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- Iowa
- Iowa State University
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- Physics 221
- Prell
- Lecture 09 - Newton's third law. Free body diagrams.pdf

Jeff N.

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1 I SOMETIMES WISH NEWTON HAD STOPPED AT TWO LAWS. Lecture 9 : Newton’s Third and Free Body Diagrams Newton’s Third Law For every force, or action, there is an equal but opposite force, or reaction. Forces ALWAYS happen in pairs. BAAB FF G G −= EXAMPLES N person ,wall N wall, person Person leaning against the wallPerson lea t the wall On exhaust particles, by rocket On rocket, by exhaust particles, On bullet, by gun On gun, by bullet On Saturn, by Mars On Mars, by Saturn On the car, by magnet On magnet, by car 2 Gravitation: You attract the Earth! F g you,Earth = F g Earth, you F gyou,Earth F gEarth, you But the acceleration that this produces on the Earth is nothing to be too proud of…) 2 22 2 24 Earth your weight (70 kg)(10 m/s ) 10 m/s , 610 kg a M − =≈ ≈ × Newton’s third law implies that if a rocket accelerates forwards, something must be pushed backwards. In outer space, there isn’t much else around besides its own fuel. rocket DEMO: What if roads where not properly fixed to the ground? Two carts are put back-to-back on a track. Cart A has a spring- loaded piston; cart B, which has twice the mass of cart A, is entirely passive. When the piston is released, it pushes against cart B, and the carts move apart. Which of the two forces exerted by the two carts on each other has a larger magnitude? 1. The force exerted by A. 2. The two forces have equal magnitude. 3. The force exerted by B. A B ACT: Carts with spring It’s a third law pair!! DEMO: Skateboards A Book on a Table Normal on book by table 3 Normal on book by table Normal on table by book Book on Table – The full story Normal force between book and table N BT = –N TB W BE N BT N TE N TB W TE W ET W EB N ET The book does not accelerate W BE +N BT =0 The table does not accelerate W TE +N TB +N TE =0 Does the earth accelerate? Gravitational force between book and earth W BE = –W EB Gravitational force between table and earth W TE = –W ET Normal force between table and earth N TE = –N ET Action-Reaction Pairs Setting limits to Newton’s Laws Galilean transformations for accelerations: P,A P,B B,A aaa=+ G G G When system B is accelerated in relation to A, funny things happen… Imagine an object moving in a straight line at constant speed relative to B (a P,B = 0). If B is accelerated relative to A, the object will appear to have a non-zero acceleration from the point of view of A! …and this could result in a curved trajectory!! Tricky puck on air table. Other examples: Standing in a bus that brakes sharply (passenger “falls forward”). Acceleration simulator (astronaut feels “pushed against the seat”) Inertial and Non-inertial frames of reference Non-inertial frame of reference: is accelerated with respect to an inertial frame of reference (and “funny things” happen → Newton’s laws don’t hold). Inertial frame of reference: moves at constant velocity relative to the fixed stars (Mach’s Principle: “funny things” don’t happen → Newton’s laws hold) They can be very tricky!! DEMO: Passenger on “bus” 4 Where do Newton’s Laws Work? Newton’s laws are true in Inertial Reference Frames (IRF). In a non-inertial ref. frame, you can have an acceleration without having a force → we think there’s a force (we’re applying the 2 nd law!) . These are “fictitious” forces (the most popular one: the “centrifugal” force). π ω ⎛⎞ == ⎜⎟ ⎝⎠ =× ≈× 2 2 Ames 4 6 Earth 2 where 1 day = 8.64 10 s 6.4 10 m aRR T T R →a Ames ≈ 0.034 m/s 2 << 9.8 m/s 2 (pretty decent IRF) Is Ames a good IRF? N TE N TBW TE W BE N BT Free Body Diagram It is a diagram with all the forces acting on one object. You should always draw a free-body diagram before attempting an application of Newton’s second law!!! * * This instructor declines all responsibility for a failed question and will disregard any whining if a free-body diagram has not been drawn. θ N B,I W B,E a Example: Apparent weight John has a mass of 100 kg and standing on a scale in an elevator which is accelerating upwards from rest at 2 m/s². What will the scale read? W on John, by Earth N on John, by scaleWhat does a scale measure? The magnitude of the normal force on the scale by John, |N JS |= |N SJ | Newton’s 2nd law on John: JS JJE NW ma−= John moves with the elevator N on scale, by John (not part of John’s free body diagram) a John has a mass of 100 kg and standing on a scale in an elevator which is accelerating upwards from rest at 2 m/s². What will the scale read? W on John, by Earth N on John, by scale JS JJE NW ma− = JS J J Nmgma− = () JS J J J 22 (100 kg)(9.8 m/s 2.0 m/s ) 1180 N Nmgma mga =+ =+= = If the scale is in kg, it will read: JS 2 1180 N 120 kg 9.8 m/s N g == Check: When the elevator is at rest (a = 0), the scale must read the “correct” weight, 100 kg (980N). 5 Note: In Physics –or at least in this course-, the word “weight” refers to mg, not to what a scale reads (which we sometimes call “apparent weight” or “perceived weight”). Back to Free Fall mg If we neglect friction, only one force is acting: net Fma= G G 2 9.81 m/s mg ma ag = == Newton’s second law: All falling bodies have the same acceleration of 9.81 m/s 2 because: • The m in mg and the m in Newton’s second law are the same (Equivalence of Gravitational and Inertial Mass. This is the basis of General Relativity!). • Weight is the only force acting! NOTE: Weight will always be there –in problems near the Earth-, but most of the time, it is NOT the only force. So the acceleration will NOT be 9.81 m/s 2 . ACT: Force and acceleration (2) Two blocks of masses m and 2m are pushed together along a horizontal, frictionless surface by a force F. The magnitude of the net force on block B is: A. 1/3 F B. 2/3 F C. F == = net,all Entire system: ( ) 3 (so ) 3 F FFmama 2m F m BA == net,B Block B: 2 2 3 F Fma == net,A Block A: 3 F Fma ACT: Force and acceleration (3) Two blocks of masses m and 2m are pushed together along a horizontal, frictionless surface by a force F. The magnitude of the force on block A by block B is: A. 1/3 F B. 2/3 F C. F 2m F m BA 6 ()−= − = =− == =− ⇒ = A B net,B Block A: x: y: 0 Block B: x: 2 y: 2 0 2 We can use what we already know: 3 Or solve the system: 2( ) FNma mgN Nma mgN NF F NFN N 2 3 F m 2m BA F N on A by B N on B by A Action-reaction pair N on A by B = N on A by B ≡ N mg N on A by floor N on B by floor 2mg Example: Box on an incline A hand keeps a 35-kg box from sliding down a frictionless incline. The plane of the incline makes an angle θ = 25° with the horizontal. What is the magnitude of the force exerted by hand? θ • Draw the free-body diagram • Choose axes (draw them!) •Use Newton’s 2 nd law in the x and y-directions. A. 35 N B. 311 N C. 343 N D. 145 N E. 100 N W B,E x y θ θ = mg sinθ mg cosθ = N B,I F B,hand W x W y A hand keeps a 35-kg box from sliding down a frictionless incline. The plane of the incline makes an angle θ = 25° with the horizontal. What is the magnitude of the force exerted by hand? F = mg sinθ = (35 kg)(9.8 m/s 2 )sin(25°) = 145 N (Answer D) ( )θ −= =sin 0 0 x mg F a ( ) θ−=cos 0 0 y Nmg a Paula Microsoft PowerPoint - Lecture 09 - Newton's third law. Free body diagrams.

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