Home_work_set_05 - *Problem 15-4 The baseball has a...

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Unformatted text preview: *Problem 15-4 The baseball has a horizontal speed v, when it is struck by the bat B. If it then travels away at an angle Bfrom the horizontal and reaches a maximum height h, measured from the height of the bat, determine the magnitude of the net impulse of the bat on the ba11.The ball has a mass M. Neglect the weight of the ball during the time the bat strikes the ball. Given: M = 0.4 kg / v1 = 35 3 {3; S .4; h = 50 m a. , 0 = 60 deg ' F); g = 9.81 E s2 wan 303 Solution: Guesses V2 = 201“- 1me = lN-s Impy = 10 NS Given 1 3M(v2 sin(0))2 = Mgh —Mv1 + Impx = Mv2 cos(6) 0 + Impy = Mvg sin(t9) v2 , 1 F' d( 1 1 ) 362 m 1me 21'2 N mP=1nv,m,m v:.— = -s x 2 p" py 2 s Impy 12.5 Impy Impx = 24.7N-s Impy *Problem 15—16 If it takes time t, for the tugboat of mass m, to increase its speed uniformly to v, starting from rest, determine the force of the rope on the tugboat. The propeller provides the propulsion force F which gives the tugboat forward motion, whereas the barge moves freely. Also, determine the force F acting on the tugboat. The barge has mass of mb. Units Used: Mg 2 1000 kg kN = 103 N Given: t1 = 35 s m, = 50 Mg km v1 = 25 — hr mb = 75 Mg Solution: The barge alone meI ' 0+Tt1=mbv1 T: t T=14.88kN 1 The barge and the tug mt + m1, v1 0+Ft1 = (m,+mb)v1 F = (——) F= 24.80kN *Problem 15-28 Block A has weight WA and block B has weight W3. If B is moving downward with a velocity v3, at t = 0, determine the velocity of A when t = t ,. The coefficient of kinetic friction between the horizontal plane and block A is ,uk Given: A ”A 51; B . fi Solutlon: sA + 253 = L vA = —2-vB Guess vA2 = 1 — T = 1 lb S . Given WA WA BlockA — 2vB1+Tt1—,ukWAt1= — vA2 g g -WB -WB VAZ BlockB ( )VBI + 2Tt1 — WBt1= —E— g g 2 VA2 fl [ J = Find(vA2,T) T = 1.501b vA2 = 6.00— T S Problem 15-41 A bullet of weight W 1 traveling at v, strikes the wooden block of weight W2 and exits the other side at v2 as shown. Determine the speed of the block just after the bullet exits the block. Also, determine the average normal force on the block if the bullet passes through it in time At, and the time the block slides before it stops. The coefficient of kinetic friction between the block and the surface is ,uk Units Used: ms = 10—3 5 Given: W1=O.03lb a=3ft W2 =101b b = 4 ft ,uk=0.5 c= 5ft At = 1 ms d =12 ft ft ft v1=1300— v2=50— s 5 Solution: W1 d W2 + W1 b —V1 — = —VB —V2 -— g c2 + d2 g g a2 + b2 W1 v1d v2b V ft VB=W —— v3=3.48— 2 c + d2 a2 + b2 5 -W1 W1 v c + (N— W2)At = —v2 a g 02 + d2 g a2 + b2 WI vza VIC 03 91b = — -— + —— + W2 N: 5 .7 gAt \la2+b2 c2+d2 W2 VB —vB—/1kW2t=0 =— t=0.225 g gflk Problem 15-51 The free-rolling ramp has a weight W,. If the crate, whose weight is W6, is released from rest at A, determine the distance the ramp moves when the crate slides a distance d down the ramp and reaches the bottom B. Given: Wr=1201b a=3 WC = 80 lb b = 4 d = 15 fi ft = 32.2 — g 2 5 Solution: 0 = atan(£) b Momentum W W 0 = (—rjvr + (4](vr — vcrcos(c9)) g g Integrate W s, = c cos(6')d s, = 4.80fi Wc + W, m *Problem 15—52 The boy B jumps off the canoe at A with a velocity vBA relative to the canoe as shown. If he lands in the second canoe C, determine the final speed of both canoes after the motion. Each canoe has a mass MC. The boy’s mass is MB, and the girl D has a mass MD Both canoes are originally at rest. Given: M3 = 30 kg MD = 25 kg m VBA = 5 — s 0 = 30 deg Solution: Guesses VA = l E vc = l E s s Given 0 = Me vA + MB(vA + VBA cos(6)) ‘MB(vA + VBA cos(0)) = (MC + MB + MD)vC (:1 3342,2261? Problem 15-59 The ball A of weight WA is thrown so that when it strikes the block B of weight W3 it is traveling horizontally at speed v. If the coefficient of restitution between A and B is e, and the coefficient of kinetic friction between the plane and the block is ,uk, determine the distance block B slides before stopping. Given: WA = 1 lb ,uk = 0.4 fl WB=101b v=20— I S ft g=32.2—2 e=0.6 S Solution: ft ft Guesses vA2 = 1— sz = 1— d = 1 R s 8 WA WA WB Given — v= — V,” + — vB2 ev= vB2 — vA2 g g g 1 WB 2 {—1 V32 - #k WBd= 0 g VA2 F' d( d) W _9'09 fl d 033ft v = m V ,v , = —- = . 32 A2 32 vB2 2.91 s d —————-——————-——————_——- *Problem 15-68 The ball of mass mb is thrown at the suspended block of mass mg with a velocity of vb. If the time of impact between the ball and the block is At, determine the average normal force exerted on the block during this time. Given: kN = 103 N m m mb=2kg vb=4— g=9.81— s 2 5 m3 = 20 kg e = 0.8 At = 0.005 s Solution: In Guesses vA=1— vB=1— F=lN s 5 Given momentum mb vb = m1, vA + m3 vB restitution evb = V]; — vA momentum B 0 + FA: = mg vB VA VA —2.55 m vB = Find(vA,vB,F) = — F = 2.62kN F Problem 15-78 The box of weight Wb slides on the surface for which the coefficient of friction is ,uk The box has velocity v when it is a distance d from the plate. If it strikes the plate, which has weight W p and is held in position by an unstretched spring of stiffness k, determine the maximum compression imparted to the spring. The coefficient of restitution between the box and the plate is e. Assume that the plate slides smoothly. Given: Wb=20lb Wp=101b 1b ,uk = 0.3 k = 400 — it c! R i v = 15 — e = 0.8 5 ft d = 2 fl g = 32.2 —2 5 Solution: ft ft ft Guesses Vb] = 1— Vb2 = 1— vpz = — 5 = 1 ft 5 s s 1 Wb 1 Wb Wb Wb W Given —(—J v2 —yk Wbd= —(—] vb]2 (— vb] = —- vb2+ — vpz 2 g 2 g g g g l Wp 2 l ev = v —v — — v = -k62 b1 p2 b2 2( g] p2 2 Vb] Vb] 13.65 Vb2 . fl . = F1nd(vb1,vb2,vp2,6) vb; = 5.46 — a: 0.456ft Vp2 s Vp2 16.38 6 Problem 15-79 The billiard ball of mass M is moving with a speed v when it strikes the side of the pool table at A. If the coefficient of restitution between the ball and the side of the table is e, determine the speed of the ball just after striking the table twice, i.e., at A, then at B. Neglect the size of the ball. Given: M = 200 gm v = 2.5 2 s 0 = 45 deg e = 0.6 Solution: Guesses v2=l% 62=1deg V3=lE 03=ldeg s Given evsin(0) = v2 sin(02) vcos(6) = V; cos(02) evz cos(02) = V3 sin(03) v2 sin(t92) = v3 cos(03) V2 V V3 _ v2 2.06 m 6'2 31.0 = Fmd(v2 , v3 , 02, 63) = — = deg 02 v3 1.50 S 03 45.0 03 V3 =‘1.500 5’3 S., W *Problem 15-88 The “stone” A used in the sport of curling slides over the ice track and strikes another “stone” B as shown; If each “stone” is smooth and has weight W, and the coefficient of restitution between the “stones” is e, determine their speeds just afier collision. Initially A has .- velocity v,“ and B is at rest. Neglect friction. . fl Given: W = 47 lb vA 1 = 8 —- s e=0.8 49=30deg, Solution: Guesses VAZt = 1 — VA2n = 1 _ s S fl VB2t=1- V32n=1_ S S Given - V VA] sm(0) = vA2, 0 = V32! vA1005(0) = VAZn + Van evAI cos(l9) = Van — VA2n VAZI VAZn . = F1nd(vA2;, VA2n , VBZta VBZn) VBZ! VBZn 2 2 ft VA2 = VA2t + VA2n VA? = 4'06: 2 2 17‘ V32 = Vth + v32" V32 = 6'24— VA2t VAZn VBZI VBZn 4.00 0.69 0.00 6.24 Problem 15-98 Determine the angular momentum HP of each of the two particles about point P. Use a scalar solution. Given: mA=2kg c=1.5m m3=l.5kg d=2m m e=4m vA=15— s f=1m 10m v= -— B s 0=30deg a=5m [=3 b=4m ":4 Solution: 11 I kg-m2 H =m—v b—c —m ———v a+ H =—66.0 PA A212A()AJ2—12A(d) PA 8 n+ n+ kg-m2 Hp]; = —vaBcos(t9)(b + e) + vaBsin(t9)(a — f) HPB = —73.9 s *Problem 15-100 The two blocks A and B each have a mass M0. The blocks are fixed to the horizontal rods, and their initial velocity is v’ in the direction shown. If a couple moment of M is applied about shafi CD of the frame, determine the speed of the blocks at time t. The mass of the frame is negligible, and it is free to rotate about CD. Neglect the size of the blocks. Given: M0 = 0.4 kg a = 0.3 m m v’ = 2 — s M = 0.6 N m t = 3 5 Solution: ZaMov' + Mt = 2aM0v v=v'+ Mt 2aM0 m v 2 9.50 — Problem 15-106 The block of weight W is originally at rest on the smooth surface. It is acted upon by a radial force FR and a horizontal force F“, always directed at 0 from the tangent to the path as shown. Determine the time required to break the cord, which requires a tension T. What is the speed of the block when this occurs? Neglect the size of the block for the calculation. Given: W=101b t9=30deg FR=21b T=301b FH=7lb r=4fi ft = _ ft V1 0 S g = 32.2 — 2 5 Solution: fi Guesses t=ls v2=1— s Given W W (—jv1r+ FHcos(6)rt = (—jvgr g g FR + FHsin(9) - T II | vole h 45 N w t it ( J : Find(t,v2) v2 = 17.76— t = 0.91 5 v2 5 ...
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