PROLEM 15.105 A 5-m steel bem is lowered by mens of two cbles unwinding t the sme speed from overhed crnes. As the bem pproches the ground, the crne opertors pply brkes to slow the unwinding motion. At the instnt considered the decelertion of the cble ttched t is.5 m/s, while tht of the cble ttched t D is 1.5 m/s. Determine () the ngulr ccelertion of the bem, (b) the ccelertion of points A nd E. α = α, ω 0, rd/ = m =.5 m/s, 1.5 m/s D = ( / ) ( / ) = + + D D t D n 1.5 = [.5 ] [α + 0 + ] ( )( ) ( ) 1.5 =.5 α b A = + ( A / ) + ( A / ) () t n = 0.500 rd/s = [.5 ] + ( 1.5)( 0.5) ] + [0 ] = 3.5 m/s ( / ) ( / ) = + + E E t E n 3.5 m/s A = = [.5 ] + ( + 1.5)( 0.5) + [0 ] = 0.75 m/s 0.750 m/s E =
PROLEM 15.107 A 900-mm rod rests on horizontl tble. A force P pplied s shown produces the following ccelertions: A = 3.6 m/s to the right, = 6 rd/s counterclockwise s viewed from bove. Determine the ccelertion () of Point G, (b) of Point. () = + / = [ ] + [( AG) α ] G A GA A G = [3.6 m/s G = [3.6 m/s ] [.7 m/s ] + [(0.45 m)(6 rd/s ) ] + ] (b) = + / = [ ] + [( A) α ] A A A = [3.6 m/s = [3.6 m/s ] [5.4 m/s ] + [(0.9 m)(6 rd/s ) ] + ] = 0.9 m/s G = 1.8 m/s
PROLEM 15.109 Knowing tht t the instnt shown crnk C hs constnt ngulr velocity of 45 rpm clockwise, determine the ccelertion () of Point A, (b) of Point D. Geometry. Velocity nlysis. Let β be ngle AC. 4 in. sin β = β = 30 8 in. ω C = 45 rpm v = 4.714 rd/s v A = va = ( C) ω = (4)(4.714) = 18.8496 in./s v 18.8496 in./s C v A nd v re prllel; hence, the instntneous center of rottion of rod AD lies t infinity. Accelertion nlysis. α C = 0 Crnk C: ω = 0 v = v = 18.8496 in./s AD A ( ) = ( C) = 0 t Rod A: AD = α AD n = C ωc = ( ) ( ) (4)(4.714) = 88.87 in./s α A = A = + ( ) + ( ) A A / t A / A = [ A ] = [88.87 ] + [8α AD 30 + ] [8ω AD 60 ] Resolve into components. () : : 0 = 88.87 + 8 cos 30 + 0 = 1.81 rd/s A AD = 8 sin 30 = (8)( 1.81)sin 30 = 51.84 in./s AD AD = 51.3 in./s A
PROLEM 15.109 (Continued) (b) = + ( / ) + ( / ) D D t D t = [88.87 ] + [8α ω 30 + ] [8 [ 60 ] = [88.87 ] + [(8)( 1.81) 30 + ] 0 = [88.87 ] + [10.568 30 ] = [177.653 + 51.84 ] D = 184.9 in./s 16.1
PROLEM 15.111 An utomobile trvels to the left t constnt speed of 7 km/h. Knowing tht the dimeter of the wheel is 560 mm, determine the ccelertion () of Point, (b) of Point C, (c) of Point D. v h 1000 m A = 7 km/h 0 m/s 3600 s km = Rolling with no sliding, instntneous center is t C. Accelertion. v = ( AC) ω; 0 m/s = (0.8 m) ω A ω = 71.49 rd/s Plne motion = Trns. with A + Rottion bout A A / = CA / = DA / = rω = (0.80 m)(71.49 rd/s) = 148.6 m/s () (b) (c) = + / = 0 + 148.6 m/s A A C= A+ CA / = 0 + 148.6 m/s D= A+ DA / = 0 + 148.6 m/s 60 = 1430 m/s = 1430 m/s C D = 1430 m/s 60
PROLEM 15.115 A hevy crte is being moved short distnce using three identicl cylinders s rollers. Knowing tht t the instnt shown the crte hs velocity of 00 mm/s nd n ccelertion of 400 mm/s, both directed to the right, determine () the ngulr ccelertion of the center cylinder, (b) the ccelertion of point A on the center cylinder. Geometry. Let point C be the center of the center cylinder, its contct point with the crte, nd D its contct point with the ground. Let r be the rdius of the cylinder. r = 100 mm. Velocity nlysis. Accelertion nlysis. Since the contcts t nd D re rolling contcts without slipping, v = 00 mm/s nd v D = 0. Point D is the instntneous center of rottion. v 00 mm/s ω = = = 1 rd/s r 00 mm Point C moves on horizontl line. = [ ] = + ( ) + ( ) = [ ] + [rα D C DC / t DC / n C C C ] + [rω ] Component : 0 = C r (1) = + ( ) + ( ) = [ ] + [rα C C / t C / n C ] + [rω ] Component : Solving (1) nd () simultneously, C = 00 mm/s rα = 00 mm/s () 00 mm/s α = 100 mm 400 mm/s = C + r () =.00 rd/s
= + ( ) + ( ) = [ ] + [rα A C AC / t AC / n C PROLEM 15.115 (Continued) ] + [rω ] = [00 mm/s ] + [00 mm/s ] + [(100 mm) (1 rd/s) ] = [100 mm/s ] + [00 mm/s ] (b) A = 3.6 mm/s A = 0.4 m/s 63.4
PROLEM 15.11 Knowing tht crnk A rottes bout Point A with constnt ngulr velocity of 900 rpm clockwise, determine the ccelertion of the piston P when 10. θ = Lw of sines. sin β sin10 =, β = 16.779 0.05 0.15 Velocity nlysis. ω = 900 rpm = 30 p rd/s A v = 0.05ω = 1.5π m/s 60 A v D = v D ω = ω vd/ = 0.15ω v = v + v D D / β [ v D ] = [1.5π 60 + ] [0.15ω β ] Components : 0 = 1.5π cos 60 0.15ω cos β Accelertion nlysis. α A = 0 ω 1.5π cos 60 = = 16.4065 rd/s 0.15 cos β A π = 0.05 ω = (0.05)(30 ) = 444.13 m/s 30 D = D α = α α D/ = [0.15α A β] + [0.15ω β ] = [6α β ] + [40.376 β ] = + D D / Resolve into components.
PROLEM 15.11 (Continued) : 0 = 444.13 cos 30 + 0.15α cos β + 40.376 sin β = 597.0 rd/s : = 444.13 sin 30 (0.15)(597.0)sin β + 40.376 cos β D = 96 m/s P = D = 96 m/s P
PROLEM 15.1 In the two-cylinder ir compressor shown, the connecting rods nd E re ech 190 mm long nd crnk A rottes bout the fixed Point A with constnt ngulr velocity of 1500 rpm clockwise. Determine the ccelertion of ech piston when θ = 0. Crnk A. v = 0, = 0, ω = 1500 rpm = 157.08 rd/s, α = 0 A A A v = va + v / A = 0 + [0.05ωA 45 ] = [7.854 m/s 45 ] = + ( ) + ( ) A A / t A / n = 0 + [0.05α A 45 ] + [0.05ω A 45 ] = [(0.05)(157.08) 45 ] = 133.7 m/s 45 Rod. v D = v D 45 ω = ω v = v + v D / vd 45 = [7.854 45 + ] [0.19ω 45 ] Components 45 : 0 = 7.854 0.19ω ω = 41.337 rd/s D = D 45 = + ( ) + ( ) D D / t D / n A [ D 45 ] = [133.7 45 + ] [0.19α 45 + ] [0.19ω 45 ]
PROLEM 15.1 (Continued) Components 45 : = 133.7 + (0.19)(41.337) = 1558.4 m/s D Rod E. Since E 0.05 sin γ =, γ = 15.58, β = 45 γ = 9.74 0.19 v E = ve 45 v is prllel to v, ω = 0. E = 1558 m/s 45 D E = E 45 / E n ωe ( ) = 0.19 = 0 ( ) ( ) Drw vector ddition digrm. γ = 45 β = 15.58 E / t= E / t β E = + ( E / ) t E = tn γ = 133.7 tnγ = 336.5 m/s = 337 m/s 45 E
PROLEM 15.133 Knowing tht t the instnt shown br A hs n ngulr velocity of 4 rd/s nd n ngulr ccelertion of rd/s, both clockwise, determine the ngulr ccelertion () of br, (b) of br by using the vector pproch s is done in Smple Problem 15.16. Reltive position vectors. ra / = (0 in.) i (40 in.) j r = (40 in.) i Velocity nlysis. r A (Rottion bout A): D / r = (0 in.) i (5 in.) j / ω A = 4 rd/s = (4 rd/s) k r = (0 in.) i (40 in.) j v = ω r = ( 4 k) ( 0i 40 j ) A / A A / v = (160 in./s) i+ (80 in./s) j r (Plne motion = Trnsltion with + Rottion bout ): r (Rottion bout E): ω = ω k r = (40 in.) i ω D/ v = v + ω r = v + ( ω k) (40 i) D D/ v = (160 in/s) i+ (40ω + 80 in./s) j D r = (0 in.) i (5in.) j / = ω k D D/ E v = ω r = ( ω k) (0i 5 j) v = 0ω j+ 5ω i D Equting components of the two expression for v D, i: 160 = 5ω ω = 6.4 rd/s j: 40ω + 80 = 0ω 40ω + 80 = 0( 6.4) ω = 5. rd/s E
PROLEM 15.133 (Continued) Summry of ngulr velocities: Accelertion nlysis. r A (Rottion bout A) ω 4 rd/s ω 6.4 rd/s ω 5. rd/s A = = = A = ( rd/s ), =, = k k k = A / A ωa / A r r = ( k) ( 0i 40 j) (4) ( 0i 40 j) = (80 in./s ) i + (40 in./s ) j+ (30 in./s ) i+ (640 in./s ) j = (40 in./s ) i+ (680 in./s ) j r (Trnsltion with + Rottion bout ): D = + α D/ ω D/ α α r r = 40i+ 680 j+ α k (40 i) (5.) (40) i = 40i+ 680j+ 40α j 1081.6i = 841.60 i+ (680 + 40 ) j (1) α r (Rottion bout E): D = α D/ E ω D/ E α r r Equte like components of D expressed by Eqs. (1) nd (). = α k (0i 5 j) (6.4) (0i 5 j) = 0α j+ 5α i 819.0i+ 104j = (5α 819.0) i+ (0α + 104) j () i: j: 841.60 = 5 819.0 = 0.896 rd/s 680 + 40 = (0)( 0.896) + 104 = 8.15 rd/s () (b) Angulr ccelertion of br. Angulr ccelertion of br. α = 8.15 rd/s α = 0.896 rd/s