Chapter 5: Applying Newton s Laws

Transcription

1 Chapter 5: Appling Newton s Laws Newton s 1 st Law he 1 st law defines what the natural states of motion: rest and constant velocit. Natural states of motion are and those states are when a = 0. In essence, the first law sets up (i) the basis to define what accelerated motion is and (ii) which coordinate sstems Newton's laws hold in. Newton s nd net Law: a m Phsical Interpretation ma ma Case 1: Equilibrium or orce-free states rom Newton s nd law, there are two was to be in force-free states: a 0 rest or constant velocit 0 Phsicall, this means that ALL the forces acting on an object must sum to zero or balance out: 0 (horizontal directions) right left net i 0 0 (vertical directions) up down hat is, there must be at least two forces acting in opposite directions relative to each other. Equilibrium has two possible situations: (i) rest is static equilibrium and (ii) constant velocit is dnamic equilibrium. CAE : Nonequilibrium ituations Whenever an object accelerates, that means that there is an unbalanced force and that the object moves in the direction of the unbalanced force. hat is, a 0 f 0 (unbalanced) f net net hand hand Eample: A tossed ball is in projectile motion. When the ball reaches the peak of its trajector, is the ball in equilibrium or nonequilibrium motion? Nonequilibrium! Wh - there is an unbalanced force acting on the ball: a g 0 net mg 0 (unbalanced) ree all Motion B definition, free fall motion is defined when the onl force acting on a projectile is the force of gravit. In chapter and 4 we saw that two objects in free fall (a heav and a lighter object) hit the ground at the same time when dropped simultaneousl. Do the hit the ground simultaneousl because the same force of gravit acts on both objects? NO! he force of gravit does not act the same for both masses because g = mg is mass dependent, and both of these objects have different masses. Newton's nd law shows that m m net mg g a g m m Because it does not depend on the mass, we sa that it is mass independent. DEMO Use feather and coin in airless tube VIDEO how the video of feather and hammer on the moon Newton s 3 rd Law 5-1

2 I can t touch ou, without ou touching me in return; I can t nudge this hair without the chair in turn nudging me back. hat is, I can t eert a force on a bod without that bod in turn eerting a force on me. When a CONAC force occurs there is a INGLE INERACION requires a PAIR of ORCE between two things rd NEWON' 3 LAW (Action-Reaction) or ever action, there is an EQUAL & OPPOIE reaction: 1 1 Conceptual Eamples: Pushing on a Wall, hitting a wall with our hand and uppose I walked into a ring with Mike son: who will feel a bigger impact force: m face or Mike son s fist? I am driving m car at night and a bug smashes into m wind shield leaving a big gob. Who received the great impact force: the car or the smashed bug? Either. m m a a a large mass has a ver small acceleration a small mass has a ver large acceleration It is the acceleration (or deceleration) that causes all of the damage (if not death) in certain situations. hat is, it is the acceleration that kills a person in fatal car accidents. Interesting Point Astronauts on the ver first trip into space had major problems in turning a screw outside the space craft. Let me eplain. On earth, whenever one turns a door knob there will alwas be a force applied b the door knob on our hand in the opposite direction that ou turn it (Newton s 3 rd law). Question: wh is it that ou don t move according to this force? Gravit stops ou from rotating. Getting back to the first NAA space flight, one of the astronauts tried to turn a knob on the outside of the space craft. Because there is not friction, as he tried to turn the knob, the knob turned him. In fact, the astronaut took one hour to finall turn the knob and in the process lost 10 lbs of water due to sweating. o how did the solve this problem for future flights add foot petals so their feet would lock in place. Appling Newton s nd Law Problem olving trategies tep 0: Is the object in equilibrium or nonequilibrium? If equilibrium ma 0 If nonequilibrium ma 0 tep 1: ketch the situation and draw a ree-bod Diagram (BD). isolate the object and reduce to a center-of-mass point identif all the forces acting on this point choose a coordinate sstem convenient to ou 5-

3 tep : Breakup all forces into its components along the - and -aes, and sum the components. A force table can be convenient. orce -component -component tep 3: Appl Newton s nd law (Σ = ma, Σ = ma ), pick the acceleration direction as positive, and solve the two equations for the desired unknown quantities. Eample 5.1 he figures show BDs for an object of mass m. Write the - and -components of Newton's second law. Write our equations in terms of the magnitudes of forces 1,,... and an angles defined in the diagram. ma ma ma ma Eample 5. he three ropes are tied to a small, ver light ring. wo of these ropes are anchored to walls at right angles with the tensions shown in the figure. What are the magnitude and direction of the tension 3 in the third rope? olution We are asked to solve for the magnitude and direction (θ) of 3. tep 0: Is the ring in equilibrium or nonequilibrium? he massless ring is in static equilibrium, so all the forces must balance out; that is, forces acting on it must cancel to give a zero net force. 0 0 right left up down tep 1: ketch the situation and draw a BD. tep : Breakup all forces into its components and sum them. here are two was to usuall solve the problem: the hard wa or the eas wa. cos 0 cos sin 0 sin

4 tep 3: Appl Newton s nd law (Σ = 0, Σ = 0),solve the two equations for 3 and θ. I will first solve for the angle θ: cos cos sin sin cos aking the inverse tangent, solve for 3 1 sub this into cos 1 1 1tan tan tan N 50 80N 1 1 tan 80 Net solve for the magnitude, I use one of the equations to solve for 3 : 50 N 0 cos 94 N solve for cos 1 50N cos58 his is the usual wa that all phsics book will show ou how to solve Newton nd tpe problems. However, it is not the efficient wa to solve them. orces are vectors and therefore, follow vector rules. hat is, (50) (80) 94 N; θ tan tan Eample 5.3 a. A 0.60 kg bullfrog is on a log tilted 30 above horizontal. How large is the normal force on the log on this bullfrog? olution Is weight a scalar (a number) or is a vector? his eample will clearl show ou that the weight is a vector without doubt. How much does the frog actuall weight on a flat surface? mg 0.60kg 9.81 m/s 5.9 N mg m0.60kg tep 0: Is the frog in equilibrium or nonequilibrium? he frog is in static equilibrium, so all the forces must balance out; that is, forces acting on it must cancel to give a zero net force. 0 0 tep 1: ketch the situation and draw a BD. right left up down tep : Breakup all forces into its components and sum them. mgcos 60 f 0 f mgcos 60 N mgsin 60 0 N mgsin

5 tep 3: Appl Newton s nd law (Σ = 0, Σ = 0), solve the two equations for N. olving for the normal force using the -equations, m0.60kg N mgsin kg 9.81 m/s cos N N he normal force is supporting onl N 5.1 N 86% of the frog's weight mg 5.9 N he answer is less than the weight of the frog. Does that mean that the "rest of the weight of 0.8 N" lies along the incline? NO! Because the weight is a vector force and not just a simple number. he weight of the frog is distriubted along two directions, the - and -ais. he other part of the frog's weight is held up b the static friction force (or with weight along the -direction) and is given b W mgcos kg 9.81 m/s cos60.9 N W m0.60kg Interpret the solution: Because on the incline plane, the weight gets distributed along two aes such that mg W W N. he force on the plane acting on the object is less than if the object was on flat ground the object s weight is NO full supported b the incline. Clearl, the weight has true vector behavior! b. A 4000 kg truck is parked on a 15 slope. How big is the friction force on the truck? olution How much does the truck actuall weight on a flat surface? mg 4000 kg 9.81 m/s 39,40 N mg m4000kg tep 0: Is the frog in equilibrium or nonequilibrium? he truck is in static equilibrium, so all the forces must balance out; 0 0 tep 1: ketch the situation and draw a BD. right left up down tep : Breakup all forces into its components and sum them. mgcos 75 f 0 f mgcos 75 N mgsin75 0 N mgsin75 tep 3: Appl Newton s nd law (Σ = 0, Σ = 0), solve the two equations for f. olving for the frictional force using the -equations, m4000kg f mgcos kg 9.81 m/s cos N N f he frictional force is supporting onl N N 6% of the truck's weight mg 3940 N 5-5

6 Eample 5.4 he figure shows two 1.00 kg blocks connected b a rope. Assume the rope is massless. he entire assembl is accelerated upward at 3.00 m/s b force. a. What is? b. What is the tension of rope? olution a. he two block-sstem is accelerating upward at a 3.0 m/s. What force is accelerating this sstem? he net force Σ. tep 0: nonequilibrium Σ = ma tep 1,, 3: Draw a BD, sum the forces and appl Newton s second law: (m m )g (m m )a (m m )(a g) 5.6 N A B A B A B b. In order to determine the tension, I have to focus on the individual block's. If I focus on block A, I find that tep 0: nonequilibrium Σ = ma tep 1,, 3: Draw a BD, sum the forces and appl Newton s second law: A mag maa m A(a g) 5.6 N (1 kg) (3 9.81)m/s 1.8 N or I can do the same to Block B: B mbg mba m B(a g) (1 kg) ( ) m/s 1.8 N RICION here are two tpes of frictional forces static and kinetic friction. mbolicall the are written as f tatic frictional force & f kinetic frictional force DEMO Produce a friction curve using a heav block with a spring scale Consider a block that is being pulled to m left. When I first appl a small force, the block does not move this is because the force of friction is balancing out the force of m hand. As I appl a larger force, the block still does not move, indicating that the static frictional force increases in strength with the applied force of m hand force. In other words, the static frictional force is not a constant force. As I continue to appl an increasing force eventuall the static force reaches a certain maimum value and then the force of m hand over comes it. his is the so-called breakawa point when the block starts to moves. At this point, friction changes its nature and instead of a varing force, the frictional force is roughl constant in nature and less than the maimal value. k Interpretation this agrees with our intuition. Phsicall, it is harder to start moving an object since f increases with the pushing force. After maing out f, the object moves. It then becomes easier to move the object since f k < f. 5-6

7 rom a microscopic viewpoint, irregularities in the surfaces cause friction. his can be seen from polished stainless steel at the micro level. Properties of riction 1. rictional forces alwas oppose the direction of motion.. Eperimental facts about riction frictional force Normal force lots of constraints (temp, area, speed,...) f N he frictional force is proportional to the normal force with a proportionalit constant μ is called the coefficient of friction: coefficient of static friction coefficient of kinetic friction f N variable force f N constant force tatic kinetic k f (ma) N constant force k Materials Coefficient of static friction, Coefficient of kinetic friction, k Units: [] = 1 (unitless) Eample teel on steel Copper on steel Copper on cast iron Copper on glass glass on glass Rubber on concrete (dr) Rubber on concrete (wet) Bonnie and Clde are sliding a 300 kg bank safe across the floor to their getawa car. he safe slides with a constant speed if Clde pushes from behind with 385 N of force while Bonnie pulls forward with a rope with 350 N of force. What is the safe's coefficient of kinetic friction on the bank floor? olution ince it is sliding with constant velocit, it is in dnamic equilibrium and the kinetic friction opposes the motion b pointing to the left. etting up our notation: 350 N, 385 N, m m 300 kg Bonnie B Clde C afe

8 tep 0: equilibrium Σ = 0 tep 1,, 3: Draw a BD, sum the forces and appl Newton s second law: f N 0 B C k f B C k kkn N mg 0 o solve for the coefficient of friction, we need to focus solving for the normal force in the -equations and substitute it into the -equations to obtain μ k. f 0 f 350 N 385 N 735 N hen, for kinetic friction B C k k B C N mg 0 N mg (300 kg)(9.8 m/s ) 940 N f 735 N f N 0.5 N 940 N k k k k k Elevator Problems and Weightlessness Worked Eample An elevator with a weight of 7.8 kn is given an upward acceleration of 1. m/s b a cable. (a) Calculate the tension in the cable. (b) What is the tension when the elevator is decelerating at the rate of 1. m/s moving downward? olution a. he mass of the elevator is g /g =837 kg and the acceleration is 1. m/s. It is moving upwards to that is the positive direction. A BD and Newton s nd law sets up the equation to determine the tension: mg ma m g a 10 up up 3.13 N up b. he onl difference from part (a) is now the velocit points downward and therefore, defines positive as downward; a BD and nd law sets up to give 4 mg ma m g a 10 down down.43 N down One can see that the tension in the cable changes, depending on the direction of the acceleration. Reinterpret the solution While in an elevator at rest or moving at constant velocit, the weight scale reading would read eactl the same as the normal force of the elevator: 0 N mg 850N net scale hat is, if ou are in a elevator at rest or moving with constant velocit, the weight scale would read our weight as if ou where standing on the ground. I would "feel" normal and weight 850N. If now ou are in an accelerating elevator, there is a net force acting on ou and according to our eample above, we should feel "heavier" since the cable had a higher tension; a higher cable tension implies there is a higher normal force. hat is, m weight increases so 0 N m(g a) ( )N 955 N (15 lbs) net up It is the net force that causes the "ma" term to appear, which in turn causes the "weight" increase in this accelerating frame. Ver important: the term "ma" is NO a force acting on ou but is the result of the net force causes this term - acceleration and force are not

9 the same thing. Nonetheless ou feel the phsical sensation of this net force. If now the elevator decelerates, the normal force of the elevator acting on ou will causes m "weight" to change according to 0 N m(g a) ( )N 745 N (167 lbs) net down he phsical sensation now is that I feel "light." Question: what would the scale read if the elevator was in free-fall? ZERO! Because the scale reads zero one sas that ou are weightless. Questions: Is a force acting on ou? YE - gravit 5-9