Chapter 5 Force and Motion I

Transcription

1 Chapter 5 orce and Motion I I. ewton s irst law. II. ewton s second law. III. Particular orces: -Gravitational - Weight -ormal -riction - ension IV. ewton s third law.

2 ewton mechanics laws cannot be applied when: 1) he speed o the interacting bodies are a raction o the speed o light Einstein s special theory o relativity. ) he interacting bodies are on the scale o the atomic structure Quantum mechanics I. ewton s irst law: I no net orce acts on a body, then the body s velocity cannot change; the body cannot accelerate v = constant in magnitude and direction. - Principle o superposition: when two or more orces act on a body, the net orce can be obtained by adding the individual orces vectorially. - Inertial reerence rame: where ewton s laws hold.

3 II. ewton s second law: he net orce on a body is equal to the product o the body s mass and its acceleration. net ma (5.1) ma, net, y may, net, ma net, x x z z (5.) - he acceleration component along a given axis is caused only by the sum o the orce components along the same axis, and not by orce components along any other axis. - System: collection o bodies. - External orce: any orce on the bodies inside the system. III. Particular orces: -Gravitational: pull directed towards a second body, normally the Earth mg (5.3) g

4 - Weight: magnitude o the upward orce needed to balance the gravitational orce on the body due to an astronomical body W mg (5.4) - ormal orce: perpendicular orce on a body rom a surace against which the body presses. mg (5.5) - rictional orce: orce on a body when the body attempts to slide along a surace. It is parallel to the surace and opposite to the motion. -ension: pull on a body directed away rom the body along a massless cord.

5 IV. ewton s third law: BC CB When two bodies interact, the orces on the bodies rom each other are always equal in magnitude and opposite in direction. (5.6) QUESIOS Q. wo horizontal orces 1, pull a banana split across a rictionless counter. Without using a calculator, determine which o the vectors in the ree body diagram below best represent: a) 1,b). What is the net orce component along (c) the x-axis, (d) the y-axis? Into which quadrant do (e) the net-orce vector and () the split s acceleration vector point? 1 (3)ˆ i (4) ˆj (1 )ˆ i () ˆj net ()ˆ i (6) ˆ 1 j Same quadrant, 4 1

6 I. rictional orce Counter orce that appears when an external orce tends to slide a body along a surace. It is directed parallel to the surace and opposite to the sliding motion. -Static: ( s ) compensates the applied orce, the body does not move. s // o motion -Kinetic: ( k ) appears ater a large enough external orce is applied and the body loses its intimate contact with the surace, sliding along it. (applied orce) Acceleration Constant velocity

7 (6.1) k s,max s, max s I s body //, max slides riction coeicients k k (6.) Ater the body starts sliding, k decreases.

8 Q1. he igure below shows overhead views o our situations in which orces act on a block that lies on a rictionless loor. I the orce magnitudes are chosen properly, in which situation it is possible that the block is (a) stationary and (b) moving with constant velocity? a y 0 a=0 a=0 a y 0 net Q5. In which situations does the object acceleration have (a) an x-component, (b) a y component? (c) give the direction o a. net

9 Q. A body suspended by a rope has a weigh o 75. Is equal to, greater than, or less than 75 when the body is moving downward at (a) increasing speed and (b) decreasing speed? net g ma m( g a) Movement g (a) Increasing speed: v >v 0 a>0 < g (b) Decreasing speed: v < v 0 a<0 > g Q8. he igure below shows a train o our blocks being pulled across a rictionless loor by orce. What total mass is accelerated to the right by (a), (b) cord 3 (c) cord 1? (d) Rank the blocks according to their accelerations, greatest irst. (e) Rank the cords according to their tension, greatest irst. 1 3 (a) pulls m total = ( )kg = 0kg (b) Cord 3 3 m=(10+3+5)kg = 18kg (c) Cord 1 1 m= 10kg (d) =ma All tie, same acceleration

10 (e) - 3 =a 3 - =5a - 1 =3a 1 =10a - 3 =a =18a+a=0a 3-13a= 5a 3 =18a -10a=3a =13a 1 =10a Q. A toy box is on top o a heavier dog house, which sits on a wood loor. hese objects are represented by dots at the corresponding heights, and six vertical vectors (not to scale) are shown. Which o the vectors best represents (a) the gravitational orce on the dog house, (b) on the toy box, (c) the orce on the toy box rom the dog house, (d) the orce on the dog house rom the toy box, (e) orce on the dog house rom the loor, () the orce on the loor rom the dog house? (g) Which o the orces are equal in magnitude? Which are (h) greatest and (i) least in magnitude? (a) g on dog house: 4 or 5 (h) Greatest: 6,3 (b) g on toy box: (i) Smallest: 1,,5 (c) toy rom dog house: 1 (d) dog-house rom toy box: 4 or 5 (e) dog-house rom loor: 3 () loor rom dog house: 6 (g) Equal: 1=, 1=5, 3=6

11 5. here are two orces on the kg box in the overhead view o the igure below but only one is shown. he igure also shows the acceleration o the box. ind the second orce (a) in unit-vector notation and as (b) magnitude and (c) direction. ˆ a (1cos 40 i 1sin 40 ˆ) j m / s ( 6ˆ i ˆ) j m / s ma kg( 6ˆ i ˆ) j m / s ( 1ˆ i 0.78 ˆ) j 0ˆ i 1 x y x y 38.7 x 3 ( 3ˆ i 0.78 ˆ) j tan or

12 Rules to solve Dynamic problems - Select a reerence system. - Make a drawing o the particle system. - Isolate the particles within the system. - Draw the orces that act on each o the isolated bodies. - ind the components o the orces present. - Apply ewton s second law (=ma) to each isolated particle.

13 9. (a) A 11kg salami is supported by a cord that runs to a spring scale, which is supported by another cord rom the ceiling. What is the reading on the scale, which is marked in weigh units? (b) Here the salami is supported by a cord that runs around a pulley and to a scale. he opposite end o the scale is attached by a cord to a wall. What is the reading on the scale? (c) he wall has been replaced by a second salami on the let, and the assembly is stationary. What is the reading on the scale now? W g mg (11kg)(9.8m / s ( a) a 0 g ) ( b) a 0 g g g

14 g g ( c) a 0 g In all three cases the scale is not accelerating, which means that the two cords exert orces o equal magnitude on it. he scale reads the magnitude o either o these orces. In each case the tension orce o the cord attached to the salami must be the same in magnitude as the weigh o the salami because the salami is not accelerating.

15 3. An electron with a speed o 1.x10 7 m/s moves horizontally into a region where a constant vertical orce o 4.5x10-16 acts on it. he mass o the electron is m=9.11x10-31 kg. Determine the vertical distance the electron is delected during the time it has moved 30 mm horizontally. d y g v 0 d x =0.03m net net d x vxt 0.03m (1. 10 m / s) t t. 4ns ma y ( g kg) a y a 7 16 y ( m / s kg)(9.8m / s ) d y voyt 0.5ayt 0.5( m / s ) (.510 s) m 14 9

16 13. In the igure below, m block =8.5kg and θ=30º. ind (a) ension in the cord. (b) ormal orce acting on the block. (c) I the cord is cut, ind the magnitude o the block s acceleration. g ( a) a 0 a x 0 gx 0 gx mg sin 30 (8.5kg)(9.8m / s ) ( b) ay 0 gy 0 gy mg cos ( c) 0 gx ma a a 4.9m / s

17 55. he igure below gives as a unction o time t, the orce component x thatactsona3kgiceblock, which can move only along the x axis. At t=0, the block is moving in the positive direction o the axis, with a speed o 3m/s. What are (a) its speed and (b) direction o travel at t=11s? t 0 v 3m / s t 11s v? x dvx ax m dt v 0 dvx dt v dt otal graph area 15s 11s 15kgm / s 3m / s 8m / s 3kg 0 v 0 dt ( v x 11s 0 x m v 0 dt ) m ( v 3m / s)3kg Midterm1_extra_Spring04. wo bodies, m1= 1kg and m=kg are connected over a massless pulley. he coeicient o kinetic riction between m and the incline is 0.1. he angle θ o the incline is 0º. Calculate: (a) Acceleration o the blocks. (b) ension o the cord. g, x Block 1: m g sin 0 k g, y 6.7 m g cos 0 m g cos0 k m g m a a 0º m m g m 1 m 1 g Block : g, x m a a Adding 3a 1.6 a 0.4m / s, 9.38

18 Midterm1_Spring04. he three blocks in the igure below are connected by massless cords and pulleys. Data: m 1 =5kg, m =3kg, m 3 =kg. Assume that the incline plane is rictionless. (i) Show all the orces that act on each block. (ii) Calculate the acceleration o m 1, m, m 3. (iii) Calculate the tensions on the cords. (iv) Calculate the normal orce acting on m m gy m 3 30º gx 1 g y =m gcos30º g x =m gsin30º m 3 g m g m 1 g 1 m 1 Block 1: m 1 g- 1 =m 1 a Block : m g(sin30º) =m a Block 3: -m 3 g = m 3 a (i) Adding (1)+()+(3) g(m m -m 3 )=a(m 1 +m +m 3 ) a= 4.41m/s (ii) 1 =m 1 (g-a)= 5kg(9.8 m/s m/s ) = 6.95 (iii) =m 3 (g+a)= kg(9.8 m/s m/s )= 8.4 (iv) = g y = m gcos30º = 5.46

19 1B. (a) What should be the magnitude o in the igure below i the body o mass m=10kg is to slide up along a rictionless incline plane with constant acceleration a=1.98 m/s? (b) What is the magnitude o the ormal orce? m( a 0.5g) cos 0 mg sin 30 ma cos 0 mg cos30 sin B. Given the system plotted below, where m 1 =kg and m =6kg, calculate the orce necessary to lit up m with a constant acceleration o 0.m/s. he pulleys and cords are massless, and the table surace is rictionless. y 30º g 0º x Movement m 1 m 1 g d d 1 1 a1t d1 m 1 a g m t a 0.5a t 1 a t a 0.5a 1 0.m / s a 1 0.4m / s 0.5m ( a g) 0.5(6kg)(0. 9.8) m / s 30 m g m m a 1 1 m a 30 (kg)(0.4m / s )

20 Chapter 5 orce and Motion II I. Drag orces and terminal speed. II. Uniorm circular motion. III. on-uniorm circular motion.

21 I. Drag orce and terminal speed -luid: anything that can low. Example: gas, liquid. -Drag orce: D - Appears when there is a relative velocity between a luid and a body. - Opposes the relative motion o a body in a luid. - Points in the direction in which the luid lows. Assumptions: * luid = air. * Body is blunt (baseball). * ast relative motion turbulent air.

22 D 1 CAv (6.3) C = drag coeicient (0.4-1). ρ = air density (mass/volume). A= eective body s cross sectional area area perpendicular to v -erminal speed: v t - Reached when the acceleration o an object that experiences a vertical movement through the air becomes zero g =D D g ma i a 0 1 CAv g 0 v t g CA (6.4)

23 II. Uniorm circular motion -Centripetal acceleration: a v r (6.5) v, a = constant, but direction changes during motion. A centripetal orce accelerates a body by changing the direction o the body s velocity without changing its speed. -Centripetal orce: m v R (6.6) a, are directed toward the center o curvature o the particle s path.

24 III. on-uniorm circular motion a r a t a - A particle moves with varying speed in a circular path. - he acceleration has two components: - Radial a r =v /R - angential a t = dv/dt -a t causes the change in the speed o the particle. - In uniorm circular motion, v = constant a t =0 a=a r r r v dt v d a a a r t ˆ ˆ t r

25 49. A puck o mass m slides on a rictionless table while attached to a hanging cylinder o mass M by a cord through a hole in the table. What speed keeps the cylinder at rest? mg or M or m Mg v m r ac 0 Mg v m r v Mgr m Mg 33E. Calculate the drag orce on a missile 53cm in diameter cruising with a speed o 50m/s at low altitude, where the density o air is 1.kg/m 3. Assume C= m / s 6. k D CAv (1. kg / m ) (0.53m / ) 3. he terminal speed o a ski diver is 160 km/h in the spread eagle position and 310 km/h in the nosedive position. Assuming that the diver s drag coeicient C does not change rom one point to another, ind the ratio o the eective cross sectional area A in the slower position to that o the aster position. v t g 160km/ h CA 310km/ h g CA g CA E D A A D E A A E D 3.7

26 11P. A worker wishes to pile a cone o sand onto a circular area in his yard. he radius o the circle is R, and no sand is to spill into the surrounding area. I μ s is the static coeicient o riction between each layer o sand along the slope and the sand beneath it (along which it might slip), show that the greatest volume o sand that can be stored in this manner is πμsr 3 /3. (he volume o a cone is Ah/3, where A is the base area and h is the cone s height). - o pile the most sand without extending the radius, sand is added to make the height h as great as possible. - Eventually, the sides become so steep that sand at the surace begins to slip. - Goal: ind the greatest height (greatest slope) or which the sand does not slide. Cross section o sand s cone Static riction grain does not move R h gy mg θ gx x y gy gx mg cos mg sin gx mg sin I grain does not slide s, max s smg cos s tan he surace o the cone has the greatest slope and the height o the cone is maximum i : tan V s cone Ah 3 h R R h R ( Rs ) 3 s sr 3 3

27 1. Block B weighs 711. he coeicient o static riction between the block and the table is 0.5; assume that the cord between B and the knot is horizontal. ind the maximum weight o block A or which the system will be stationary. System stationary s, max s Block B Knot 1 y m B g 1 s,max x sin 30 cos cos 30 gb ga 3 Block A m A g sin P. wo blocks o weights 3.6 and 7., are connected by a massless string and slide down a 30º inclined plane. he coeicient o kinetic riction between the lighter block and the plane is 0.10; that between the heavier block and the plane is 0.0. Assuming that the lighter block leads, ind (a) the magnitude o the acceleration o the blocks and (b) the tension in the string. (c) Describe the motion i, instead, the heavier block leads. Block A A B Block B B k,b gxa gya ka Light block A leads gxb gyb kb A A k,a ga B gb

28 Light block A leads Block gxa A ka A ka m A gya ka A m A g cos 30 (0.1)(3.1 ) 0.31 a (3.6 ) sin a a Block B B gyb m B g cos kb kb B (0.)(6.3 ) 1.5 m a (7. ) sin gxb kb B 0.73 a.35 a 3.49 m / a s W W A W B W cos 0. A kb ka B Heavy block B leads Reversing the blocks is equivalent to switching the labels. his would give ~(μ ka -μ kb )<0 impossible!!! he above set o equations is not valid in this circumstance a A a B he blocks move independently rom each other.

29 74. A block weighing is held against a vertical wall by a horizontal orce o magnitude 60. he coeicient o static riction between the wall and the block is 0.55 and the coeicient o kinetic riction between them is A second P acting parallel to the wall is applied to the block. or the ollowing magnitudes and directions o P, determine whether the block moves, the direction o motion, and the magnitude and direction o the rictional orce acting on the block: (a) 34 up (b) 1 up, (c) 48 up, (d) 6 up, (e) 10 down, () 18 down. (a) P=34, up Without P, the block is at rest P P mg ma I we assume 34 1 down s,max s a 0 s,max 33 Block does not move k 0.55 (60 ) 33 k s 0.38 (60 ).8 =60 mg= ==60 (b) P=1, up (c) P=48, up P P mg ma up P P mg ma down s,max 33 ot moving s,max 33 ot moving (d) P=6, up P P mg 0 (*) 6 40 up s,max 33 P mg ma Block with moves k up Assumption.8 down (*) wrong

30 (e) P=10, down P P mg ma up s,max 33 ot moving () P=18, down P P mg ma up s,max k 33 moves.8 up 8. Blocks A and B have weights o 44 and, respectively. (a) Determine the minimum weight o block C to keep A rom sliding i μ s between A and the table is 0.. (b) Block C suddenly is lited o A. What is the acceleration o block A i μ k between A and the table is 0.15? (a) s,max Block A a s 0 s, max 0 s (1) W c Block B m B g 0 () W A= 44 ( 1) () s Blocks A, B W A WC WC W B = (b) C disappears m m B k g m A B a a m A 6.6 g a.a a.3m 17 / s

31 9. he two blocks (with m=16kg and m=88kg) shown in the igure below are not attached. he coeicient o static riction between the blocks is: μ s =0.38 but the surace beneath the larger block is rictionless. What is the minimum value o the horizontal orce required to keep the smaller block rom slipping down the larger block? mg Movement Mg reat both blocks as a single system sliding across a rictionless loor mtotal a a m M Small min required to keep m rom sliding down? block ' ma m m M mg 0 ' mg s s mg m M ( 1) () s M mg 488 m M M 44. An amusement park ride consists o a car moving in a vertical circle on the end o a rigid boom o negligible mass. he combined weigh o the car and riders is 5k, and the radius o the circle is 10m. What are the magnitude and the direction o the orce o the boom on the car at the top o the circle i the car s speed is (a) 5m/s (b) 1m/s? s 0 (1) () y B he orce o the boom on the car is capable o pointing any direction W B W m v R B W 1 v Rg ( a) v 5m / s B 3.7 up ( b) v 1 m / s B. 3 down