! Wen evaluating conditions for equilibrium, you need to make sure to include all forces acting on te system. In te system sown in te Figure, tere are more forces acting on te system tan te forces indicated. For example, tere sould be an upward force to balance te downward forces. Of course, te problem is ow to apply te equilibrium conditions correctly. Te force of gravity. Consider an extended rigid object tat can rotate around a specific rotation point. F ' If te rotation point coincides wit te center-of-gravity of te object, it will be in static equilibrium in any orientation. Wat is te relation between te position of te center of mass and te position of te center of gravity? r m m g Te force of gravity. F ' If te object is in equilibrium, te net torque and te net force acting on it must be equal to 0. r m Te net force acting on te object is equal to If te net force is equal to 0 N, we must require tat!! F = F '"! # = F '" g #m F ' = ( ) ĵ = (! ) ĵ = F '" m g 5

6 Te force of gravity. F ' Te condition tat F = is not sufficient for static equilibrium. We must also require tat te net torque is equal to 0 Nm. Te net torque acting on te object is equal to! "! = "{ r # ( $m g! )} = " $m r! If te net torque must be 0 Nm, we must require tat M! r cm!! g ( ) #! g = M! r cm #! g r m m g Te force of gravity. Te system will be in equilibrium if M! r cm!! g Te requires tat Te center-of-gravity is located exactly below or above te rotation axis (r cm parallel to vertical axis). F ' r m m g or Te center-of-gravity coincides wit te rotation axis (r cm ) Used to determine te location of te center-of gravity of an object. Sample problem 1. A uniform beam of lengt L wose mass is m, rest wit its ends on two digital scales. A block wose mass is M rests on te beam, its center one-fourt away from te beam s left end. Wat do te scales read? Fl L Fr If te system is in equilibrium, te net force must be 0 N:! F y = F l + F r " " 6

7 Sample problem 1. If te system is in equilibrium, te net torque must be 0 Nm. Note: te toque associated wit a force depends on te coice of te origin. Te condition tat te torque must be 0 Nm must be satisfied wit respect to any coice of origin. If we coose te left scale as our origin, te left scale force does not appear in our torque equation: Fl L Fr "! z = F l 0 + F r L # L 4 # L 2 Sample problem 1. Te force generated by te rigt scale is tus equal to F r = L 4 + L 2 L = Fl L Fr We can now use te first condition of equilibrium to determine te force generated by te left scale: F l = +! F r = Sample problem 2. A ladder wit lengt L and mass m rests against a wall. Its upper end is a distance above te ground. Te center of gravity of te ladder is one-tird of te way up te ladder. A firefigter wit mass M climbs alfway up te ladder. Assume tat te wall, but not te ground, is frictionless. Wat is te force exerted on te ladder by te wall and by te ground? Fw a Fgy Fgx a/3 a/2 7

8 Sample problem 2. Forces exerted by te wall and te floor: Te wall exerts a orizontal force (normal force). Fw Te floor exerts a vertical force (normal force) and a orizontal force (friction force). Fgy Fgx Note: te friction force must be present in order to ensure tat te net force in te orizontal direction add up to 0 N. a a/3 a/2 Sample Problem 2. Te first condition for equilibrium requires tat Fw! F x = F W " F gx and! F y = F gy " " Two equations wit tree unknown. We need more information! But we still ave te second condition for equilibrium. a Fgy Fgx a/3 a/2 Sample Problem 2. Te second condition for equilibrium requires: "! = F W # a 2 # a 3 Note: we ave used to resting point on te ground as out reference point. Te torque due to te two forces acting on tis point do not contribute to te torque wit tis coice of reference point. We can now determine F W easily: Fw a Fgy Fgx a/3 a/2 F W = a 2 + a 3 = ga! 1 2 M + 1 " # 3 m $ % & 8

9 By examining te net force in te orizontal direction, we can determine te friction force: F gx = F W = ga Sample Problem 2.! 1 2 M + 1 " # 3 m $ % & Note: te frictional force depends on te position of te firefigter Fgy and increases wen te fire Fgx figter climbs te ladder. a/3 Since te frictional force must be a/2 less tan µ s F gy, tere may be a a maximum eigt tat can be reaces by te fire figter above wic te ladder will slip. Fw Let s test our understanding of te basic aspects of equilibrium by working on te following concept problems: Q19.1 Q19.2 Stress and strain. Te effect of applied forces. Wen we apply a force to an object tat is kept fixed at one end, its dimensions can cange. If te force is below a maximum value, te cange in dimension is proportional to te applied force. Tis is called Hooke s law: F = k ΔL Tis force region is called te elastic region. 9

10 Stress and strain. Te effect of applied forces. Wen te applied force increases beyond te elastic limit, te material enters te plastic region. Te elongation of te material depends not only on te applied force F, but also on te type of material, its lengt, and its cross-sectional area. In te plastic region, te material does not return to its original sape (lengt) wen te applied force is removed. Stress and strain. Te effect of applied forces. Te elongation ΔL can be specified as follows:!l = 1 F E A L 0 were L 0 = original lengt A = cross sectional area E = Young s modulus Stress is defined as te force per unit area (= F/A). Strain is defined as te fractional cange in lengt (ΔL 0 /L 0 ). Note: te ratio of stress to strain is equal to te Young s Modulus. Stress and strain. Direction matters. 10

11 Stress and strain. A simple calculation could ave prevented te deat of 114 people. Stress and strain. A simple calculation could ave prevented te deat of 114 people. Initial Design Actual Design Credit: ttp:// Done for today! On Tursday: armonic motion! 11