hwsol4 - Applying Newton's Second Law Part A: Consider...

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Unformatted text preview: Applying Newton's Second Law Part A: Consider another special case in which the inclined plane is vertical (0 = 1r/2). In this case, for what value of ml would the acceleration of the two blocks be equal to zero? m, :::IY1 ~ Part B: Which of the four drawings is a correct force diagram for this problem? B;=s OY\ uorre..c.t. bof~ TeYtsion J ()...s e.xerrs we)( lX. fot"c,~ bl'fk bloc/is C4..~ BrA.V,'l--y. t N v 111211 "'2g illIg I III III block I bl'fk a block I t NT v T I:;~ck m~ h 1Il~ h ~T mig A"ormo..l Force. o-C,h an bloc,X J.. blOCkrm~an 2 I ... ~ "'l.S: c Part C: Given the criteria just described, what orientation of the coordinate axes should you use in this problem? block 2. 11/,1I block J '" 1(/ IIIht: - lII(a 11120~ d L e. \I e,;1 f. 0 r h l 0 c I~ 1- Owncl. j-;Jl-uJ (~lon3 ~. rite /nGII'Vl.e.) for b lOG/'i Part D: What is F2.r, the sum of the x components of the forces acting on block 2? Take forces acting up the incline to be positive. L ().s, \ 'i\ 0 -(t'~ ~ Z F.~x ::: T ------_.. PARTE: rn a. () <J s"';') t7 2-F Iy =-T-n13 I ..since.. Cl-+ - P6.f' l- F: -I-hey +-h e, ()...f"f!.- a.H-O-cneJ ra.. rL. . 9 j -rhey I Cl.CceleYC:)..-re s {A..fl' e.. 0-;).1' ':::-Q Y - tYl, D., fa.rf- Gt: (Y1 :lCl~X ::.. -r - n1.;1.~.sr" >' ..:: - T + J'Yl, ~ yMlt-m~ r. -~.s i t'\G -T mJ. M, +f) -=7 I:=. II ' ) ~-r.slt\G5+ ~rm:t. Newton's 3rd Law Discussed Part A: Every force has one and only one 3rd law pair force. -rrve.~ -;:. W'e>() lcl.JI'\ VV r t> ~ CA- pair j f f here... a.-S -m 0 I" e.. t- fleA- Y) 1-. Part B: The two forces in each pair act in opposite directions. .::...JL- -r (' v e., Part C: The two forces in each pair can either both act on the same body or they can act on different bodies. .r O..!Sf- .. - i +j u s+ Wa 0 v / c! h '" + YYl a f{ e. ~ enS e. W l't A fle... pocf...y.' Part D: The two forces in each pair may have different physical origins (for instance, one of the forces could be due to gravity, and its pair force could be due to friction or electric charge). FoJsc... . '. ()fP os e (L{rn01Njh eft,-Ch 3ra..v if-y q.. +rr'ction (!> +-Ae-r _.' -rhe- ?-O'C-t? ,'S ;tl(),Y nO I- et300-1 O-ncl oppos;+e::.. F;z. (plu~ SOvrr'le.....hCJoL-y() Part E: The two forces of a 3rd law pair always act on different bodies. r rue.- .1 Fs (f\ , Ct, ::; yY1;;)..Cl~ c:- J... 60 eLI' e.s Part F: Given that two bodies interact via some force, the accelerations of these two bodies have the same magnitude but opposite directions. (Assume no other forces act on either body.) false.. . ; of fA,e h Ck- V a..-f-io11 a.~ Oll ~ ~ ~ o..c.c e, leI" W,. [I d.; ff~reJ' f!-X" ~ +- t>1O-SS es he.. cJ...J'fferen tp I e) (.fC5r Part G: According to Newton's 3rd law, the force on the (smaller) moon due to the (larger) earth is equal in magnitude but anti parallel to the force on the earth due to the moon. O ~ - ~()..r' tA b" m. ~O1\ en e..D..rt h ~ -0 mO()/) ~ J1 -l- ( I' cArC4.[' eI V~c.. +OY'S Free-Body Diagrams Suppose that you are asked to solve the following problem: Chadwick is pushing a piano across a level floor. The piano can slide across the floor without friction. If Chadwick applies a horizontal force to the piano, what is the piano's acceleration? To solve this problem you should start by drawing a free-body diagram. Part A: Determine the object of interest for the situation described in the problem introduction. The Piano!!! () Part B: Identify the forces acting on the object of interest. From the list below, select the forces that act on the piano. <l f o )< ().,Yl 0 "s vVe,( ~ h t not'yt\O-I t:O r(., e.. 0 t-ore e. P Gho.cl w; (,./-\0 Y1 ? ,'0-- n (; Part C: Which of the following statements best describes the correct directions and relative magnitudes of the forces involved? FpU<h"! 1 f FtJ J Pw 1::;; J Fw ) F..., Part D: Determine the object of interest for this situation. (Jjo.-no , r:fJ ~ r:'IIV ~o- Part E: Which diagram accurately represents the free-body diagram for the piano? 9 ro.. v,' +-y Problem 4.10 A dockworker applies a constant horizontal force of Pto a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance xin a time t1. Part A: What is the mass of the block ofice? -> r.~~ m. 0... ~.l;{).t-A -'> -/ Cl.:= ~ t~ I rrt::. F ()... -' Ft< ?,.X Part B: If the worker stops pushing at the end of tJ, how far does the block move in the next t2? V,t. :. a.. t-, == ~ ~ ~ )(?. c( ==- V? tJ.::: ~X f. t J Problem 4.32 You have just landed on Planet X. You take out a ball of mass 111, release it from rest from a height of h and measure that it takes a time of tto reach the ground. You can ignore any force on the ball from the atmosphere of the planet. How much does the ball weigh on the surface of Planet X? We, 111\0 vi )ve,i~ h r ::::- CAm W~o.+ fs -rhe. 0-Cce/era+-/~n? h := ( t':t /:t tL +A..e,re, t-O r e., ~ (A..::; t2.. ~~ lAS we... s e. e.. We,;8A-I- :::; Problem 4.36 An advertisement claims that a particular automobile can "stop on a dime." What net force would actually be necessary to stop an automobile of mass 11ltraveling initially at a speed of vin a distance equal to the diameter of a dime, which is d? P::;- mo... - Wh~+ 15 fhe. Tn e.t 3 n; +ud-e t) f 0.. cceler~}ion ? uS'{. Vi ')..=, V,'~ . r lo... 0 X ~x:;.ot Vl! ( ol ,.(J..fYl e-tet"' <:> I of. ,..J?1 e.) =:.. 6 ~. ::::V V -:l. ::= ~ 0... ol v?, ~ O\.=:- clot. There-fore- IFI == ~J: ~ Problem 4.42 A spacecraft descends vertically near the surface of Planet X. An upward thrust of 25.0 kN from its engines slows it down at a rate of up at a rate of 0.80 m/52 1.20 m/s2 , but if an upward thrust of only 10.0 kN is applied, it speeds . Part A: What is the direction of the acceleration of the spacecraft in the case of25.0 kN thrust? Si h e e.- rh e.., $f ~d is c1..ecre,cu/.rt .:J f he her ~ c.eel er o..:f-; "11 f ~ Uf/ ~ _ oPo: OJ ~ Part B: What is the direction of the acceleration of the spacecraft in the case of 10.0 kN thrust? .s " h <: e.. t- h fAt- fl.. s:p e-~ I~5 t'(lC r e. a...s i\~9 J ne.+ a..Lc..el-e..ro..h~1I /s ~. ~ 0.- tN'~ \AI ()..rck Part C: Apply Newton's second law to each case, slowing down or speedzng up, and use this to find the spacecraft's weight near the surface of Planet X. cto s IN ~ or J~ Yl (!) KI f h a.. + + hen l-orLe W ==- L -I- a.. c. ~ {> Y' (J.. -I- /0 f1 n e-fFJ - m a.. r F~ ~ f- \AI ;:::= m~":). we- ol. i vi&.. e- ..f- h e.s e... +- wo e-~ U().-tic ns (;A I ... ~:l -- F, - ltV =- ~o..l - -:::- I.;;'() ~~-w::;:;. ritA...;). o-..4h":- - O. SO ~5 ;) solv;n~ t~(' w w -<: ~. P, -a.;r~ ..... C\,-~';t. wh e.h ~.(, plu'() in fJUmb-ers: yv ;:::: J (; . () fA. tv Problem 4.44 A gymnast of mass mclimbs a vertical rope attached to the ceiling. You can ignore the weight of the rope. Part A: Calculate the tension in the rope if the gymnast climbs at a constant rate. Y) if )5 she c. I; rn hs no a, + 8- C. 6 S t-o. 1') + 1"0.. I-e. +Aer~ n ef f-o r(e. 8n +- A e CbY nln cts j-r:::;. 1?'\5 Part B: Calculate the tension in the rope if the gymnast hangs motionless on the rope. 0... e, A i n I;;:' yrl 0 Part C: Calculate the tension in the rope magnitude (1. if the gymnast accelerates up the rope with an acceleration of w~ J'" nO t.A/ I-w::::-m~ .so ,:::. m(et r<j) Part D: Calculate the tension in the rope acceleration of magnitude a. if the gymnast slides down the rope with a downward 1-W':::-rn~ 1= (11 (Cj -c...) Problem 4.47 A 4.9-N hammer head is stopped from an initial downward velocity of 3.2 mls in a distance of 0.45 cm by a nail in a pine board. In addition to its weight, there is a l5-N downward force on the hammer head applied by the person using the hammer. Assume that the acceleration of the hammer head is constant while it is in contact with the nail and moving downward. Part A: Calculate the downwardforce exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward. -:> F F::::-}'Y\D.- Wh",,+- is rAe... ~a.s5 of file ho..Vl"~er? '-1.9 AI - m ~ - W~()J- 1 s the. ~cce/eY'Qf-t(jY\ ~tA. X V :l.. == /1 _ LI'-- ~ =. '15 em ::= 'I. 5x /0 -) yY) v: 3. J.J11/s v'J... ~ :J..X +Aere-I-or e.. F::.. Lt. 9 <:> Ii . X- :>. -l- J 5 1\1 J7' Part B: Suppose the nail is in hardwood and the distance the hammer head travels in coming to rest is only 0.12 em. The downwardforces on the hammer head are the same as on part (A). What then is the force exerted by the hammer head on the nail while the hammer head is in contact with the nail and moving downward? ~ Jot ~ MdS.5 jS F s f j II. . . <-t. 9 ~ ---=.om tho...t C,htU19e.5. ... rv )( is rhe... only vd.r;()..ble v"" t\.~-;lX' F ~ 4.QN 3 - ~ ;;}.y v~ -rlstJ Problem 4.50 An athlete whose mass is m.is performing weight-lifting exercises. Starting from the rest position, he lifts, with constant acceleration, a barbell that weighs w. He lifts the barbell a distance of Iin a time of t. Part A: Use Newton 's laws to find the total force that his feet exert on the ground as he lifts the barbell. F {:'l ()() r :::-F()..t h Ie r~ -t+hefr"tA-SS "'" _ ~ be>.f' be 1/ -f ..v1 I" b C{ 01vV rh~ bo..r bell IS '" b -' .-- S ~ So F-floor rn ~ +- W -I- ..0L 0- Y '/1;1. 0.( ').... ~~ t?~x ~:I, ~ J..)( F,-}1 ex 1)1 'F flo 0"":::; yYl ~ t- \IV' ... \AI -- Problem 4.52 A student tries to raise a chain consisting of three identical links. Each link has a mass of m. The threepiece chain is connected to a string and then suspended vertically, with the student holding the upper end of the string and pulling upward. Because of the student's pull, an upward force of F is applied to the chain by the string. Use Newton's laws to answer the following questions. Part A: Find the acceleration of the chain. F - 3t"'fj == fA ~ 3 m~ rT\fj tF j" 3 tfl ~ '= - m 3 3 Part B: Find the force exerted by the top link on the middle link. P+,oto..l ~ F -rap -t ",,\olJI ~(Y\ > ~ 3rYl - J..F .----. 3 - Problem 4.53 The position of a 2.75 x Ht 1'/ training helicopter under test is given by r .> 6 0.0 d- !!1- t .,I\.,.r' S3 3 ..(. JWt/s ~ 1:' J A- Find the net force on the helicopter at t =5.0 s. - O. 6 6 ~:t S t~ f{ F::: rn a. ty\~ J.. 76;</0 ~ 5 N -;'"t".> (J.=- ~ cAr JJ cl f-.. -=:.. :;;. 3.:t. 0.6;) ? t ( m A .f-O; - :l". 0 -~ 6 111 s~ X /' 0.:::" 0./::1 ~ S3 t./'J - O. J~ ? J~ . ~ . .A... 0.* .-. 1:;; / yn b I; J Q e C" Acts m 0.. ~ 0, Si A. - O. ) J S). t< Fy ~ ..> J. 7 5x/o 5 N . o. 6 ?3-~ :2- lj F - ..1.. 5 ;< /~ 5N . ().,). /5 7 z. J Problem 5.16 A block of ice of mass 8.00-kg, released from rest at the top of a frictionless ramp of a 1.50-m-long slides downhill, reaching a speed of 2.50 m/s at the bottom. What is the angle between the ramp and the horizontal? W~ e..CD f j ncJrio 11.- (!1....ccel e-r""a.t ion U51~ Y\ 9 + h - .( am; liar fJ 0. VI- :;. ;t ex..n X '1, e-. WI -e.- 11 n a IN' 5 ';Is) ::: '1. 2. ().. (t . 5 m;s:l. 100 m) mOV/I1!j dowfl 0... D.. ~ I.. .0 g /-0' 0-- J.. 'I ro...Yl'1p. st'Y\ G- -j _ 0- G.::: ()..r c s ," f\ 'J . 0 ( Cj. 8 8 ""/s ~ 1 1"'1/~;; I -rh e r~.f 0 rtL.o () ~ I J...:l5 Problem 5.56 In the figure a worker lifts a weight Wby pulling down on a rope with a force . The upper pulley is attached to the ceiling by a chain, and the lower pulley is attached to the weight by another chain. The weight is lifted at constant speed. Assume that the rope, pulleys, and chains all have negligible weights. F F Part A: In terms of lV, find the tension in the lower chain. T J..enS\6n i t1 I 0 We-f C ho.(',., ;.W ; Part B: In terms of lV, find the tension in upper chain. S:)'l(.- +heY\ rn ClS -.S 0 f 1- he. c A ~ j n /05 ~ ee liB ; b Ielen SI 0" 0fJP ~:::: vv Part C: In terms of w, find the magnitude of the force F if the weight is lifted at constant speed. F;:= ~Ofe.. - Problem 5.58 A Rope with Mass. In most problems in this book, the ropes, cords, or cables have so little mass compared to other objects in the problem that their mass can safely be neglected. But if the rope is the only object in the problem, then clearly its mass cannot be neglected. For example, suppose we have a clothesline attached to two poles. The clothesline has a AI 0 . mass , and each end makes an angle wIth the horizontal. Part A: What is the tension at the ends of the clothesline? iW/" Y LeA" }h~ '(x ()..Y\ ~ ) ~ be.X 4- {) ~ r()..1\(j ~ -r y -== UJ / ).. r ::--L~ -S -,. ~f ,. elvl::' .y:!A n(j-g-) w~ J{ (j 1he-re..Pore-; is ; The- f-ef1$Jon 0.. +- -f he, e.t1 c,J.. 0 f 1'l r), e--c ItJ fA e s II i1 e- -r ,'C{, YL--l-- ;::;-:$ j (B ) ; I end-. :: :L S,'n ~ Part B: What is the tension at the lowest point? tA.+- t~-e- low e..s+ fJOit1r / rne. , J-e.1'\sj(J n IS d.Ue- S"o/e-ly f-O IX rx k ~~ d.. -ra n e ,...e CO U 1"',e.0 a..-J1 hoI i C-t.. if i 1\ of /' n" f- e.. WCJolJ. fhe,o r~ f-,' CO {fy C> T? +e n -s }- .;: tJ Y' [):::. ...
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