Download Fundemental Problems-Advanced Engineering Dynamics-Assignment Solution and more Exercises Engineering Dynamics in PDF only on Docsity!
F u n d a m e nta l P ro b l e m s
P a rti a l S o l uti o n s A n d Answers
Chapter 1 2 Fl2-1. v = Vo + act 10 = 35 + ac(15) ac = -1.67 mjs2^ = 1.67 mjs2 �
Fl2-2. s = So + vot + �act o = 0 + 15t + � (-9.81)t t = 3.06 s
F12-3. ds = v dt
isds = i
t (4t - 3t2)dt
s = (2t2 - t3) m s = 2(42) - 43 = -32 m = 32 m �
Fl2-4. (^) a = (^) '!if = (^) 1ft (0.5t3 - 8t)
a = (1.5t2 - 8) mjs When t = 2 s, a = 1.5(22) - 8 = -2 mjs2 = 2 mjs2^ �
F12-S. v = 'fit = 1ft (2t2 - 8t^ +^ 6) = (4t - 8) mjs v = 0 = (4t - 8)
Ans.
Ans.
Ans.
Ans.
t = 2 s A�
F12-6.
slt=o = 2(02) - 8(0) + 6 = 6 m slt= 2 =^ 2(22) - 8(2) + 6 = -2 m slt= 3 = 2(32) - 8(3)^ +^ 6 = 0 m (ilshot = 8 m + 2 m = 10 m
j v dv = j a ds
I
v
v dv = r (10 - 0.2s)ds
5 m/s^ Jo v = ( V20s - 0.2s2^ + 25) mjs At s =^ 10 m,
Ans.
v =^ V20(1O) - 0.2(102) + 25 =^ 14.3 mjs Ans.
F12-7. v = j(4t2 - 2) dt
v = � t3 - 2t + e 6 8 0
s = j (� t3 - 2t + el ) dt
s = � t4 - t2^ + elt + e
t = 0, s = -2, e 2 = - t = 2, s = -20, e 1 = -9. t = 4, s = 28.7 m F12-S. a = v 'flf = (20 - 0.05s2)(-0.ls) At s = 15 m, a = -13.1 mjs2 = 13.1 mjs2^ � F12-9. v = 'fit = 1ft ( 0.5t3) = 1.5t vlt=6 s = 1.5(62) = 54 mjs v = 'fit = 1ft (108) = 0 F12-10. ds = v dt 1" ds = i
t (-4t + 80) dt s = -2t2^ + 80t s = -2(2of + 80(20) = 800 ft
Ans.
Ans.
Ans.
a = !b!dt = !Ldt^ (-4t + 80) = -4 ftjs2 = 4 ftjs2^ � Also, a = Ll. vLl.t = 0 20 - s 80 ftfs 0 = -4 ftjs F12-11. a ds = v dv a = � = 0.25sfs (0.25s) = 0.0625s als=40m = (^) 0.0625(40 m) = 2.5 mjs2 (^) --> F12-12. 0 ,,; t < 5 s, v = 'fit = 1ft (3t2) = (6t) mjs vlt=5 s = 6(5) = 30 mjs 5 s < t ,,; 10 s, v = 'fit = 1ft (30t - 75) = 30 mjs v = Ll.Ll.ts = 22510 mm^ - -^755 m^ m^ = 30 mjs o ,,; t < 5 s, a = '!if = 1ft (6t) = 6 mN 5 s < t ,,; 10 s, a = '!if = 1ft (30) = 0 0 ,,; t < 5 s, a = ilvjilt = 6 mjs 5 s < t ,,; 10 s, a = ilvj M = 0
F12-13. 0 � t < 5 s,
dv =^ a dt l
v dv = 1
20 dt v = (20t) m/s v = 20(5) = 100 m/s 5 s < t � t', (�) dv = a dt lv dv = 11 -10 dt 100 mls 5s v = (150 - 1Ot) mis, o = 150 - lOt' t' = 15 s Also, Liv = 0 = Area under the a-t graph 0 = (20 mN)(5 s) + [-(10 m/s)(t' - 5) s] t' = 15 s Fl2-14. 0 � t � 5 s,
ds = v dt sl& =^ 15t21b s = (15t2) m
r t
io^
ds =
io^
30t dt
s = 15(52) =^ 375 m 5 s < t � 15 s, ds = v dt; r^ ds = 11 (-15t + 225)dt
i 375 m 5s
S = (-7.5t2^ + 225t - 562.5) m s = (-7.5)(15)2^ + 225(15) - 562.5 m = 1125 m Ans. Also, Lis = Area under the v-t graph = (^) � (150 m/s)(15 s) = 1 125 m
Fl2-15. l
x dx = 1
32t dt
x = (16t2) m
lY dy = 11 S dt t = (^) � Substituting Eq. (2) into Eq. (1), i =^ 4x
Fl2-16. Y = 0.75(St) = 6t Vx = x = % = 1ft (St) = S m/s ->
Ans.
Ans.
FUNDAMENTAL PROBLEMS 6 8 1
Vy = y = '!l; = 1ft (6t) = 6 m/s i The magnitude of the particle's velocity is v = Vv; + v; = V(S m/s)2 + (6 m/s) = 10 m/s Ans. F12-17. y = (4t2) m Vx = x = 1ft (4t^4 ) = (16t3) m/s -> Vy =^ Y^ =^ 1ft (4t2) = (St) m/s i When t = 0.5 s, v = Vv; + v; =^ V(2 m/s)2^ +^ (4 m/s) = 4.47 m/s Ans. ax = Vx = 1ft (16t3) = (4St2) mN ay = Vy = 1ft (St) = S m/s When t =^ 0.5 s, a =^ Va; + a; =^ V(12 m/s2f + (S mN) = 14.4 m/s2 Ans. F12-18. y = 0.5x y = 0.5.: Vy = t When t = 4 s, Vx = 32 m/s Vy = 16 m/s v = Vv; + v; =^ 35.S m/s ax = Vx = 4t ay = Vy = 2t When t = 4 s, ax = 16 m/s2^ ay = S m/s
Ans.
a = Va; + a; = V162^ + S2^ = 17.9 mN Ans. F12-19. Y = (t4) m Vx = x = (4t) m/s When t = 2 s, Vx = S m/s Vy = 32 m/s v = Vv; + v; = 33.0 m/s ax =^ Vx = 4 m/s ay =^ Vy =^ (12t2) m/s When t = 2 s, ax = 4 m/s2 ay = 4S m/s
Ans.
a = Va; + a; = V42^ + 4S2^ = 4S.2 m/s2 Ans.
F12-30.
Fl2-31.
tan 8 = ¥X = fx (f4 x2) = 12 x
8 = tan-1(i2 x) I
x�lO ft
= tan-1^ (f¥) = 39.81° = 39.8° Ans.
p =^ [1 +^ �dY/d:)2P/^2 = [1^ +^ (�4P/2^1
Id y/dx I 1121 x�lO ft = 26.468 ft _ (^) v2 _ (20 ft/S)2^ _ 2
an - p - 26.468 ft - 15.11 ft/s
a = �--�--------� V(at)2 +^ (a,i^ = V(6 ft/S2^ f^ + (15.11 ft/S2) = 16.3 ft/s2^ Ans.
(aB)t = -O.OO1s = (-0.001)(300 m)(I rad) mN
= -0.4712 m/s
v dv = at ds
t Bv dv = t^ �^ �^. ;01s ds
J 2 5 m/s^ Jo
VB = 20.07 m/s VB2^ (20.07 m/s? (^2)
(aB)n = p = 300 m = 1.343 m/s
aB = V(aB)t2 + (aB)n
= V( -0.4712 m/s2)2^ + ( 1.343 m/s2) = 1.42 m/s2^ Ans.
Fl2-32. at ds = v dv
at = v !flf = (0.2s)(0.2) = (0.04s) m/s
at = 0.04(50 m) = 2 m/s
v = 0.2 (50 m) = 10 m/s
a = - =v2^ (10 m/s)2= 0.2 m/s n p (^) 500 m a = Va? + a� = V(2 m/s2)2 + (0.2 mN) = 2.01 m/s2 Ans.
Fl2-33. vr = r = 0
ve = r8 = (4008) ft/s v = Vv; + vJ 55 ft/s = V02^ �^ --+ [(4008) ft/S]2--�-------,;: 8 = 0.1375 rad/s Ans.
FUNDAMENTAL PROBLEMS 6 8 3
F12-34. r = 0.lt3It�1.5 s = 0.3375 m r = 0.3t2It�1.5 s = 0.675 m/s r = (^) 0.6tlt�1.5 s = 0.900 m/s 8 = (^) 4t3/2It�1.5 s = 7.348 rad 8 =^ 6tl/2It�1.5 s = 7.348 rad/s 8 =^ 3t-l/2It�1.5 s = 2.449 rad/s Vr = r = 0.675 m/s ve = r8 = (0.3375 m)(7.348 rad/s) = 2.480 m/s ar = r - r8 2 = (0.900 mN) - (0.3375 m)(7.348 rad/s) = -17.325 m/s ae = r8 + 2i8 = (0.3375 m)(2.449 rad/s2)
- 2(0.675 m/s)(7.348 rad/s)^ =^ 10.747 m/s v = VVr 2 + Ve
= V(0.675 m/s)2 + (2.480 m/s)
= 2.57 m/s a = � = V( - 17.325 m/s2)2 + ( 10.747 m/s2)
Ans.
= 20.4 m/s2^ Ans. F12-3S. r = 28 r = 28 r = 28 At 8 = 1T/4 rad, r = 2(*) = I r = 2(3 rad/s) = 6 ft/s r = 2(1 rad/s) = 2 ftN
a,. = r - riP = 2 ftN - (I ft)(3 rad/s)
= -12.14 ft/S ae = r8 + 2rO = (I ft)(1 rad/s2) + 2(6 ft/s)(3 rad/s) = (^) 37.57 ft/S
a = Va,.^2 + ae
= V(-12.14 ft/s2)2 + (37.57 ft/S2) = 39.5 ftN Ans.
6 8 4 PART I A L S O L U T I O N S A N D A N S W E R S
Fl2-36. r =^ eO r = eOiJ r = e°(j + eOiJ ar = r - riJ2^ = (e°(j + eOiJ2) - (eOiJ2 = e1T/4(4) = 8.77 m/s2^ Ans. ao = rO + U O =^ (e°(j) + (2(e°iJ)iJ)^ =^ eO(O + 2(2) = e1T/4(4 + 2(2)2) = 26.3 m/s
Fl2-37. r = [0.2(1 + cos e)] mlo� 300 = 0.3732 m
r = [ -0.2 (sin e)O 1 m/sle� 30 °
= -0.2 sin 30°(3 rad/s) = -0.3 m/s Vr = r = -0.3 m/s Ve = riJ = (0.3732 m)(3 rad/s) = 1.120 m/s
Ans.
v = Yv; + vJ = Y( -0.3 m/s)2 + (1.120 m/s) = 1.16m F12-38. 30 m = r sin e
r = (��n�) = (30 csc e) m
r = (30 csc e)le�450 = 42.426 m
r = -30 csc e ctn e elo�450 = - (42.4260) m/s
Vr = r = -(42.4260) m/s ve = rO = (42.4268) m/s v =^ YV; + vJ 2 = Y( -42.4268 )2 + (42.4268 ) o =^ 0.0333 rad/s Fl2-39. IT = 3SD (^) + SA o = 3VD + VA 0 = 3VD + 3 m/s VD = -1 m/s = 1 m/s i Fl2-40. SB + 2sA + 2h = I VB + 2VA =^0 6 + 2VA = 0 VA = -3 ft/s = 3 ft/s i Fl2-41. 3SA + SB = I 3VA + VA = 0
Ans.
Ans.
Ans.
Ans.
3VA + 1.5 = 0 VA = -0.5 m/s = 0.5 m/s i Ans. F12-42. IT = 4 SA + SF o = 4 VA + VF o = 4 VA + 3 m/s VA = -0.75 m/s = 0.75 m/s^ i Ans.
F12-43. SA + 2(SA^ -^ a) +^ (SA^ - sp) =^ I 4sA - SP = I + 2a 4VA - Vp = 0 4VA - 4 = 0 VA = 1 m/s F12-44. Sc + SB = ICED (SA^ - sc) +^ (SB^ - sc) +^ SB^ =^ IACDF SA + 2sB - 2sc =^ IACDF Thus Vc + VB = 0 VA + 2VB - 2vc = 0 Eliminating vc, VA + 4VB = 0 Thus, 4 ft/s + 4VB = 0 VB = -1 ft/s = 1 ft/s i F12-4S. VB = VA + VB/A 100i = 80j + VB/A VB/A = 100i - 80j VB/A = Y(VB/A);^ ,------0---------,;: +^ (VB/A); = Y(100 km/h)2^ + (-80 km/h) = 128 km/h
Ans. (1)
(2)
Ans.
Ans.
e = tan-1[
(VB/A)Y]
= tan-1(^
80 km/h )
= 38.r� Ans. (VB/A)x 100 km/h F12-46. vB = VA + VB/A (-40Oi - 692.82j) = (650i) + V B/A VB/A = [-105Oi - 692.82j] km/h VB/A = Y(VB/A); + (^) (VB/A); = Y( 1050 km/h)2 + (692.82 km/h) = 1258 km/h Ans.
_l [(VB/A)Y] _1 (692.82 km/h)
e = tan -- = tan = 33.4° ;p-
(VB/A)x 1050 km/h
F12-47. VB = VA + VB/A (5i + 8.660j)^ =^ (12.99i + 7.5j) +^ vB/A vB/A^ = [-7.99Oi^ + 1.160j] m/s VB/A = Y(-7.990 m/s)2^ + (1.160 m/s)
Ans.
= 8.074 m/s Ans. dAB = VB/At = (8.074 m/s)(4 s) = 32.3 m Ans.
6 8 6 PART I A L S O L U T I O N S A N D A N S W E R S
(15 m/s) Fn = (500 kg) (^) 200 m =^ 562.5 N I.E;^ =^ mat; E; = (500 kg)(1.5 m/s2) = 750 N F = YF,; + E;2 = \1(562.5 N)2^ �--� +- (750 N)2-� = 938 N
F13-13. ar = r - riP = 0 - (1.5 m + (8 m)sin 45°) = (-7.157 02) m/s I.Fz = maz; T COS 45° - m(9.81) = m(O) T =^ 13.87^ m I.Fr = mar; -(13.87m) sin 45° = m( -7.157 / F)
Ans.
8 = 1.17 rad/s Ans.
F13-14. 8 = 7Tt^21 ,=o.5 s = (7T/4) rad
o = 27Ttl,=o.5 s = 7T rad/s 8 = 27T rad/s r = 0.6 sin 810=7T/4 rad =^ 0.4243 m ; = 0.6 (cos 8)Olo=7T/4 rad = 1.3329 m/s r = 0.6 (cos 8)0 - (sin8)0210=7T/4 rad =^ -1.5216 mN ar = r - riP = -1.5216 mN - (0.4243 m)(7T rad/s) = -5.7089 m/s
F13-15.
ae = r 8 + 2fe = 0.4243 m(27T radN)
- 2(1.3329 m/s)(7T rad/s) = 1 1 .0404m/s I.Fr = mar; Fcos 45° - N cos 45° -0.2(9.81)cos 45° = 0.2(-5.7089) I.Fo =^ mao; F sin45° + N sin45° -0.2(9.81)sin 45° = 0.2(11.0404) N = 2.37 N F^ =^ 2.72 N^ Ans.
r = 50e2elo=7T/6 rad = [50e2(7T/6) 1 m = 142.48 m
r =.^ 50 2e( 20 ' 8 ) = 100e 20 ' 8 0=7T/6 1 rad
= [100e2(7T/6)(0.05) 1 = 14.248 m/s
;: = 100( (2e200)0 + e20(O) )le=7T/6 rad
= 100[ 2e2(7r/6) (0.052) + e2(7r/6)(0.01) 1
= 4.274 m/s
ar = r - rez = 4.274 m/s2 - 142.48 m(0.05 rad/s) = 3.918 mN ao = ':0+ 2;8 = 142.48 m(O.01 radN)
- 2(14.248 m/s)(0.05 rad/s) = 2.850 mN I.Fr = mar; Fr = (2000 kg)(3.918 m/s2) = 7836.55 N I.Fe =^ mae; Fe = (2000 kg)(2.850 m/s2) = 5699.31 N F = \IF? + Fl = (^) \1(7836.55 N)2 (^) + (5699.31 N) = 9689.87 N = 9.69 kN F13-16. r = (0.6 cos 28) mlo=oo = [0.6 cos 2COO)] m = 0.6 m f = (-1.2 sin2(0) m/slo=oo
= [ -1.2 sin2(00)( -3) 1 m/s = 0
r = - 1 .2( sin280 + 2cos2(02) mNle=oo = -21.6 m/s Thus, ar = r - r02 = -21 .6 m/s2 - 0.6 m(-3 rad/s) = -27 m/s ao = r8 + 2f 8= 0.6 m(O) + 2(0)( -3 rad/s) = 0 I.Fe = mae; F^ - 0.2(9.81) N^ =^ 0.2 kg(O) F = 1.96 N Ans. Chapter 1 4 Fl4-1. (^) Tl + (^) I.UI-2 = (^) T 2
Fl4-2.
o + m(500 N)(0.5 m) - !(500 N/m)(0.5 m) = !(10 kg)V v = 5.24 m/s Ans. I.Fy = may; NA - 20(9.81) N cos 30° = 0 NA = 169.91 N TI + I.UI -2 =^ T 2 o + 300 N(lO m) - 0.3(169.91 N) (10 m)
- 20(9.81)N (10 m) sin 30° = !(20 kg)V v = 12.3 m/s Ans.
Fl4-3. (^) Tl + 2:,Ul - 2 = T 2
Fl4-4.
Fl4-5.
0 + 2[ 115
m (600 + 2S2) N dS] - 100(9.81) N(15 m) = �(100 kg)V v = 12.5 m/s Ans. Tl +^ 2:,Ul-2 =^ T 2 �(1800 kg)(125 m/s)2 - [C50000N^ �^20 000 N)^ (400 m)] = �(1 800 kg)V v = 8.33 m/s Ans. Tl + (^) 2:, Ul- 2 = T 2 �(10 kg)(5 m/sf + 100 Ns' + [10(9.81) N] s' sin 30° -�(200 N/m)(s'f^ =^0 s' = 2.09 m s = 0.6 m + 2.09 m = 2.69 m Ans.
Fl4-6. (^) TA + 2:,UA-B = TB
Fl4-7.
Fl4-8.
Fl4-9.
Consider difference in cord length AC - BC, which is distance F moves. 0 + 10 Ib(V(3 ft)2^ + (4 ft)2 - 3 ft
VB =^ 16.0 ft/s �2:,Fx =^ max; 30 (�) = 20 a v = Vo + act
_- l"2^ ( 5 I ^2
- 2 S^ U^ gJ^ VB
a = 1 .2 m/s2 ---->
v = 0 + 1.2(4) = 4.8 m/s P = F · v = F (cos (J)v
= 1 1 5 W
�2:,Fx^ =^ max; lOs = 20 a a = 0.5s m/s2 ----> vdv = ads
1v 15m
o V dv = 0 0.5 s ds V = 3.536 m/s P =^ F · v = 10(5)(3.536) = 177 W (+i)2:,Fy =^ 0; Tl - 100 lb = 0 (+ i)2:,Fy =^ 0;
Tl = 100 lb
100 lb + 100 lb - T 2 = 0 T 2 = 200 lb
Ans.
Ans.
Ans.
FUNDAMENTAL PROBLEMS 6 8 7
Pout =^ TB • VB = (200 Ib)(3 ft/s) = (^) 1.091 hp Pout 1.091 hp fin^ =^ ----;- (^) 0.8 =^ 1.36 hp^ Ans. F14-10. (^) 2:,Fy' = may'; N - 20(9.81) cos 30° = 20(0) N = 169.91 N 2:,Fx' =^ max,; F - 20(9.81 ) sin 30° - 0.2(169.91 ) = 0 F = 132.08 N P = F · v = 132.08(5) = 660 W F14-11. + i 2:,Fy = may; T - 50(9.81)^ =^ 50(0)^ T^ =^ 490.5 N Pout = T · v = (^) 490.5(1.5) = 735.75 W fin = Pout (^) = 735.75 (^) = 920 W E; 0. Fl4-12. 2sA + sp = I 2 aA + ap = 0 2 aA + 6 = 0 aA = -3 m/s2^ = 3 m/s2^ i 2:,Fy = may; TA-490.5N = (50 kg)(3m/s2) TA =^ 640.5N Pout = T · v = (640.5N/2)(12) = 3843 W
Ans.
Ans.
fin = --;-^ Paut^ = o:s^3843 = 4803.75 W = 4.80 kW Ans.
Fl4-13. TA + VA =^ TB + VB o + 2(9.81)(1.5)^ =^ �(2)(VB)2^ +^0 VB = 5.42 m/s Ans.
- i 2:,Fn =^ man; T - 2(9.81) = 2 (--(5.42)2^ )
T =^ 58.9 N^ Ans. Fl4-14. (^) TA + VA = (^) TB + VB � mAd + mghA =^ � mBvJ^ + mghB
[�(2 kg)(l m/s)2] +^ [2 (9.81) N(4 m)]
= [�(2 kg)vJ] + [0] VB = 8.915 m/s = 8.92 m/s Ans.
- i 2:,Fn = (^) man; NB - 2(9.81)N = (2 kg) (^ (8.915 m/s)2) 2 m NB = 99.1 N Ans.
FlS-6. Block B:
FlS-7.
( (^) + l ) (^) mVI (^) + J F dt = mVl
o + 8(5) - T(5) = 3 �. 1 (1)
T = 7.95 1b Block A: C::.) (^) mVl + J F dt = (^) mVl o (^) + 7.95(5) - J.Lk(10)(5) = 3 �Ol(1) J.Lk = 0.
Ans.
Ans. (:::.) mAvA)1 +^ mB(vB)1 = mAvAh +^ mB(VBh (20(1 03) kg)(3^ m/s)^ + (1 5(1 03) kg)(^ -1 .5^ m/s) = (20(103) kg)(VA)l + (15(103) kg)(2 m/s) (VAh =^ 0.375^ m/s^ --+
J
11 (:::.) m(vBh + L^ F dt^ = m(vBh II (15(103) kg)(^ - 1.5^ m/s) +^ Favg(0.5^ s) = (^) (15(103) kg(2 m/s) Favg = 105(103) N = 105 kN
Ans.
Ans.
FlS-S. (^) (:::.) (^) mp[(vph1x + mcl(vch1x = (mp (^) + mc)vl
5[ 1O(�) ] (^) + 0 = (^) (5 (^) + 20)Vl Vl = 1.6 m/s
FlS-9. Tl +^ VI =^ Tl^ + Vz �mA(VA)r + (vg)1 =^ � mAvA)� + (Vg)l �(5)(5f^ +^ 5(9.81)(1.5) = �(5)(VA)? (vAh = 7.378 m/s (:::.) mA(vAh + mB(vB)l = (mA + mB)v 5(7.378) (^) + 0 = (5 + 8)v
Ans.
v = 2.84^ m/s^ Ans.
FlS-lO. (^) (:::.) mA(vAh + mB(vBh = mAvAh +^ mA(vB)l 0 + 0 = 10(vAh + 15(vBh (1) TI^ +^ VI =^ Tl + Vz � mA(VA)r +^ � mB(vB)l^ +^ (V.)l = (^) �mA(vA)� + (^) �mB(vB)� + (^) (V.h 0 + 0 + H5( 103)] (0.21) = �(10)( (^) vA)� + �(15)( vB)? + 0 5(VA)? + 7.5 (VB)� = 100 Solving Eqs. (^) (1) and (^) (2),
(2)
(VB)l =^ 2.31 m/s --+ Ans. (vAh =^ - 3.464^ m/s^ =^ 3.46^ m/s^ <-^ Ans.
FUNDAMENTAL PROBLEMS 6 8 9
FlS-ll. (t) (^) mA(VAh + mB(vBh = (mA + mB)vl o + 10(15) = (15 + 10)Vl Vl = 6 m/s T1 + V1 = Tl + Vl �(mA + mB)v?^ + (V.)l =^ � (mA + mB)vj + (Ve) 3
�(15^ + 10) (61) + 0 = 0 + H 1O( 103) lS�ax
smax = 0.3 m = 300 mm
FlS-12. (^) (:::.) 0 + 0 = (^) mp(vp)x - mcVe o = (^) (20 kg) (^) (vp)x - (250 kg)ve (vp)x = 12.5 ve (1) Vp = Ve (^) + Vpje (vp») + (vp)yj = -ve i (^) + [(400 m/s) cos 300i
- (400^ m/s) sin^ 300j (vpLi + (vp)yj = (346.41 - ve)i + 200j (vp)x = 346.41 - ve (vp)y = 200^ m/s (vpL = 320.75 m/s Ve = 25.66 m/s Vp =^ V (vp); + (vp); = V (^) (320.75 m/s)l + (^) (200 m/s)l = 378 m/s
FlS-13. (^) (:::.) e = (VB)l - (vAh (VA)! - (VB)! (9 m/s) - (1 m/s) = 0. (8^ m/s)^ - (-2 m/s)
Ans.
Ans.
FlS-14. (^) (:::.) (^) mAvAh + mB(vB)l = mAvAh + mB(vBh [15(103) kg1(5 m/s)^ + [25(103)1(-7^ m/s) = (^) [15(103) kg1(vA) 1 + [25(103)1(vBh 15(vAh + 25(VB) 1 = - 100 (^) (1) Using the coefficient of restitution equation, (vBh - (vAh (:::.) e =^ - ,-----------,-----,-- (VA)I - (VB)I 0 6 =
(VB)l - (vAh
. (^5) m/s (^) - ( -7 m/s) (VB)l - (vAh = 7.2 (^) (2) Solving, (vBh =^ 0.2^ m/s --+ (VA)l = -7 m/s = 7 m/s <-
Ans. Ans.
6 9 0 PART I A L S O L U T I O N S A N D A N S W E R S
FlS-lS. Tl + V, = T 2 +^ V � m(VA)t + mg(hAh =^ � m(VA)� + mg(hA) H3�0 2 slug) (5 ft/s)2^ +^ (30 Ib)(lOft) = (^) H3�0 2 slug)(vA)� + 0 (v Ah =^ 25.87 ft/s^ +--- (:':-)^ mA(vAh + mB(vBh =^ mAvAh +^ mB(vBh (3�02 slug) (25.87 ft/s)^ +^0 = (^) (3�0 2 slug)(vA)3 + (^) (3�0 2 slug)(vBh 30(VA)3 + 80(VBh = 775.95 (^) (1)
(:':-) e^ =^ (vBh - (VA) (vAh - (vBh 0.6 (^) = (VB)3 - (VA) 25.87 ft/s - 0 (VB)3 - (vAh = 15.52 (2) Solving Eqs. (1) and (2), yields (VB)3 = 11.3 ft/s +--- (vAh =^ - 4.23 ft/s^ =^ 4.23 ft/s^ -->^ Ans. FlS-16. After collision: Tl + "LU1 -2 = T 2
�Ci. 2 ) (VA)� - 0.2(5)(?z) = 0 (vAh =^ 1.465 ft/s H3h0 2 ) (VB)� - 0.2(10)(lz) = 0 (vB)2 =^ 1.794 ftls "Lmv1 = "Lmv 3i. 2 (vAh +^0 =^ 3i. 2 (1.465)^ +^ 3;° 2 (1.794) (VA)1 =^ 5. e (^) = (vB) 2 - (vA) 2 1.794 - 1. (VA)1 - (VB)1 5.054 - 0 = 0.0652 Ans.
FlS-17. ( + i ) m[(vb)11y = (^) m[(vbhh
[(vbh1y = [(vb)1 1y = (20 m/s)sin30° = 10 m/s (^) i
(:::.) e^ = (vwh - [(vbh1x [(vbh1x - (vwh 0.75 =^ 0 - [(vb)21x (20 m/s)cos 30° - 0 [(vbh1x = - 1 2.99 m/s^ = 1 2.99 m/s^ +--- (vbh =^ V[(vbh1; +^ [(vbh1; = V( 12.99 m/s)2^ + (10 m/s) = 16.4 m/s Ans.
() (^) = tan-1( [(Vb)21y) = tan-1( 1 0 m/s^ ) [(vbh1x 12.99 m/s = 37.6°^ Ans. F1S-18. "Lm(vx)1 =^ "Lm(vx) 2 o + 0 = (^) 3i 2 (1) + (^) 3�^12 (VBx) 2 (vBxh = - 0.1818 ft/s "Lm( Vy)1 =^ "Lm(^ vyh 3i2 (3)^ +^0 =^0 +^ 3�^12 (VBy) 2 (VBy) 2 = 0.545 ft/s �------
(vBh = V( -0.1818)2^ + (0.545)
= 0.575 ft/s^ Ans. F1S-19. Ho = "Lmvd; Ho = (^) [2(10)(�) l(4) - [2(10)m l(3) = 28 kg · m2/s ) F1S-20. Hp = "Lmvd; Hp = (^) [2(1 5) sin 30°1(2) - [2(15) cos 30°1(5) = -99.9 kg · m2/s^ =^ 99.9 kg · m2/s )
F1S-21. (^) (Hz)1 + "L J Mz dt = (Hzh 5(2)(1.5) + 5(1.5)(3) = 5v(1.5) v = 5 m/s Ans.
F1S-22. (Hz)1 + "L J Mz dt^ =^ (Hzh
o + (^) 14 s(10t)(�) (1.5)dt = 5v(1.5) v =^ 12.8 m/s^ Ans.
F1S-23. (^) (Hzh + "L J Mz dt = (Hzh
o + (^) 15 sO.9t2 dt = 2v(0.6) v =^ 31.2 m/s^ Ans.
F1S-24. (Hzh + "L J Mz dt = (Hzh
o + (^) 14 S8tdt + 2(10)(0.5)(4) =^ 2[10v(0.5) v =^ 10.4 m/s^ Ans.
6 9 2 PART I A L S O L U T I O N S A N D A N S W E R S
F16-10. VA = WOA X rA = (12 rad/s)k X (0.3 m)j = [-3.6i] m/s VB^ =^ VA^ +^ wAB X rB/A VBj = (-3.6 m/s)i
- (WABk) X (0.6 cos 300i - 0.6 sin 30°j) m VBj = [WAB(0.6 sin 30°) - 3.6]i + WAB(0.6 cos 300)j o = WAB(0.6 sin 30°) - 3.6 (^) (1) VB = WAB(0.6 cos 30°)^ (2) WAB = 12 rad/s^ VB = 6.24 m/s^ i F16-11. Vc =^ VB^ +^ WBC X rCjB
F16-12.
F16-13.
vd = (-60i) ft/s
- (-wBck) X (-2.5 cos 300i^ +^ 2.5 sin 300j)^ ft vd = (-60)i^ +^ 2.165wBd^ +^ 1.25wBC i o = -60 + 1.25wBC Vc =^ 2.165 WBC WBC = 48 rad/s Vc = 104 ft/s VB^ =^ V A + W X rB/A -VB cos 30° i + VB sin 30° j = (-3 m/s)j + (-wk) X (-2 sin 45°i - 2 cos 45°j) m -0.8660VBi + 0.5vBj = -1.4142wi + (1.4142w - 3)j -0.8660VB = -1.4142w 0.5VB = 1.4142w - 3 W = 5.02 rad/s VB = 8.20 m/s VA 6 WAB = -- = - = 2 rad/s rA/IC 3 rCjIC =^ V1.52^ + 22 =^ 2.5 m cp = tan-1C^2 s ) = 53.13° Vc =^ WAB rCjIC = 2(2.5) = (^) 5 m/s (J = 90° - cp = 90° - 53.13° = 36.9° �
Ans.
Ans. Ans.
Ans. Ans.
F16-14. (^) VB = WAB rB/A = 1 2(0.6) = 7.2 m/s �
F16-15.
Vc = 0 VB (^) 7. WBC = -- = - = 6 rad/s rB/IC 1. Vo 6 W =^ -- = - = 20 rad/s rO/lc 0. rA/IC =^ VO.32^ + 0.62^ = (^) 0.6708 m
cp = tan-l (�:�) = 26.57°
VA = wrA/IC = 20(0.6708) = 13.4 m/s (J = 90° - cp = 90°- 26.57° = 63.4°d
Ans. Ans.
Ans.
Ans. Ans.
F16-16. The location of Ie can be determined using similar triangles. 0.5 - rCjlc rCjlc 3 1 .5 rCjIC^ =^ 0.1667 m W = --Vc = -- =1.5 9 rad/s rCjlc 0. Also, rO/lc = 0.3 - rCjlc = 0.3 - 0. = 0.1333 m.
Ans.
Vo = wro/lc = 9(0.1333) = (^) 1.20 m/s Ans. F16-17. VB = wrB/A = 6(0.2) = 1 .2 m/s rB/IC =^ 0.8 tan 60° =^ 1.3856 m rCjlc =^ cos 600^ 0.8^ = 1 6. m VB 1. WBC = -- =^3856 = 0.8660rad/s rB/IC 1. = 0.866 rad/s Then, Vc =^ WBC rCjlc = 0.8660(1.6) = 1.39 m/s
F16-18. (^) VB = wAB rB/A = 10(0.2) = 2 m/s Vc = WCD rCjD = wCD(0.2) ---->
rB/IC =^ co�·�0^0 = 0.4619 m
F16-19.
rCjlc = 0.4 tan 30° = 0.2309 m WBC VB^2 =^ --rB/IC^ =^ 0 461 9 =^ 4.330 rad/s . = (^) 4.33 rad/s Vc = WBC rCjlc wCD(0.2) = 4.330(0.2309) WCD = 5 rad/s
W = -- = - = 2 rad/sVA^6 rA/IC 3 3B = 3A^ +^ a X rB/A - w2rB/A
aBi = -5j + (ak ) X (3i - 4j) - 22(3i - 4j)
aBi = (4a - 12)i + (3a + l l )j
aB = 4a - 12 0 = 3a + 1 1 a = -3.67 rad/s aB = -26.7 m/s
F16-20. (^) 3A = 30 + a X rA/O - w2rA/ = 1 .8i + (-6k) X (0.3i) - 1 22(0.3j) = 13.6i - 43.2j) m/s
Ans.
Ans.
Ans.
Ans.
Ans. Ans.
Ans.
Ans.
Fl6-21. (^) 3A =^ 3B + a^ X^ 'A/B - W2 'A/B 3i = aBj + (-ak) X 0.3j - 202(0.3j) 3i = 0.3ai + (aB - 120)j 3 = 0.3a a = 10 rad/s2^ Ans. 3A^ =^ 30 + a^ X^ 'A/o - w2'A/ = 3i + (- 10k)^ X^ (-0.6i) - 202(^ -0.6j) = 1243i + 6j} m/s2 Ans. rAIIC Fl6-22. (^3)
0.5 - rAI1C 1.5 rAIIC^ =^ 0.3333 m w = -- =VA^3 rAIIC 0 3333.^ =^ 9 rad/s 3A = 3C +^ a X^ 'A/c - w2'A/c 1.5i - (aA)nj = -0.75i + (adnj
- (-ak) X 0.5j - 92(0.5j) 1.5i - (aA)nj = (0.5a - 0.75)i (^) + [(adn - 40.5] j 1.5 = 0.5a - 0. a = 4.5 rad/s
Fl6-23. VB = W rBIA = 12(0.3) = 3.6 m/s VB 3. WBC = -- = - =rBIIC^ 3 rad/s
3B - a^ -^ X^ 'BIA - W 'BIA^2 = (-6k) X (0.3i) - 122(0.3i) = 1 -43.2i - 1 .8j} m/s 3C =^ 3B + aBC X 'C;B - WJC'C;B aci = (-43.2i - 1.8j)
- (aBC k)^ X^ (1.2i) - 32(1.2i) aci = -54i + (1.2aBC - 1.8)j ac = -54 m/s2 = 54 m/s2 � o = 1 .2aBc - 1.8 aBC = 1.5 rad/s
Fl6-24. (^) VB = (^) W rBIA = 6(0.2) = 1.2 m/s -->
rBIIC =^ 0.8 tan 60°^ =^ 1.3856 m WBC = -- = -VB^ 1. rBIIC 1.3 58 6 = 0.8660 rad/s 3B - a^ -^ X^ 'BIA - W 'BIA^2 = ( -3k) X (0.2j) - 62(0.2j) = (^) [0.6i - 7.2j] m/s 3C = 3B + aBC X 'C;B - W2'C;B ac cos 300i + ac sin 300j
Ans.
Ans. Ans.
= (0.6i - 7.2j) + (aBC k (^) X 0.8i) - 0.86602(0.8i) 0.8660aci + 0.5ad = (0.8aBC - 7.2)j
FUNDAMENTAL PROBLEMS 6 9 3
0.8660ac = 0 0.5ac = 0.8aBC - 7. aC = 0 aBC = 9 rad/s
Chapter 1 7 F17-1. �"2,Fx = m(ac)x; 100(�) = 100a a = 0.8 m/s2 -->
- j "2, Fy =^ m(ac)y; NA + NB - 1 00m - 100(9.81 ) = 0 C + "2,Mc^ =^ 0; NA(0.6) + 1 00(�)(0.7)
- NB(O.4) - 100(�)(0.7) = 0 NA = 430.4 N = 430 N NB = 61 0.6 N = 611 N F17-2. "2,Fx' = m(ac)x'; 80(9.81) sin 15° = 80a a = 2.54 m/s "2,Fy' = m(ac)y'; lV.4 (^) + NB - 80(9.81) cos 15° = 0 C + "2, Mc = 0; NAO.5) - NB(0.5) = 0 NA = NB = 379 N F17-3. (^) C + "2,MA = (^) "2,(Mk)A; lOm(7) = 3 �o 2 a(3.5) a = 19.3 ft/s � "2, Fx = m(ac)x; Ax + 1O(�) = 3�o 2 (1 9.32) Ax = 6 lb
- j "2, Fy = m(ac)y; Ay - 20 + 1O(�) = 0 Ay = 12 lb F17-4. FA = f.LsNA = 0.2NA FB = f.LsNB = 0.2NB � "2, Fx = m(ac)x ;
Ans.
Ans.
(1)
(2) Ans. Ans.
Ans.
(1)
(2) Ans.
Ans.
Ans.
Ans.
0.2NA + 0.2NB = 100a^ (1)
- j "2, Fy =^ m(ac)y; NA + NB - 100(9.81) = 0 (2) C + "2, Mc =^ 0; 0.2lV.4(0.75) + lV.4(0.9) + 0.2NB (0.75)
- NB (0.6) = 0 (3) Solving Eqs. (1), (2), and (3), NA = 294.3 N = 294 N NB = 686.7 N = 687 N a = 1 .96 m/s2 Ans. Since NA is positive, the table will indeed slide before it tips.
FI7-13. (^) lc = rzml2 = rz(60)(32) = 45 kg · m
- j 2:Fy = m(ac)y; SO - 20 = 60ac ac = 1 m/s2j C + 2: Mc =^ lca; SO(l) (^) + 20(0.75) = 45a a = 2.11 [ad/s2^ Ans.
FI7-14. (^) C + 2: MA = (Mk)A; -200(0.3) = -100ac(0.3) - 4.5a 30ac + 4.5a = 60 (1) ac = ar = a(0.3) (2) a = 4044 rad/s2^ ac = 1.33 m/s2 ---->
FI7-15. (^) + j 2:Fy = m(ac)y; N - 20(9.S1) = (^0) N = 196.2 N � 2:Fx =^ m(acL; 0.5(196.2) = 20ao ao = 4.905 m/s2^ ----> C + 2: Mo = loa; 0.5(196.2)(004) - 100 = -1.Sa a = 33.S rad/s
FI7-16. (^) C + 2: MA = (Mk)A; 20(9.S1)sin30o (0.15) = O.lSa + (20ac)(0.15) O.lSa (^) + 3ac = 1 4. ac = ar = a(0.15) a = 23.36 rad/s2^ = 2304 radN ac = 3.504 m/s2^ = 3.50 m/s
FI7-17. (^) + j 2:Fy = m(ac)y;
N - 200(9.S1 ) = 0 N^ =^1962 N � 2:Fx =^ m(acL; T - 0.2(1 962) = 200ac C + 2: MA = (Mk)A; 450 - 0.2(1962)(1) = lSa (^) + 200ac(OA)
3C = 3A^ +^ £I' X^ rC/A - w2rC/A aci = - aAj (^) + ak X (OAj) - w2( -OAj) aci = OAai + (OAw2 - aA)j ac = OAa Solving Eqs. (1), (2), and (3), a = 1.15 rad/s2 ac = 00461 m/s T = 4S5^ N
Ans.
Ans.
Ans.
Ans. Ans.
(1)
(2)
(3)
Ans.
FUNDAMENTAL PROBLEMS 6 9 5
FI7-18. � 2: Fx = m(ac)x; 0 = 1 2(ac)x (ac)x = 0 C + 2:MA = (Mk)A -12(9.S1 )(0.3) = 1 2(ac)y(0.3) - rz(12)(0.6)2a 0.36a - 3.6(ac)y = 35.316 (1) w = O 3C =^ 3A^ +^ £I' X^ rC/A - w2rC/A (ac)y j = aAi + (-ak) (^) X (0.3i) - 0 (ac)y j = (aA)i - 0.3 j aA = 0 (ac)y = -0.3a Solving Eqs. (1) and (2) a = 24.5 rad/s (ac)y = -7.36 m/s2 (^) = 7.36 mN� Chapter 1 8 FI8-1. 10 = mk8 = So( 0042) = 12.S kg · m 2 Tl = 0 T 2 = (^)! low2 = (^)! (12.S)w2 = 6Aw S = Or = 20(27T)(0.6) = 247T m
Tl + 2:Ul-2 =^ T 2
o (^) + 50(247T) = 6Aw W = 24.3 rad/s FI8-2. (^) Tl = 0
T 2 = ! m(vc)� +! lcw�
= (^)! (3�^02 slug)(2.5w 2 )
- (^)! [M3�0 2 slug)(5 ft)21w� T 2 = 6A700w� Or, 10 = � ml2 = � (^) (3�0 2 slug)(5 ft) = 12.9400 slug·ft So that T 2 = � low� = � (12.9400 slug · ft2)W 2 2 = 6A700wi Tl + 2:Ul-2 = T 2 Tl + [-Wyc^ + MO] = T 2 o + [-(50 lb)(2.5 ft)^ +^ (100 lb·ft)(^ �)] = 6A700w� W 2 = 2.23^ rad/s
Ans. (2)
Ans.
Ans.
Ans.
6 9 6
FIS-3.
FIS-4.
FIS-S.
PART I A L S O L U T I O N S A N D A N S W E R S
(VG)Z = wZrG/lC = wz(2.S) IG = fi mZZ = fi (SO)(sZ) = 104.17 kg · mz Tl =^0 T 2 = � m(vG)� + � IGw� = � (SO)[ wz(2.S) (^) F + �(104.17)w� = (^) 208.33w� Up = PSp = 600(3) = 1800 J Uw = -Wh = -SO(9.81)(2.S - 2) = -24S.2S J Tl + (^) "LU1-Z = Tz o + 1800 + (-24S.2S) = 208.33w� Wz = 2.732 rad/s = 2.73 rad/s T = � mvO 2 + (^) � 10wZ = (^) � (SO kg)(OAw)2 + (^) � [SO kg(0.3 m)2]wZ = 6.2Swz J Or, T = � llcwz = (^) � [SO kg(0.3 mf + SO kg(OA m)Z]wZ = 6.2Swz J So = Or = 10(21T rad)(OA m) = 81T m Tl + "LU1-Z =^ Tz TJ + P cos 30° So = T 2 o + (SO N)cos 300(81T m) = 6.2Sw2 J w = 13.2 rad/s 1.G (^) = .1. mZ2 = 12 .1.l (^) (30)( 3Z) = 22 S kg ' mZ Z. Tl =^0 Tz = � mVG Z^ + � IGWz = � (30)[w(0.S)j2 (^) + �(22.S)w2 = lSw 10 = IG + mdz = fi (30)(32) + 30(0.SZ) = 30 kg ' mz Or, T 2 = � 10wz = � (30)wz = lSwz Sl =^ Orl =^ 81T(0.S) =^ 41T m S 2 = Or 2 = 81T(1.S) = 121Tm
Up, = PZs 2 = 20(121T) = 2401T J UM =^ MO^ = 20[4(21T)] = 1601T J Tl + (^) "LU1-Z = Tz o + 1201T + 2401T + 1601T = lSwz w = 10044 rad/s = lOA rad/s
Ans.
Ans.
Ans.
FIS-6. Vo = wr = w(OA) 10 = mkoz = 20(0.32) = 1 .8 kg · m TI = 0 Tz = � mVG Z + � IGwz = � (20)[ w(OA) (^) F + �(1.8)wz = 2.Swz
UM = MO = M C:) = SO(�.�) = 2S00 J
Tl + "LU1-Z = Tz o + 2S00 = 2.Sw w = 31.62 rad/s = 31.6 rad/s Ans. FIS-7. VG = wr = w(0.3) IG =^ � mr2^ = (^) � (30)( 0.3Z) = (^) 1.3S kg · mZ TJ = 0 Tz = � m(vG)�^ +^ � IGw� = (^) � (30)[W 2 (0.3)]2 (^) + �(1.3S)wi = (^) 2.02Sw� (Vg)l = WYl = 0 (Vg)z =^ -WY2 =^ -30(9.81)(0.3)^ =^ -88.92 J Tl + � = T 2 + V 2 o + 0 = 2.02Sw1 + (-88.29) W 2 = 6.603 rad/s = 6.60 rad/s Ans. FIS-S. Vo = wro/lc = w(0.2) 10 = mkOZ = SO( 0.32) = 4.S kg ' mZ Tl =^0 T 2 = (^) � m(vo)i + �low�
= hSO)[ wz(0.2) F + � (4.S)w�
= 3.2Sw (Vg)l = WYl = 0 (Vg)z = -Wyz = (^) -SO(9.81)(6 sin 30°) = -1471 .SJ TI + � = T 2 + V 2 o + 0 = 3.2Swi + (-1471.S) W 2 = 21.28 rad/s = 21.3 rad/s Ans.
6 9 8 PART I A L S O L U T I O N S A N D A N S W E R S
F19-3. VA = WA'A/IC = WA (0.15)
C +^ "LMo =^ 0; 9 - AI(0.45)^ =^0 AI =^ 20 N
C +(Hch +^ "L j
l Mcdt = (Hc) 2 I, o + 20(5) = 10wAO.15)
- [1O(0.12) ]wA WA = 46.2 rad/s Ans.
F19-4. fA = mkl = 10 (0.082) = 0.064 kg · m
IB =^ mk�^ =^50 (0.152)^ =^ 1 .1 25 kg · m
('B ) (0.2)
WA =^ 'A WB =^ 0:1 WB =^ 2WB
(5 S
Jo^ Fdt^ =^ 500 - 1.28(WB)^2
(5 S
Jo^ Fdt^
= (^) 5.625(wBh Equating Eqs. (1) and (2), 500 - 1.28(wBh^ =^ 5.625(WB) 2
(WB) 2 =^ 72.41 rad/s^ =^ 72.4 rad/s^ Ans.
F19-S. (�) m[ (voL]1 + "L J Fx dt = m[ (voLh
o + (150 N)(3 s) + FA(3 s) = (50 kg)(0.3w 2 )
( +fCWI + "L J Mc dt = Icw 2
o + (150 N)(0.2 m)(3 s) - FAO.3 m)(3 s) = [(50 kg)(0.175 mf]w 2 W 2 = 37.3 rad/s Ans. FA = 36.53 N Also,
hcwi +^ "L^ J Mlc dt^ =^ Ilcw 2
o + [(150 N)(0.2 + 0.3) m](3 s) = (^) [(50 kg)(0.175 m)2 (^) + (50 kg)(0.3 mf]w 2 W 2 = 37.3 rad/s Ans.
F19-6. (+i) m[(vc)I]Y + "L J Fy dt = m[(vc)2]Y
o + NA(3 s) - (150 Ib)(3 s) = 0 NA = 150 lb
( +^ (Hlch +^ "L J Mlc dt^ =^ (Hlch
o + (25 Ib · ft)(3 s) - [0.15(150 Ib)(3 s)](0.5 ft) = W.� slug(1.25 ft)2]W 2 + w.� slug) [w 2 (1 ft)](l ft) W 2 = 3.46 rad/s Ans.
An swe rs to S e l e cted P ro b l e m s
Chapter 1 2 12-1. v2^ =^ VB^ +^ 2ac(^ S - so) ac = 0.5625 m/s v = Vo + act t = 26.7 s 12-2. (^) v = 0 + (^1) (30) = 30 m/s S = 450 m 12-3. t = 3 s S = 22.5 ft 12-5. (^) dv = a dt v =^ (^6 t2 - 2t3/2)^ ft/s ds = (^) v dt S = (2t3 - � t5/2 + (^) 1 5) ft 12-6. h^ =^127 ft v =^ -90.6^ ft/s^ =^ 90.^6 ft^ /^ s^ � 12-7. (^) v = 13 m/s �s = 76 m t = 8.33 s 12-9. dt = dva v = Vr-2k-t-+-v--" 6 12-10. SA = 3200 ft 12-11. a = -24 m/s �s = -880 m ST = 91 2 m 12-13. �s = 2 m ST = 6 m vavg = 0.333 m/s (vsp)avg = 1 m/s 12-14. vavg = 0.222 m/s (vsp)avg = 2.22 m/s 12-15. d = 517 ft d = 616 ft 12-17. h = 5t' - 4.905(t,)2 + 1 0 h = 19.81t' - 4.905(t,)2 - 14. t' = 1. 682 m h = 4.54 m 12-18. S = 1708 m vavg = 22.3 m/s 12-19. al H = 1.06 m/s 12-21. VA = (^) ( 3 t^2 - 3t) ft/s VB = (4t3 - 8t) ft/s t = 0 s and = 1 s B stops
12-22.
12-23.
12-25.
12-26.
12-27. 12-29.
t = 0 s t = Yz s SABl r=4 s = 152 ft (ST)A = 41 ft (ST)B = 200 ft Choose the root greater than 10 m S = 11.9 m v = 0.250 m/s v = (20e-2r) m/s a = ( -40e-2r) m/s S = 1O(1 - e-2r) m
1 (g + kV02 )
S = 2k In g + kv 1 hmax = 2k In ( 1 + :gYok^ 2) v =^ 4.11^ m/s a = 4.13 m/s v =^ 1.29^ m/s S Ir=6s = -27.0 ft v =^ 4.50t2 - 27.0t^ +^ 22. The times when the particle stops are t = 1 s and t = 5 s. Stot = 69.0 ft
12-30.^ S^ =^ �(^1 - e-kr)
12-31. 12-33.
12-34.
12-35.
12-37.
a = -kvoe-kr
t =
vf In
(Vf^ +^ V^ )
2g vf - v Distance between motorcycle and car 5541.67 ft t = 77.6 s Sm = 3.67(1W (^) ft a = 80 km/s t = 6.93 ms vavg =^10 m/s^ <- aavg = 6 m/s ball A h = vot' - !r t,2 2 VA = Vo - gt' h = vo (t' - t) - !r et' - t)2 2 VB = Vo - get' - t) 2vo + gt t' = ----''----- 2g--=--
699
