This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Problem 13—13 Determine the normal force the crate A of mass M exerts on the smooth cart if the cart is
given an acceleration a down the plane. Also, what is the acceleration of the crate? Given:
M = 10 kg
a = 2 £1
2
7 s
6 = 30 deg
Solution: N— Mg = —M(a) sin(0) “crate = (a) sin( 6) Problem 1315 The driver attempts to tow the crate using a rope that has a tensile strength me. If the crate is originally at rest and has weight. W, determine the greatest acceleration it can have if the
coefﬁcient of static friction between the crate and the road is gs and the coefﬁcient of kinetic friction is ,uk. Given: T max = 200 lb
W = 500 lb
as = 0.4
,uk = 0.3 ft
g = 32.2 —2 S .
0 = 30 deg _ Solution: Equilibrium : In order to slide the crate, the
towing force must overcome. static friction. ‘ 3%! Initial guesses F N = 100 lb T = 50 lb Given Tcos(€) — ,uSFN = 0 FN+ Tsin(6) — W T If T = 187.613 lb > T max = 200 lb then the truck will not be able to pull the create without breaking
the rope. I If T = 187.613 1b < T max = 2001b then the truck will be able to pull the create without breaking the rope and we will now calculate the acceleration for this case. 0' '[FN] =IFin<i1>(FN, T) ft ,
Initial guesses F N = 1001b a = l —2 Require T = T max
3 ' v ' . ,
W FN
Given Tcos(0) — #kFN = —a FN+ Ts»in(0) — W: 0 [ j = Find(FN,a)
g a _ .
* 7. ' ‘ ﬂaunt
a = 3.426—
1» 5f“: '17‘2 S *Problem 1324 At a given instant block A of weight WA is moving downward with speed VAo Determine its speed
at a later time t. Block B has a weight W3 and the coefﬁcient of kinetic friction between it and the horizontal plane is ,uk Neglect the mass of the pulleys and cord. ‘ Given: V . ,
' '  21"“
WA = 10 lb _
VA0 = 6 —
S a
t = 2 S “:1
W3 = 41b7 ‘ E _ W1; ,
#k = 02 e ' G}? > IT
ft , ' ' ' 112%
g = 32.2 — > . ,
, 2, . i e
S V . > WE . Solution: L = SB + 2sA . , ft V ft
Guesses aA=lE aB=1—2 T=llb
s s
‘WB
Given T —— pk WB = a3
g
”WA
2T— WA = [ jaA
g
‘ 0 = a3 + 2aA
, T , .
,  “A 10.403 '11
em = Find(T,aA,aB) T= 3.3851b ' = ~—
'  a3 —20.806 52
dB ‘ .
, 1:.7: 1’;
VA =’ VA0 + “A1 , VA 2263+ ﬂ Problem 1333 The collar C of mass mc is free to slide along the smooth shaft AB. Determine the acceleration of
collar C if collar A is subjected to an upward acceleration a. The collar moves in the plane. Given: B
mc = 2 kg
m
a = 4 E mfg rm“
, ' g
m
g = 9.81 7 = é/ﬂdﬂaﬁ
s
6 = 45 deg
Solution: The collar accelerates along the rod and the rod accelerates upward. émcgcosw) = mdaCA — (a)cos(¢9): “04 v: (g1+ a)cos(0) aCA 891(9) . 6—905 m 9 m
ac = ac— = .— lacl = 7.491 3 “aCA cos(0) + a —"2. 905 'SZ *"3’ 7 Problem 1335 The block A of mass m A rests on the plate B of mass m B in the position shown. Neglecting the mass of the rope and pulley, and using the coefﬁcients of kinetic friction indicated, determine the
time needed for block A to slide a distance 3’ on the plate when the system is released from rest. Given:
mA = 10 kg
m3 = 50 kg. V
» s’ = 0.5 m '
#AB = 02
.UBC = 011
9 = 30 deg In
S Solution:
SA + SB = L Guesses 53.4% In
aA=1—2 aB=l S m
2
S
T=1N NA=1N 'N3 = 1 N
Given ' 7 . ' r
"0147+ (13 = 0
NA — mA_gcos(€) = 0
NB  NA — m3gcos(6) = 0
T  #ABNA  mAgsin(0) = mA 0A T + #ABNA 4E #BCNB _ Tagsin(6) = mBaB aA
a3 ‘ T T = Find(aA,aB, T,NA,NB) NA' =
NA. V ’ NB
NB ‘ “BA = “B — aA “BA = 3.708 E; t = — t"=f0,._5i9 s *Problem 13—44 Each of the three plates has mass M If the coefﬁcients of static and kinetic friction at each surface of
contact are ,us and ,uk respectively, determine the acceleration of each plate when the three horizontal forces are applied. Given:
M = 10 kg
,us = 0.3 ,
,uk = 0.2
FB = 15 N
FC = 100 N '
' FD = 18 N
m
g = 9.81 —2
5
Solution: Case 1: Assume that no slipping occurs anywhere.
FABmax = #s(3Mg) . FBCmax = [15(2Mg) FCDmax = #s(Mg) Guesses FAB=1N Fgc=1N FCD=1N Given —FD+.FCD=0 Fc—FCD—Fgc=0 —FBV—FAB+FB(}=0
FAB , FAB ' 677 FABmax 88.29
FBC = Find(FAB.FBc,Fcp) FBC =. 82 N ' , FBCmax = 58.86 N F CD 7 ' F CD 18 F CDmax 2943 If FAB = 67N < FABmax = 88.29N and FBC = 82N > FBCmax = 58.86N and
F CD = 18N < F CDmax = 29.43 N then nothing moves and there is no acceleration. Case 2: If FAB = 67N < FABmax = 88.29N and FBC = 82N > FBCmax = 58.86N and
F CD = 18N < F CmeC = 29.43 N then slipping occurs between B and C. We will assume that no slipping occurs at the other 2 surfaces. iSet Fgc=,uk(2Mg) (:3 =0 ac=dD=a
Guesses FAB=1N FCD=1N (1:195
' 3
Given , —FD+FCD=Ma FcFCD—FBC=Ma —FB—FAB+FBC2=0
'FAB . . . .
F F' d(F F ) * FAB 2424 N‘ 7 2138 m
l = In 5 5a . = a="' _
CD AB CD FCD 39.38 ; '  52. a ac=a aD=a If FAB = 24.24N <' FABmax ; 88.29N and FCD = 39.38N ' ‘> FCDmax = 29.43N then . , m ' ' m
we have the correct answer and the accelerations are (13 = 0 , ac = 2.138 —2 , aD = 2.138 —2
s s Case 3: If FAB = 24.24N < FABma'x = 88.29N and 'FCD = 39.38N > FCDmax = 29.43N then slipping occurs between C and D as well as between B and C. We will assume that no slipping
occurs at the other surface. Set FBC = #k(2Mg) FCD = #k(Mg)
Guesses FAB=1N _ aC=IE; 0113:122
‘ . s . . s
Given —FD+FCD=MaD Fc—FCD—FBc=MaC —FB—FAB+FBc=0
CF' d(F ) F 24 24 N ' lac ‘ ~ $44114 7 m“
0C = 1n . AB,a 9a =  t. =.. I .‘_'
C D AB .110 * . 0.162 , ':S2:
“D, , 5', If FAB = 24.24 N < F A Bmax = 88.29 N then we have the correct answer and the accelerations I aC=41143§ aD‘: 0.162% ' There are other permutations of this prOblems depending on the numbers that one chooses. W Problem 1345 Crate B has a mass m and is released from rest when it is on top of cart A, which has a
mass 3m. Determine the tension in cord CD needed to hold the cart from moving while B is sliding down A. Neglect ﬁiction.
Solution: BlockB:
NB — mgcos(¢9) = 0‘ N3 = mgcos(0) I Cart:
—T+ NBsin(0) = 0 T: mgsm(é)cos(0) ‘5 Problem 1373 The pendulum bob B of mass M is released from rest when 0 = 0°. Determine the initial tension
in the cordiand al'so'at the instant the bob reaches point D, 0 = (9,. Neglect the size of the bob; Given: V M = 5kg 0,: 45 deg L=2m g=9.8122 A
» 7 s Solution: '
Initially, iv = Oso an = o T== 0
At D We have Mgcos(o91) = Mat at = gcos(01)' at = 6.937 22 , s
2 .
. Mv T .
TD—Mgsm(t91)= L "5“ r
. ‘ :54 Now ﬁnd the veloc1ty v > , 3 Min,
Guess v = 1 E 7 I S a, '
Given [V v dv : J, gcos(t9)L d0
' 0 A 0 v = Find(v) V v = 5.268 3 S 2 V TD => Mgsin(01) + :47] TD % 104.1 N' *Problem 1376 The spool S of mass M ﬁts loosely on the inclined rod for which the coefﬁcient of static friction
is ,us. If the spool is located a distance d from A, determine the maximumconstant speed the spool can have so that it does not slip up the rod. Given: "7
. Vyksfs
M = 2 kg e = 3 . . _ e
f ,
,us=0.2 f=4 , ﬁnk '5  if d=0.25m g=9.8132 ‘3 
V s »
Solution: :6 / . E IE—FN n
f ' ’ 
p = —f— ‘ i I _ N
{ 32+f2] .. ' S Guesses NS = 1 N v = 1 0 H N .
[ s] = Find(Ns,v) NS = 21.326 N v = 0.969 2 ,,
V S *Problem 1380 The block has weight W and it is free to move along the smooth slot in the rotating disk. The
spring has stiffness k and an unstretched length 6. Determine the force of the spring on the block
and the tangential component of force which the slot exerts on the side of the block, when the
block is at rest with respect to the disk and is traveling with constant speed v. Given:
W = 2 1b
k = 2.5 l—b
ft
6 = 1.25 ﬁ
it
v = 12 —
5
Solution:
2
W v
2:Fn=man; Fs = k(p _ 6) = [—]
, g P Choosing the positive root,
1
p = F [kg6'+ (y’kzgzéz +4kvi2)] p = 2.617ﬁ
g F3 = k(p — 5) F52: 3.41911: 2F, = ma,; 2F, = ma,; F, = 0 Problem 13—87 The spool of mass M slides along the rotating rod. At the instant shown, the angular rate of
rotation of the rod is 6', which is increasing at 6". At this same instant, the spool is moving
outwardralong the rod at r’ which is increasing at r" at r. Determine the radial ﬁ'ictional force
and the normal force of the rod on the spool at this instant. Given:
M = 4 kg r = 0.5 m
d
g = 6 B— r' i 3 E
S S '
0, = 2 1d r" = 1 2
2 2
s s
m
= 9.81 —
g 2
s
Solution:
 2
a, = r” — r67 a9 = r67'+ 2r'0
Fr = Marv F0 = Mae
Fz = Mg' { Fr; Fr
F, = 458.0 N 75‘ F92 +15? = 153.1 N m *Problem 13—92 The partiCle has maSs M and is conﬁned to move
along the smooth horizontal slot due to the
rotation of the arm 0A. Determine the force of
the rod on the particle and the normal force of the slot on the particle when 6: 0,. The rod is rotating with angular velocity 0' and angular
acceleration 6". Assume the particle contacts only
one side of the slot at any instant. M Given:
M = 0.5 kg
61 = 30 deg
6’ = 2 E h = 0.5 m
s
m
€,=3ﬂ g=9.81—2
2 s
s
Solution:
0 = 01 h = rcos(€) r = h r : 0.577 m
cos(t9)
0 = r’cos((9) — rsin(t9)t9' , r' = (rsm(6)]g r’ = 0.667 E
cos(6) s
2 0 = r"COS(t9) — 21"Sin(0)‘9 — ”05(9) ‘9 _ rsin(t9)6" r" = 2r’0’tan(t9) + r92 + rtan(0)0' r” = 4.849 32
' s
2 r” — r02
(FN— Mg)cos(6) = M(r” — r9 ) ’ FN = Mg + FN : 6.371 N
cos(6?) —F + (FN— Mg)sin(0) = —M(r€’ + 2r'6’) F = (FN— Mg)sin(0) + M(r6" + 2r’0') F = 2.932 N» Problem 13106 Using air pressure, the ball of mass M is forced to move through the tube lying in the horizontal plane
and having the shape of a logarithmic spiral. If the tangential force exerted on the ball due to the air is F, determine the rate of increase in the ball’s speed at the instant 6': 6', .What direction does it act in? Given: M=0.5kg a=0.2m b=0.1 61:35 F=6N
2 Solution: .
r aebg 1
“MW = — = —b6_ = Z
d_r abe '
d6
1
y/ = atan(zj y/ = 84.289 deg
F .
F=Mv’ v'=— v’=12n—1
M S2 ...
View
Full Document
 Spring '08
 Priezjev

Click to edit the document details