PROBLEM deceleration of the cable attached at B is 2.5 m/s, while that + ] ( )( ) = 2.5 2α. a = rad/s. a 3.25 m/s. = 3.

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1 PROLEM 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 ] ( )( ) ( ) 1.5 =.5 α b A = + ( A / ) + ( A / ) () t n = 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 m/s E =

2 PROLEM 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

3 PROLEM 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 = rd/s v A = va = ( C) ω = (4)(4.714) = in./s v 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 = in./s AD A ( ) = ( C) = 0 t Rod A: AD = α AD n = C ωc = ( ) ( ) (4)(4.714) = in./s α A = A = + ( ) + ( ) A A / t A / A = [ A ] = [88.87 ] + [8α AD 30 + ] [8ω AD 60 ] Resolve into components. () : : 0 = cos = 1.81 rd/s A AD = 8 sin 30 = (8)( 1.81)sin 30 = in./s AD AD = 51.3 in./s A

4 PROLEM (Continued) (b) = + ( / ) + ( / ) D D t D t = [88.87 ] + [8α ω 30 + ] [8 [ 60 ] = [88.87 ] + [(8)( 1.81) 30 + ] 0 = [88.87 ] + [ ] = [ ] D = in./s 16.1

5 PROLEM 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 = rd/s Plne motion = Trns. with A + Rottion bout A A / = CA / = DA / = rω = (0.80 m)(71.49 rd/s) = m/s () (b) (c) = + / = m/s A A C= A+ CA / = m/s D= A+ DA / = m/s 60 = 1430 m/s = 1430 m/s C D = 1430 m/s 60

6 PROLEM 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

7 = + ( ) + ( ) = [ ] + [rα A C AC / t AC / n C PROLEM (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

8 PROLEM 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 = 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 ω cos β Accelertion nlysis. α A = 0 ω 1.5π cos 60 = = rd/s 0.15 cos β A π = 0.05 ω = (0.05)(30 ) = m/s 30 D = D α = α α D/ = [0.15α A β] + [0.15ω β ] = [6α β ] + [ β ] = + D D / Resolve into components.

9 PROLEM (Continued) : 0 = cos α cos β sin β = rd/s : = sin 30 (0.15)(597.0)sin β cos β D = 96 m/s P = D = 96 m/s P

10 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 = 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 ] = m/s 45 Rod. v D = v D 45 ω = ω v = v + v D / vd 45 = [ ] [0.19ω 45 ] Components 45 : 0 = ω ω = rd/s D = D 45 = + ( ) + ( ) D D / t D / n A [ D 45 ] = [ ] [0.19α 45 + ] [0.19ω 45 ]

11 PROLEM 15.1 (Continued) Components 45 : = (0.19)(41.337) = m/s D Rod E. Since E 0.05 sin γ =, γ = 15.58, β = 45 γ = 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 = E / t= E / t β E = + ( E / ) t E = tn γ = tnγ = m/s = 337 m/s 45 E

12 PROLEM 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 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

13 PROLEM (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 i = i+ ( ) 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: = = rd/s = (0)( 0.896) = 8.15 rd/s () (b) Angulr ccelertion of br. Angulr ccelertion of br. α = 8.15 rd/s α = rd/s

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