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Unformatted text preview: Problem 14—2
The crate of weight W has a velocity VA when it is at A. Determine its velocity after it slides down
the plane to s = s'. The coefﬁcient of kinetic friction between the crate and the plane is ,uk. Given: W=201b a=3 ft :
VA 2 12 — b 4
s
s’ = 6 ft #k = .2 ‘7 ‘8 I
XVIIr, ,1}
Solution: \
Akit‘
a
0 = atan(;) NC = Wcos(6) F = ,ukNC
m
Guess v’ = 1—
s
1 W 1 W ft
Given — — vAz + Wsin(6)s'—Fs' = — — v’2 v’ = Find(v’) v' = 17.72—
2 g 2 g s _——————_—I——_n_——————___—_ *Pr0blem 1412 The force F, acting in a constant direction on the block of mass M, has a magnitude which
varies with position x of the block. Determine the speed of the block after it slides a distance d ,. When x = 0, the block is moving to the right at vo‘The coefﬁcient of kinetic friction between the
block and surface is ,uk Given: M=20kg c=3 d1 : 3 m d = 4
m N
v0 = 2 — k = 50 —
s 2
m
,uk 2 0.3 g = 9.81 —
Solution:
62 “,2
Guess v1 = 2r—n
5
Given
1 d I 1
c
—Mv02+ —— kxzdx—ykMng —,uk kxzde—MVIZ
2 2 d2 0 2 d2 2
c + c + V] = Find(v1) v1 = 3.774 31
S *Problem 1416 The block A of weight WA rests on a surface for which the coefﬁcient of kinetic friction is yk.
Determine the distance the cylinder B of weight W3 must descend so that A has a speed VA starting from rest. ft R 1 I?
v = — ___ g
A S *‘x f \“ mgm
AN '
Solution:
W1“.
L = SA + 233
Guesses d = 1 ft
Given
2
1 2 VA
WBd—ykWA2d= 2— WAvA + W3 7
g d = Find(d) d: 0.313 ft Problem 1421 The crash cushion for a highway barrier consists of a nest of barrels ﬁlled with an
impact—absorbing material. The barrier stopping force is measured versus the vehicle penetration
into the barrier. Determine the distance a car having weight W will penetrate the barrier if it is
originally traveling at speed v0 when it strikes the ﬁrst barrel. Units Used: kip = 103 1b
Given:
W = 4000 lb
ft
v0 = 55 —
s
 32 2 ﬂ
g —  2
5
Solution: 2 g 1(W) 2
Area=— — v0
2 g — — v0 —Area=0 Harrier stopping fume [kiwi : ﬁ 111% 35 In 25 Vehicle mnumtinn 1ft] Area = 187.888 kip ft We must produce this much work with the
barrels. Assume that 5 ft < x < 15 ft Area = (2 ft)(9 kip) + (3 ft)(18 kip) + (x — 5 ft)(27 kip) x _ Area — 72 kipft
— 27 kip + 5 ft x = 9.292 ft Check that the assumption is corrrect! Problem 1423 The block of weight W is released from rest at A and slides down the smooth circular surface
AB. It then continues to slide along the horizontal rough surface until it strikes the spring.
Determine how far it compresses the spring before stopping. Given:
W = 5 lb yk = 0.2
a = 3 ft :9 = 90 deg l
b = 2 ft k = 40 —b
ft
Solution:
Guess d : 1 ft
Given 1
Wa—ykW(b+d)—§kd2=0 d = Find(d) d: 0.782 ft Problem 14—26 Cy1inder A has weight WA and block B has weight W8. Determine the distance A must descend from
rest before it obtains speed VA. Also, what is the tension in the cord supporting block A? Neglect the
mass of the cord and pulleys. Given:
ft
WA 2601b VA =8—
5
ft T“
WB=101b g=32.2—2
s
Solution:
L=2sA+sB 0=2vA+vB
W
System ’1
1 WA 1 W3
0 + WAd— WBZd = [—j VA2 + —(—j (2VA)2
2 g 2 g
WA + 4W3 2
_ VA
2g r
d = d = 2.484 ft
WA — 2W3 Block A alone 2 1 WA 2 WAVA
0+WAd—TdZE—VA T=WA— T=361b
g *Problem 14—36 A block of weight W rests on the smooth semicylindrical surface. An elastic cord having a
stiffness k is attached to the block at B and to the base of the semicylinder at point C. If the block is released from rest at A( 0: 0°), determine the unstretched length of the cord so the
block begins to leave the semicylinder at the instant 6 = 6,. Neglect the size of the block. Given
W = 2 lb
lb
k = 2 —
ft
61 2 45 deg 4
a = 1.5 ft G
g = £2 txxx a;
s x 55:”.
 . F. i f"
Solution. 5 H «ex
ft {\“x ?
Guess 6 : 1 ft v1 = l — x j
S “"ef
Given N5 : U
W v12 ‘/
Wsin(61) = /
g a In!
1 1 W V]2
2 2
—k7za—6 ——k 7r—6 a—6 —Wasinl9 = ——
2< >2[( 1)] (ngz
V1 . ft
[ j = Find(v1,5) v1 = 5.843— a = 2.77 ft
5 s ——————————_—_—_ Problem 1475 The bob of the pendulum has a mass M and is released
from rest when it is in the horizontal position shown.
Determine its speed and the tension in the cord at the
instant the bob passes through its lowest position. Given: M = 0.2 kg m r = 0.75 m g = 9.81 —2 S is Solution:
Datum at initial position: T1+V1=T2+V2 1
0 + 0 = EMvg2 —Mgr v2=\/Er 2F" = Ma" T: 5.89 N Problem 1477 The collar of weight W is released from rest at A and travels along the smooth guide. Determine
its speed when its center reaches point C and the normal force it exerts on the rod at this point.
The spring has an unstretched length L, and point C is located just before the end of the curved
portion of the rod. Given:
W = 5 lb
L = 12 in
W
h = 10in
lb
k = 2 .—
1n
— 32 2 ft
g —  2
5
Solution: TA+VA=Tc+VC 2
k k ft
vC = 2gL + (—gjhz — (—th/Lz + h2 — L) vC = 12.556 — W W Narmtlﬁ] ital] 2
W vc kL
NC = —[—] — L2 + h2 — L) NC = 18.919lb *Problem 1480 The rollercoaster car has mass M, including its passenger, and starts from the top of the hill A
with a speed VA. Determine the minimum height h of the hill crest so that the car travels around both inside loops without leaving the track. Neglect friction, the mass of the wheels, and the size
of the car. What is the normal reaction on the car when the car is at B and when it is at C? Units Used: kN = 103 N Given: m
M=800kg vA=0— r3 = 10 m
ft
rC = 7 m g = 32.2 —2
5
Solution: Check the loop at B ﬁrst We require that N B = 0
V32
in
_N _ M = _ — v : r V = —
B g VB B \[g B B S
l 2 1 2
TA+VA=TB+VB EMvA +Mgh=§MvB +Mg2rB
V32 — VA2
h = —— + ZrB h = 25 m
2g
Now check the loop at C
1 2 l 2
TA+VA=Tc+VC EMVA +Mgh=EMvC +Mg2rC vC = J vA2 + 2g(h — 2rC) vC = 14.694 3
S NC  Mg VC
"C NC: 2
VC rC —Mg NC: 16.825kN Since NC > 0 then the coaster successfully
passes through 100p C. Problem 1485 The bob of mass M of a pendulum is ﬁred from rest at position A. If the spring is compressed to a distance 5 and released, determine (a) its stiffness k so that the speed of the bob is zero when it
reaches point B, where the radius of curvature is still r, and (b) the stiffness k so that when the bob
reaches point C the tension in the cord is zero. Units Used: kN = 103 N Given: “ ﬂ f,
m M = 0.75 kg g = 9.81 —2 s y, (3: 50mm r=0.6m i 3,,
Solution: A AtB: ,V M 1 52 p33 4444 4 H —k = Mgr 3 2 ‘E 2Mgr kN k = k = 3.53 —— E
52 m z
At C ‘
vcz
—Mg= 2— VC=V2g1
r
l l M kN
E1:52 = Mg3r+ —MvC2 k = —(6gr+ VCZ) k = 14.13— 2 62 m Problem 1491 The ride at an amusement park consists of a
gondola which is lifted to a height h at A. If
it is released from rest and falls along the
parabolic track, determine the speed at the
instant y = 0'. Also determine the normal
reaction of the tracks on the gondola at this
instant. The gondola and passenger have a
total weight W. Neglect the effects of
friction. Given: W=5001b d=20ft h=120ft a=260ft Solution: 2 X
,v(x) = — p(x) = a X
y'(x) = 2
a (I + y'(x)2)3 y"(x) 6(x) = atan(y'(x)) Guesses
x2 = 1 ft
Given
1
Wk 2 —
2
x2
V2 g ft
v2 =10: (W) 2
— v2 +Wd = Find(x2 , v2 , FN) FN H 2
y (x) = 
a
FN =11b
d=y(x2)
x2 = 72.1ft I," {f \
1 FN — Wcos(6(x2)) 2 ft
V2 = —80.2 : K
g p(x2) FN = 952 lb *Problem 1496 The doublespring bumper is used to stop the steel billet of weight W in the rolling mill.
Determine the maximum deﬂection of the plate A caused by the billet if it strikes the plate with
a speed v. Neglect the mass of the springs, rollers and the plates A and B. Given: lb
W = 1500 lb k1 = 3000 — ft ft lb
v = 8 — k2 = 4500 — 3 ft Solution:
ka1 = k2x2
1 W 1 1
— — v2 2 —k1x12 + —k2x22
2 g 2 2
2 1 W 2 1 2 1 km
 — v = k1x1 + —k2
2 g 2 2 kg 2 2
(W) 2 k1 x1 2
— v = k1+ x1
g k2 x1 = 0.235 m ...
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 Spring '08
 Priezjev

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