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Transcript
Physics 151: Lecture 8 Reaminder: Homework #3 : (Problems from Chapter 5) due Fri. (Sept. 22) by 5.00 PM Today’s Topics : Review of Newton’s Laws 1 and 2 - Ch. 5.1-4 Newton’s third law: action and reaction - Ch. 5.6 Physics 151: Lecture 8, Pg 1 See text: 5.1-4 Review Newton’s Laws 1 and 2 Isaac Newton (1643 - 1727) published Principia Mathematica in 1687. In this work, he proposed three “laws” of motion: Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame. Law 2: For any object, FNET = F = ma Physics 151: Lecture 8, Pg 2 See text: 5-6 Newton’s Third Law: If object 1 exerts a force on object 2 (F2,1 ) then object 2 exerts an equal and opposite force on object 1 (F1,2) F1,2 = -F2,1 For every “action” there is an equal and opposite “reaction” This is among the most abused concepts in physics. REMEMBER: Newton’s 3rd law concerns force pairs which act on two different objects (not on the same object) ! Physics 151: Lecture 8, Pg 3 An Example Consider the forces on an object undergoing projectile motion FB,E = - mB g FB,E = - mB g FE,B = mB g FE,B = mB g EARTH Physics 151: Lecture 8, Pg 4 Normal Forces Certain forces act to keep an object in place. These have what ever force needed to balance all others (until a breaking point). FB,T FT,B Physics 151: Lecture 8, Pg 5 Force Pairs Newton’s 3rd law concerns force pairs. Two members of a force pair cannot act on the same object. Don’t confuse gravity (the force of the earth on an object) and normal forces. It’s an extra part of the problem. FB,T FB,E = -mg FT,B FE,B = mg Physics 151: Lecture 8, Pg 6 An Example Consider the following two cases Physics 151: Lecture 8, Pg 7 An Example The Free Body Diagrams mg FB,T= N mg Ball Falls For Static Situation N = mg Physics 151: Lecture 8, Pg 8 An Example The action/reaction pair forces FB,E = -mg FB,T= N FT,B= -N FE,B = mg FB,E = -mg FE,B = mg Physics 151: Lecture 8, Pg 9 Lecture 8, Act 1 Newton’s 3rd Law Two blocks are being pushed by a finger on a horizontal frictionless floor. How many action-reaction pairs of forces are present in this system? a (a) 2 (b) 4 b (c) 6 Physics 151: Lecture 8, Pg 10 Lecture 8, Act 1 Solution: Fa,f Fb,a Ff,a a FE,a Fa,b bF Fg,a Fg,b Fa,g Fb,g Fa,E E,b Fb,E 2 4 6 Is Fa,f = Fb,a? (A) YES (B) NO Fb,a = Fa,f [mb/(mb+ma)] Physics 151: Lecture 8, Pg 11 Lecture 8, Act 2 You are going to pull two blocks (mA=4 kg and mB=6 kg) at constant acceleration (a= 2.5 m/s2) on a horizontal frictionless floor, as shown below. The rope connecting the two blocks can stand tension of only 9.0 N. Would the rope break ? (A) YES (B) CAN’T TELL A rope (C) NO a= 2.5 m/s2 B Physics 151: Lecture 8, Pg 12 Lecture 8, Act 2 Solution: What are the relevant forces ? mB mA mA rope T Fapp = a (mA + mB) Fapp = 2.5m/s2( 4kg+6kg) = 25 N a= 2.5 m/s2 Fapp aA = a = 2.5 m/s2 T -T -T T = a mA T = 2.5m/s2 4kg = 10 N T > 9 N, rope will brake ANSWER (A) a = 2.5 m/s2 a = 2.5 m/s2 mB total mass ! Fapp Fapp - T = a mB T = 25N - 2.5m/s2 6kg=10N T > 9 N, rope will brake THE SAME ANSWER -> (A) Physics 151: Lecture 8, Pg 13 See text: Example 5.7 Exercise: Inclined plane A block of mass m slides down a frictionless ramp that makes angle with respect to horizontal. What is its acceleration a ? m a Physics 151: Lecture 8, Pg 14 See text: Example 5.7 Inclined plane... Define convenient axes parallel and perpendicular to plane: Acceleration a is in x direction only. j m a i Physics 151: Lecture 8, Pg 15 See text: Example 5.7 Inclined plane... Consider x and y components separately: i: mg sin = ma a = g sin j: N - mg cos =0. m N = mg cos ma j mg sin N mg cos mg i Physics 151: Lecture 8, Pg 16 See text: Example 5.7 Angles of an Inclined plane m a = g sin N mg Physics 151: Lecture 8, Pg 17 Free Body Diagram A heavy sign is hung between two poles by a rope at each corner extending to the poles. Eat at Bob’s What are the forces on the sign ? Physics 151: Lecture 8, Pg 18 Free Body Diagram T2 T1 2 1 Eat at Bob’s mg Add vectors : Vertical (y): mg = T1sin1 + T2sin2 Horizontal (x) : T1cos1 = T2cos2 y T2 T1 1 2 x mg Physics 151: Lecture 8, Pg 19 Example-1 with pulley Two masses M1 and M2 are connected by a rope over the pulley as shown. Assume the pulley is massless and frictionless. Assume the rope massless. If M1 > M2 find : Acceleration of M1 ? Acceleration of M2 ? Tension on the rope ? T1 T2 Video M2 Animation M1 a Free-body diagram for each object Physics 151: Lecture 8, Pg 20 Example-2 with pulley A mass M is held in place by a force F. Find the tension in each segment of the rope and the magnitude of F. Assume the pulleys massless and frictionless. Assume the rope massless. T4 T1 We use the 5 step method. F Draw a picture: what are we looking for ? What physics idea are applicable ? Draw a diagram and list known and unknown variables. Newton’s 2nd law : F=ma T3 T2 < T5 M Free-body diagram for each object Physics 151: Lecture 8, Pg 21 Pulleys: continued FBD for all objects T4 T2 T3 T4 T1 T3 T2 5 F < T T5 T3 F=T1 T5 M T2 M Mg Physics 151: Lecture 8, Pg 22 Pulleys: finally Step 3: Plan the solution (what are the relevant equations) F=ma , static (no acceleration: mass is held in place) T5 T5=Mg M T2 T1+T2+T3=T4 T4 Mg F=T1 T3 T2+T3=T5 T3 F=T1 T 5 T2 Physics 151: Lecture 8, Pg 23 Pulleys: really finally! Step 4: execute the plan (solve in terms of variables) We have (from FBD): F=T1 T5=Mg T2+T3=T5 T1+T2+T3=T4 Pulleys are massless and frictionless T1=T3 T4 T2=T3 T2+T3=T5 gives T5=2T2=Mg T1 T2=Mg/2 T2 F T1=T2=T3=Mg/2 and T4=3Mg/2 T5=Mg T3 and F=T1=Mg/2 Step 5: evaluate the answer (here, dimensions are OK and no numerical values) < T5 M Physics 151: Lecture 8, Pg 24 Lecture 9, ACT 1 Gravity and Normal Forces A woman in an elevator is accelerating upwards The normal force exerted by the elevator on the woman is, A) greater than B) the same as C) less than the force due to gravity acting on the woman Physics 151: Lecture 8, Pg 25 Lecture 9, ACT 1 Gravity and Normal Forces The free body diagram is, N mg For the woman to accelerate upwards, the normal force on the woman must be A) greater than the force due to gravity acting on the woman Note, both of these forces act on the woman, they cannot be an action/reaction pair Physics 151: Lecture 8, Pg 26 Lecture 9, ACT 1b Gravity and Normal Forces A woman in an elevator is accelerating upwards The normal force exerted by the elevator on the woman is, A) greater than B) the same as C) less than the force the woman exerts on the elevator. Physics 151: Lecture 8, Pg 27 Lecture 9, ACT 1b Gravity and Normal Forces The action/reaction force diagram for the woman and elevator is, N = FW,E FE,W By Newton’s third law these must be (B) equal. Physics 151: Lecture 8, Pg 28 Recap of today’s lecture Newton’s 3 Laws Free Body Diagrams Action/Reaction Force pairs Reading for Friday, Ch 5.7-8, pp. 123-139 Applications of Newton’s Laws and Friction Homework #3 (due next Wed. / Sept. 21 by 11:59 pm) Physics 151: Lecture 8, Pg 29