Lesson 4: Relative motion, Forces, Newton s laws (sections )

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1 Lesson 4: Relate moton, Forces, Newton s laws (sectons ) We start wth a projectle problem. A olf ball s ht from the round at 35 m/s at an anle of 55º. The round s leel.. How lon s the ball n the ar?. What s the maxmum heht of the ball? 3. How far from the launchn pont does the ball ht the round? 4. What s the ball s poston after.0 seconds? When does t reach ths heht aan? 5. When s the ball 0 m aboe the round? In the plot below, all unts are n meters = 55 o x Soluton:. When the ball hts the round, = 0. t 0 t( a ( t) a t) Snce the onl wa a product can equal zero s when one of the factors equals zero, t 0 or ( a t) 0 Lesson 4, pae

2 Lesson 4: Relate moton, Forces, Newton s laws (sectons ) The frst condton tells us that the olf ball starts from the round. The second es us the tme of flht, tf ( a t) 0 a t t f a Ths s more useful f we use a = and = sn (see the daram below the trajector plot). t f a sn (35m/s)sn m/s 5.85s What anle maxmzes the tme of flht? The anle that maxmzes sn. The larest sne can be s and that occurs at 90º. Ht the ball straht up!. At the maxmum heht, f = 0. f 0 sn t t h a t h sn But ths s ½ the tme of flht. When the ball s shot oer leel round half of the tme the ball s on up, the other half of the tme t s on down. It takes half the total tme to reach the hhest pont. The heht at ths tme s h t sn t sn a ( t) h ( t sn h ) sn Lesson 4, pae

3 Lesson 4: Relate moton, Forces, Newton s laws (sectons ) h sn (35m/s) sn 55 (9.8 m/s ) 4.9 m sn sn What anle maxmzes the heht? We need to fnd the maxmum of sn. The maxmum of sn occurs at the maxmum of sn, whch aan s 90º. Ht t straht up. 3. The ball hts the round when t = tf. The horzontal dsplacement from the launch pont to where the ball hts the round s called the rane (R). x t x R cost sn cos sn cos Ths can be rewrtten usn the dentt sn = sn cos, sn R (35 m/s) sn[(55 )] 9.8m/s 7 m What anle maxmzes the rane? Ths tme we want to fnd the maxmum of sn. The maxmum of sne occurs at 90º. Ths tme = 90º or = 45º. The anle requred for maxmum rane oer leel round s 45º. Also the rane s smmetrc about 45º. For some anle, and sn[ (45 )] sn(90 ) cos sn[ (45 )] sn(90 ) cos( ) cos f Lesson 4, pae 3

4 Lesson 4: Relate moton, Forces, Newton s laws (sectons ) The rane for = 45º + s the same for = 45º. Another wa s to sa f the ranes are the same! (45 ) (45 ) Where s the ball at.0 seconds? Its horzontal poston s found usn Its ertcal poston s t 37.7 m x snt x t cost (35m/s) cos m a ( t) ( t) (35m/s)sn 55 (.0s) f s (9.8 m/s )(.0s) When s t at ths heht aan? There are man was to fnd ths. Frst, note that the tme of flht of the ball s 5.85 s from part. Snce the trajector s smmetrc, f t reaches ths heht.0 s after launch, t wll reach t aan.0 s before t lands, t 5.85s.00s 3.85s Lesson 4, pae 4

5 Lesson 4: Relate moton, Forces, Newton s laws (sectons ) Another wa s to use the smmetr of the -component of the eloct. The - components of the eloctes at the same hehts hae the same mantudes but opposte sns. At.0 s f f a t sn t (35m/s)sn 55 (9.8m/s )(s) 9.07 m/s When s the speed 9.07 m/s? f f a sn t t t f 9.07 m/s (35m/s) sn m/s 3.85s sn Fnall, the worst wa s to just sole for t usn the quadratc formula, (9.8m/s )( t) (4.9 m/s )( t) (35m/s)sn 55 t 37.7 m 0 (8.67m/s) t 37.7 m 0 t snt 37.7 m (35m/s)sn 55 t b t ( 8.67) s, 3.86s a ( t) ( t) b 4ac a (8.67) (4.9) (9.8m/s )( t) 4(4.9)(37.7) The frst answer s when t reaches 37.7 m whle ascendn (we knew t would be.0 s), the second s when t reaches 37.7 m whle descendn. (I qut wrtn unts n the quadratc because t makes the equaton een more unweld.) Lesson 4, pae 5

6 Lesson 4: Relate moton, Forces, Newton s laws (sectons ) 5. To fnd when the ball reaches 0 m we use the quadratc equaton aan, (9.8m/s )( t) (4.9 m/s )( t) (35m/s)sn 55 t 0m 0 (8.67m/s) t 0m 0 t snt 0m (35m/s)sn 55 t b t a ( t) b 4ac a ( 8.67) s, 5.04s ( t) (8.67) (4.9) Notce that the sum of these tmes s 5.85 s, the tme of flht. (9.8m/s )( t) 4(4.9)(0) Summar: Dered equatons for a projectle launched from leel round wth ntal eloct at an anle aboe the round: Tme of flht Tme to maxmum heht Maxmum heht Rane t f sn sn th h sn sn R t f If the round s not leel, for example thrown a ball from the top of a buldn, these equatons wll not appl (unless the ntal and fnal hehts are the same)! The projectle traels n a parabolc path as lon as we nelect ar resstance. The moton s smmetrc about the maxmum heht (the ertex of the parabola). Relate eloct s a reat example of addn ectors. Hae ou eer had ths happen to ou? Whle sttn n our car at a red traffc lht, the car besde ou slowl drfts forward. You mash on the brake to stop our car from rolln backwards, but our car s not mon. Lesson 4, pae 6

7 Lesson 4: Relate moton, Forces, Newton s laws (sectons ) Wthn our enronment, there s no wa to dstnush between our car mon backwards and the car besdes ou mon forward. The eloct s relate. We need a reference frame (the traffc lht, for example) to defne who s mon. The tran moes at 0 m/s and Wanda can walk at m/s. How fast wll Gre see Wanda walk? Wanda s eloct relate to Gre s the sum of the eloct of the Wanda relate to the tran plus the eloct of tran relate to Gre. WG Notce the order of the subscrpts. We hae the Ts cancelln from the two terms on the rht. Ths equaton wll alwas hold, but how do we use t? What s our rule about ectors? WT WE DO NOT DEAL WITH VECTORS. WE DEAL WITH THEIR COMPONENTS. Take the x-component: TG WGx WTx TGx ( m/s) ( 0 m/s) m/s Gre sees Wanda walkn to the rht at m/s. What happens when she walks back to her seat? WGx WTx TGx ( m/s) ( 0 m/s) 9m/s Accordn to Gre, Wanda s walkn at 9 m/s to the rht. Lesson 4, pae 7

8 Lesson 4: Relate moton, Forces, Newton s laws (sectons ) Hopefull, ths s prett eas. But what about ths? From Example 3.0. Jack wants to row drectl across the rer from the east shore to a pont on the west shore. The current 0.6 m/s and Jack can row at.6 m/s. What drecton must he pont the boat and what s hs eloct across the rer? The eloct of the rowboat relate to the shore s equal to the eloct of the rowboat relate to the water plus the eloct of the water relate to the shore. RS WS The rowboat s to head drectl to the west. Take components. RSx and x WSx RS WS The daram s the ke to soln relate eloct problems. For the x-component, RSx RS x WSx cos 0 cos The -component, RS 0 WS WS sn sn Lesson 4, pae 8 WS

9 Lesson 4: Relate moton, Forces, Newton s laws (sectons ) Our unknowns are and RS. From the -component equaton, From the x-component equaton. WS sn WS 3.7 sn 0.6m/s.6 m/s RS cos m/s 0.56 (.6 m/s) cos 3.7 The boat must pont 3.7º N of W upstream. Its speed across the water s 0.99 m/s. Chapter 4 Force and Newton s Laws of Moton We can descrbe moton, but wh do thns moe? Forces: Objects nteract throuh forces. A force s a push or pull. Forces can be lon rane (rat, electrc, manetc, etc.) or contact (normal force, tenson, etc.). F Obousl, forces are ector quanttes snce ther effect depends on the drectons of the forces. The net force s the ector sum of all forces actn on an object. Lesson 4, pae 9

10 Lesson 4: Relate moton, Forces, Newton s laws (sectons ) F net F F F F n A free-bod daram (FBD) s an essental tool for fndn the net force actn on an object. (See pae 96.) Draw the object n a smplfed wa Identf all the forces that are exerted on the object. Draw ector arrows representn all the forces on the object. Examples. Freel falln object.. Object hann from a rope. 3. Object sttn on a horzontal table. 4. Object sttn on a horzontal table ben pulled b a rope. T N N T W W W W Drawn the free-bod daram s the ke to soln problems. Newton s Frst Law (law of nerta): An object s eloct ector remans constant f and onl f the net force actn on the object s zero (pae 97). An object mon at constant eloct has no net force! A reolutonar dea. An object mon at constant eloct s sad to be n translatonal equlbrum. That eloct could be zero. Lesson 4, pae 0

11 Lesson 4: Relate moton, Forces, Newton s laws (sectons ) Inerta s the resstance to chanes n eloct. Newton s Second law: The rate of chane of an object s eloct s proportonal to the net force actn on t and nersel proportonal to ts mass (pae 0). F m a Recall our rule: we neer deal wth ectors, we deal wth ther components. A far more useful form of Newton s second law wll be F x ma x F ma The left hand sde s suppled b the free-bod daram. The rht hand sde s suppled b our knowlede of the moton. The SI unt of force s the newton. N = k m/s. What s mass? Mass s a measure of nerta. Mass s not the same as weht. When an object s dropped t s pulled down b ts weht and ts acceleraton s downward. Appln Newton s second law es W Lesson 4, pae

12 Lesson 4: Relate moton, Forces, Newton s laws (sectons ) F W ma m and the relatonshp between weht and mass s W = m. Newton s Thrd Law: In an nteracton between two objects, each object exerts a force on the other. These forces are equal n mantude and opposte n drecton (pae 03). If two objects A and B are exertn forces on each other, F BA A B F AB F AB F BA The forces are equal n mantude and opposte n drecton. Newton s Laws of Moton (paes 97-03). An object s eloct ector remans constant f and onl f the net force actn on the object s zero.. When a nonzero net force acts on an object, the object s eloct chanes. The object s acceleraton f proportonal to the net force actn on t and nersel proportonal to ts mass. F x max F ma 3. In an nteracton between two objects, each object exerts a force on the other. These forces are equal n mantude and opposte n drecton. F AB F BA Lesson 4, pae