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Unformatted text preview: *Problem 154 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 = lNs 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
= 24.7Ns
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 1528 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 = —2vB 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 1541 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 1551 The freerolling 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 1559 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 1568 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 1578 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 1579 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 1588 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 1598 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 kgm2
H =m—v b—c —m ———v a+ H =—66.0
PA A212A()AJ2—12A(d) PA 8
n+ n+
kgm2 Hp]; = —vaBcos(t9)(b + e) + vaBsin(t9)(a — f) HPB = —73.9
s *Problem 15100 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 15106 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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 Spring '08
 Priezjev

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