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# hw6 - PRDB LEM'i 3.9 The Iii-1h blockA is released from...

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Unformatted text preview: PRDB LEM'i 3.9 The Iii-1h blockA is released from rest in the pesitien shewn. Neglecting the effect elf frietien and the masses of the pulleys. determine the velocity 4 I1 {\Uﬁuutl ef the bleek aﬁer it has moved 2 ﬁ up the incline. SOLUTION Given: EleekA is released from rest and moves up incline 2 ft. Friction and other masses are neglected. 5‘ . ”3/ Find: Velecity efthe bleak after 2 e. v -' " . "3'13 ._-:~l . _ Z From Law of Cesines . ' . 'L 4 ‘3 4. r1 2 . * =1.— ﬁrt'll": - (9/. d1 = {4}2 + (21) — 2(4}{2}cusis° .93 = 4.9452 fr" 4’ = 2.1919 n U3.) = WE {distance pulley C is lewered] = 31') Ib|:2{4— 2. [319)] ﬁ = +23.ﬂ21ﬁ41b U”, = —lﬁlh{\$in15°(2 ft} = —3.2822 ﬁ-ih U=T2—?i =U3IJ—Ule i W4]v2-D=Um—Um K ={23n21— 2.2322} 9-19 92 = 99.449 -.- = 2.9124 PROBLEM 13.12 Boxes are transported by a conveyor belt with a veioeity v“, to a ﬁxed incline at A where they slide and eventually fall off at 3. Knowing that pk = H.401 determine the velocity of the conveyor belt if the boxes are to have a zero velocity at B. T4 = in”; T3 = l-l UH =[Wsin15° — “one :11} \ EF = o N — Weesli” = :1: 1'5" = Ill/eoslﬁﬂ UA--B = WIsinlS“ — ﬁ.4ﬂcos15°]{6 In} UA—s = —[ﬂ.?6531]W = 4.76531mg TA + Use-e = Te %mv§ — 0.?653img = 13 pg = [2}{U.?6531}[9.31 ml] v.3: = 15.015 v“ = 3.87" mfe 4 Down to the left. PROBLEM 13.29 A Ell-lb block is attached to spring A and connected to spring 3 by a cord and pulley. The block is held in the position shown with both springs mistretohed when the support is removed and the block is released with no initial velocity. Knowing that the constant of each spring is 12 lbfin" determine {a} the velocity of the block after it has moved down 2 in_, {h} the maximum velocity aehjeved by the block. 1 2 l 1 Ul" 2 = H1214} _ Tl] {rill _ Elsixsl gravity spring A spring 3 U. 2 = {2!} leEf‘; ﬂ] _ £[144lbfﬁ}[% ﬂ]: 1 1 +5144 II::.I1=t}[E ﬂ] [ 2o1b J": 32.2 rvs2 3.3333—2—05 = i[ 2” l»: v: 1.638 ms! 2 22.2,! {h} Let x = Iﬁstanee moved down by the 20 lb bleek “Ii-2 2 Wi-tl _ ihi-tl: - %kﬂ[i = 20 -144{ }- l—:i{lx} = 2n — [144 + 35p: .1 = {Ll lllt] ﬂ (= 1.33333in.} For x = [1.11 1111. U = 2.2222 — osssss — r222222 = £3202le vm = 1.39143 pm = 1.392 fb’si PROBLEM 13.44 A section of track for a roller coaster consists of two circular arcs AB and CD joined by a straight portion BC. The radius of AB is 2'! m and the radius of CD is T2 to. The car and its occupants, of total mass 25D kg, reach point A with practicall}.r no velocity and then drop freely along the track. Determine the maximum and minimum values of the normal force exerted by the track on the car as the car travels from A to D. Ignore air resistance and rolling resistance. 50 LUTIDN Nommtjorce at 3 See solution to Problem 13.43, N“ = ELI] N Neurons Low From B to C [car moves in a straight line} Bi.” ‘75 Ni: \Imo. to“ C c +/ N5; — Weos4ii“ = [I it; = [25:] kg 1: 9.31m53]cos4m = mam Ar Candﬂ [car in the curve at C'} “chasm: .L mﬂn=mit Ez‘tlm EL w: iLLHQE-h maﬂ=mﬂs ii PROBLEM 13.44 CONTINUED v2 ND = [2511 x 9.31} 1+ :1? Since v1] :- V1? and cost? c I, N 1:: :> NC Work and energyﬁ'om A m D l a 1rd = 11,13; = 1':- TD =Em1§=125v31 UH} = 1121211 + 13} = [251:1 kg}(9.31w53}{45 m} = [1133615 1",, + UH, = TD 11 “11:13:12.5 = 1251229 pg, = 332.911 332.90 129.31} = 5513.1N N = N3 = Tammm = ND = 55211114 PROBLEM 13.59 r— tuu mm —- anemia—1 A E—kg seller can slide without friction along a horizontal red and is in equilibrium at A when it is pushed 25 mm tn the right and released item 1 B k = sun Wm .'..—.-E_ rest. The springs are undefﬂnnecl when the collar is atA and the constant of each spring is Sﬂﬂ Ich’m. Determine the maximum trelucitgrr cf the cellar. SOLUTION B t: [nisf + {ans}2 = assist: an Stretch = ﬂ.2?ﬂ4 - 11.25 31 = 0.02ﬂ416 at new] '7! t: {hits}2 +{n15} =c.23e49 Stretch = E12304? — 0.25 .51 = 41.01951 M i . :5 = game; =% V. = %[snc,cec}(sﬁ + 322] I tq = 25neeu[u.cuntats} = 199.37? N-m I = u e 1' e+e=e+£ v; =199.3s v2 =14.12mls~1 SOLU‘I'IDM PROBLEM 13.66 An S—lh collar A can slide without friction along a vertical rod and is released ﬁorn rest in the position shown with the Springs undeformed. Knowing that the constant of each spring is 211 lbi'ft... determine the velocity of the collar after it has moved (a) 4111.,{11} "L6 in. {:1} Calculate spring lengths after deﬂection. Original springr length: 2.5 ft: Collar moved 41n. = [1.333 ft 1 s ‘1, E=Vt=ll e=——v- 2 32.2} 4 Vt = —Elb -fr =—2.oﬁ’ift-lh -g t e 1 V3, = %[zc}[[2.174 — 2.5}2 + [2.5 — 2.24312] =1.41113ﬁ-1h I, +V. =13+V1: c=l[i] 2—2.ero+1.4113 2 32.2 v = 3.1394 ﬁts v = 3.13ft1’s '1 {h} Ca1cu1ate spring lengths after deﬂection cellar moved "Loin. = 0.153311 V21: = {—3 mg? n] = _—5.ﬂeeT 11-11: yin, =é12c][{3nscss — 2.5}2 + {2.5 — 2.112923]? = 5.031411“: '1", + P] = 13 + V3: o = i352}: — sees? + 5.11314 v = £15331]? ﬁts v = [1.533 ﬁts '1 ...
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hw6 - PRDB LEM'i 3.9 The Iii-1h blockA is released from...

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