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Home_work_set_03 - Problem 13—13 Determine the normal...

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Unformatted text preview: Problem 13—13 Determine the normal force the crate A of mass M exerts on the smooth cart if the cart is given an acceleration a down the plane. Also, what is the acceleration of the crate? Given: M = 10 kg a = 2 £1- 2 7 s 6 = 30 deg Solution: N— Mg = —M(a) sin(0) “crate = (a) sin( 6) Problem 13-15 The driver attempts to tow the crate using a rope that has a tensile strength me. If the crate is originally at rest and has weight. W, determine the greatest acceleration it can have if the coefficient of static friction between the crate and the road is gs and the coefficient of kinetic friction is ,uk. Given: T max = 200 lb W = 500 lb as = 0.4 ,uk = 0.3 ft g = 32.2 —2 S . 0 = 30 deg _ Solution: Equilibrium : In order to slide the crate, the towing force must overcome. static friction. ‘ 3%! Initial guesses F N = 100 lb T = 50 lb Given Tcos(€) — ,uSFN = 0 FN+ Tsin(6) — W T If T = 187.613 lb > T max = 200 lb then the truck will not be able to pull the create without breaking the rope. I If T = 187.613 1b < T max = 2001b then the truck will be able to pull the create without breaking the rope and we will now calculate the acceleration for this case. 0' '[FN] =IFin<i1>(FN, T) ft , Initial guesses F N = 1001b a = l —2 Require T = T max 3 ' v ' . , W FN Given Tcos(0) — #kFN = —a FN+ Ts»in(0) — W: 0 [ j = Find(FN,a) g a _ . * 7. ' ‘ flaunt a = 3.426— 1» 5f“: '17‘2 S *Problem 13-24 At a given instant block A of weight WA is moving downward with speed VAo- Determine its speed at a later time t. Block B has a weight W3 and the coefficient of kinetic friction between it and the horizontal plane is ,uk Neglect the mass of the pulleys and cord. ‘ Given: V . , ' ' - 21"“ WA = 10 lb _ VA0 = 6 — S a t = 2 S “:1 W3 = 41b7 ‘ E _ W1; , #k = 02 e ' G}? > IT ft , ' ' ' 112% g = 32.2 — > . , , 2, . i e S V . > WE . Solution: L = SB + 2sA . , ft V ft Guesses aA=l-E aB=1—2 T=llb s s ‘WB Given T —— pk WB = a3 g ”WA 2T— WA = [ jaA g ‘ 0 = a3 + 2aA , T , . , - “A 10.403 '11 em = Find(T,aA,aB) T= 3.3851b ' = ~— ' - a3 —20.806 52 dB ‘ . , 1:.7: 1’; VA =’ VA0 + “A1 , VA 2263+ fl Problem 13-33 The collar C of mass mc is free to slide along the smooth shaft AB. Determine the acceleration of collar C if collar A is subjected to an upward acceleration a. The collar moves in the plane. Given: B mc = 2 kg m a = 4 E mfg rm“ , ' g m g = 9.81 7 = é/fldflafi s 6 = 45 deg Solution: The collar accelerates along the rod and the rod accelerates upward. émcgcosw) = mdaCA — (a)cos(¢9):| “04 v: (g1+ a)cos(0) -aCA 891(9) . 6—905 m 9 m ac = ac— = -.— lacl = 7.491 3 “aCA cos(0) + a —"2. 905 'SZ *"3’ 7 Problem 13-35 The block A of mass m A rests on the plate B of mass m B in the position shown. Neglecting the mass of the rope and pulley, and using the coefficients of kinetic friction indicated, determine the time needed for block A to slide a distance 3’ on the plate when the system is released from rest. Given: mA = 10 kg m3 = 50 kg. V » s’ = 0.5 m ' #AB = 0-2 .UBC = 011 9 = 30 deg In S Solution: SA + SB = L Guesses 53.4% In aA=1—2 aB=l S m 2 S T=1N NA=1N 'N3 = 1 N Given ' 7 . ' r "0147+ (13 = 0 NA — mA_gcos(€) = 0 NB - NA — m3gcos(6) = 0 T - #ABNA - mAgsin(0) = -mA 0A T + #ABNA 4E #BCNB -_ Tagsin(6) = -mBaB aA a3 ‘ T T = Find(aA,aB, T,NA,NB) NA' = NA. V ’ NB NB ‘ “BA = “B — aA “BA = 3.708 E; t = — t"=f0,._5i9 s *Problem 13—44 Each of the three plates has mass M If the coefficients of static and kinetic friction at each surface of contact are ,us and ,uk respectively, determine the acceleration of each plate when the three horizontal forces are applied. Given: M = 10 kg ,us = 0.3 , ,uk = 0.2 FB = 15 N FC = 100 N ' ' FD = 18 N m g = 9.81 —2- 5 Solution: Case 1: Assume that no slipping occurs anywhere. FABmax = #s(3Mg) . FBCmax = [15(2Mg) FCDmax = #s(Mg) Guesses FAB=1N Fgc=1N FCD=1N Given —FD+.FCD=0 Fc—FCD—Fgc=0 —FBV—FAB+FB(}=0 FAB , FAB ' 677 FABmax 88.29 FBC = Find(FAB.FBc,Fcp) FBC =. 82 N ' , FBCmax = 58.86 N F CD 7 ' F CD 18 F CDmax 29-43 If FAB = 67N < FABmax = 88.29N and FBC = 82N > FBCmax = 58.86N and F CD = 18N < F CDmax = 29.43 N then nothing moves and there is no acceleration. Case 2: If FAB = 67N -< FABmax = 88.29N and FBC = 82N > FBCmax = 58.86N and F CD = 18N < F CmeC = 29.43 N then slipping occurs between B and C. We will assume that no slipping occurs at the other 2 surfaces. iSet Fgc=,uk(2Mg) (:3 =0 ac=dD=a Guesses FAB=1N FCD=1N (1:195 ' 3 Given , —FD+FCD=Ma Fc-FCD—FBC=Ma —FB—FAB+FBC2=0 'FAB . . . . F F' d(F F ) * FAB 2424 N‘ 7 2138 m l = In 5 5a . = a="' _ CD AB CD FCD 39.38 ; ' - 52. a ac=a aD=a If FAB = 24.24N <' FABmax ; 88.29N and FCD = 39.38N ' ‘> FCDmax = 29.43N then . , m ' ' m we have the correct answer and the accelerations are (13 = 0 , ac = 2.138 -—2 , aD = 2.138 —2 s s Case 3: If FAB = 24.24N < FABma'x = 88.29N and 'FCD = 39.38N > FCDmax = 29.43N then slipping occurs between C and D as well as between B and C. We will assume that no slipping occurs at the other surface. Set FBC = #k(2Mg) FCD = #k(Mg) Guesses FAB=1N _ aC=IE; 0113:122 ‘ . s . . s Given —FD+FCD=MaD Fc—FCD—FBc=MaC —FB—FAB+FBc=0 CF' d(F ) F 24 24 N ' lac ‘ ~ $44114 7 m“ 0C = 1n . AB,a 9a = - t. =.. I .‘_' C D AB .110 * . 0.162 , ':S2: “D, , 5', If FAB = 24.24 N < F A Bmax = 88.29 N then we have the correct answer and the accelerations I aC=41143§ aD‘: 0.162% ' There are other permutations of this prOblems depending on the numbers that one chooses. W Problem 13-45 Crate B has a mass m and is released from rest when it is on top of cart A, which has a mass 3m. Determine the tension in cord CD needed to hold the cart from moving while B is sliding down A. Neglect fiiction. Solution: BlockB: NB — mgcos(¢9) = 0‘ N3 = mgcos(0) I Cart: —T+ NBsin(0) = 0 T: mgsm(é)cos(0) ‘5 Problem 13-73 The pendulum bob B of mass M is released from rest when 0 = 0°. Determine the initial tension in the cordiand al'so'at the instant the bob reaches point D, 0 = (9,. Neglect the size of the bob; Given: V M = 5kg 0,: 45 deg L=2m g=9.8122 A » 7 s Solution: ' Initially, iv = Oso an =- o T== 0 At D We have Mgcos(o91) = Mat at = gcos(01)' at = 6.937 22 , s 2 . . Mv T . TD—Mgsm(t91)= L "5“ r . ‘ :54 Now find the veloc1ty v > , 3 Min, Guess v = 1 E 7 I S a, ' Given [V v dv : J, gcos(t9)L d0 ' 0 A 0 v = Find(v) V v = 5.268 3 S 2 V TD => Mgsin(01) + :47] TD % 104.1 N' *Problem 13-76 The spool S of mass M fits loosely on the inclined rod for which the coefficient of static friction is ,us. If the spool is located a distance d from A, determine the maximumconstant speed the spool can have so that it does not slip up the rod. Given: "7 . Vyksfs M = 2 kg e = 3 . . _ e f , ,us=0.2 f=4 , fink '5 - if d=0.25m g=9.8132 ‘3 - V s » Solution: :6 / . E IE—FN n f ' ’ - p = —f— ‘ i I _ N { 32+f2] .. ' S Guesses NS = 1 N v = 1 0 H N . [ s] = Find(Ns,v) NS = 21.326 N v = 0.969 2 ,, V S *Problem 13-80 The block has weight W and it is free to move along the smooth slot in the rotating disk. The spring has stiffness k and an unstretched length 6. Determine the force of the spring on the block and the tangential component of force which the slot exerts on the side of the block, when the block is at rest with respect to the disk and is traveling with constant speed v. Given: W = 2 1b k = 2.5 l—b ft 6 = 1.25 fi it v = 12 — 5 Solution: 2 W v 2:Fn=man; Fs = k(p _ 6) = -[—] , g P Choosing the positive root, 1 p = F [kg6'+ (y’kzgzéz +4kvi2)] p = 2.617fi g F3 = k(p — 5) F52: 3.41911: 2F, = ma,; 2F, = ma,; F, = 0 Problem 13—87 The spool of mass M slides along the rotating rod. At the instant shown, the angular rate of rotation of the rod is 6', which is increasing at 6". At this same instant, the spool is moving outwardralong the rod at r’ which is increasing at r" at r. Determine the radial fi'ictional force and the normal force of the rod on the spool at this instant. Given: M = 4 kg r = 0.5 m d g = 6 B— r' i 3 E S S ' 0, = 2 1d- r" = 1 2 2 2 s s m = 9.81 — g 2 s Solution: - 2 a,- = r” — r67 a9 = r67'+ 2r'0 Fr = Marv F0 = Mae Fz = Mg' { Fr; Fr F, = 458.0 N 75‘ F92 +15? = 153.1 N m *Problem 13—92 The partiCle has maSs M and is confined to move along the smooth horizontal slot due to the rotation of the arm 0A. Determine the force of the rod on the particle and the normal force of the slot on the particle when 6: 0,. The rod is rotating with angular velocity 0' and angular acceleration 6". Assume the particle contacts only one side of the slot at any instant. M Given: M = 0.5 kg 61 = 30 deg 6’ = 2 E h = 0.5 m s m €,=3fl g=9.81—2 2 s s Solution: 0 = 01 h = rcos(€) r = h r : 0.577 m cos(t9) 0 = r’cos((9) — rsin(t9)t9' , r' = (rsm(6)]g r’ = 0.667 E cos(6) s 2 0 = r"COS(t9) — 21"Sin(0)‘9 — ”05(9) ‘9 _ rsin(t9)6" r" = 2r’0’tan(t9) + r92 + rtan(0)0' r” = 4.849 32 ' s 2 r” — r02 (FN— Mg)cos(6) = M(r” — r9 ) ’ FN = Mg + FN : 6.371 N cos(6?) —F + (FN— Mg)sin(0) = —-M(r€’ + 2r'6’) F = (FN— Mg)sin(0) + M(r6" + 2r’0') F = 2.932 N» Problem 13-106 Using air pressure, the ball of mass M is forced to move through the tube lying in the horizontal plane and having the shape of a logarithmic spiral. If the tangential force exerted on the ball due to the air is F, determine the rate of increase in the ball’s speed at the instant 6': 6', .What direction does it act in? Given: M=0.5kg a=0.2m b=0.1 61:35 F=6N 2 Solution: . r aebg 1 “MW = — = —b6_ = Z d_r abe ' d6 1 y/ = atan(zj y/ = 84.289 deg F . F=Mv’ v'=— v’=12-n—1 M S2 ...
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