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Unformatted text preview: PROBLEM 13.11 ' Boxes are transported by a conveyor belt with a velocity v0 to a ﬁxed
incline at A where they slide and eventually fall off at B. Knowing that
,uk = 0.40 , determine the velocity of the conveyor belt if the boxes leave
the incline at B with a velocity of 2 m/s. Given: At A, For AB, At B, Find: v0 UH = (Wsi1115° — ,uk N)(6 m)
\2F=o N—Wcos15°=0 N = W00515°
UH = W(si1115° — 0.40cos15°)(6 m)
U A_B = —(0.76531)W = —0.76531mg
TA + UA—B = TB gmvg — 0.7653lmg = 2 m v3 = (2)(2 +(0.76531)(9.81m/52)) v3 = 19.0154 v0 = 4.36 m/s 4 PRGBLEM 13.23 Four 3kg packages are held in place by rﬁictionon a conveyor which is
disengaged from its drive motor. When the system is released from rest,
packager I'leaves the belt at A just as package 4 comes onto the inclined
portion of the belt at B. Determine (a) the velocity of package 2 as it
leaves the belt at A, (b) the velocity‘of package 3 as it leaves the belt at A.
Neglect the mass of the belt and rollers. SOLUTION Given: Conveyor is disengaged, packages held by friction and system is released from rest. Neglect mass of
belt and rollers. Package 1 leaves the belt as package 4 comes onto the belt. Find: (a) Velocity of package 2 as it leaves the belt at’A.
(b) Velocity of package 3 as it leaves the belt at A.
(a) Package 1 falls off the belt, and 2, 3, 4 move down. U1_2 = (3)(W)(0.8) = (3)(3 kg) x 9.81 m/s2 (0.8)
U1_2 = 70.632 J
TI + Ul_2 = T2 0 + 70.632 = 4.5v§ ) «v3 = 15.696 v2 = 3.9618 v2 = 3.96 m/s4 PROBLEM 13.23 CONTINUED (b) Package 2 falls off the belt and its energy is lost to the system, and 3 and 4 move down 2 ft. T2' = (2)[%mv§] T2’ = (3 kg)(15.696) g=Mmm
T3 = (2)[§mv§] = (3 kg>(v%)
T3 = 3v?
U2_3 = 2(W)(0.8) = (2)(3 kg x 9.81 m/s2)(0.8 m)
U2_3 = 47.088j
T2' + U2_3 = T3 = 47.088 + 47.088 ‘=, 3v32
v32 = 31.392 v3 = 5.6029
V3 = m/S ‘ PROBLEM_I13.'28 A 3kg collar C slides on a frictionless vertical rod. It is pushed up into ‘ i the position sh’OWn; compreSsing the upper spring by 50 mm and
released. Determine (a) the maximum deﬂection of the lower spring,
(b) the maximum velocity of the Collar. “‘27 11— 300 mm A 't a 2 N/mm SOLUTION ' /
(a) Spring constants ’ 3 N/mm = 3000 N/m 2 N/mm = 2000 N/m
Max deﬂection at 2 when velocity of C = 0
v1=0, T1=o, v2=0,‘,‘T2=0 Ul_2 = Ue + Ug( (0 05 my _ (2000 N/m)
' 2 2
+(3 kg)(9.81m/s2)§;£ym) 4.
= 3.750 1000(y,,,)2 + 4' 529430,") U646 T1 + U1_2 = T2: 0 —1~000(y,,,)2 + 29.43(y,,,) +9549: 0
ym =127.8mm{ _ (3000 N/m)
1—2 ‘ ——_ . ym = 0 m
Q53 . 9 10C?) 10 .
0‘: M (b) Maximum velocity occurs as the mwer spring is compressec‘i a?” w. am, Okla distance y'
i; .7 We“. 30 Bu; 7] = 0; T2 =Vémcv2 = %(3 kg)v2 = (1.5)v2 a A: #11“ /
I ‘5 $er $ “51+ U1—2 = T2; 0  (1000)(y')2 + 29.43(y') + 12.579 = (1.5)v2 EU tr substitute y' = 0.014715 m 2
% = 0 40000,!) + 2943 = o; y' = 0.014715 m
' ﬂJﬁlra" —0.0002165 + 0.43306 +M= 1.5v2 v = 2.95 m/s‘ " PROBLEM 13.43“ a A section of track for a‘ roilercoasterconsists of two circular arcs AB and
CD: joined a‘istraight portion BC. The radius of AB is 27 m and the
radius of ‘ CD is ‘72 m. The‘ca‘r and its occupants, of total mass 250 kg,
‘ reachipoint A with praCtically no Velocity and then drop freely along the
track: Determine the manual force exerted by the track on the car as the
carreach’es point B. Ignore air resistance and rolling resistance. ' SOLUTION TB = émvg = %(250 kg)v§ =125v§ (1250 kg) x (9.81m/s2) UH = W(27)(1g—.cos40°)
“UH =~r(250 kg x 9:81m/s2)‘(2’7 m)(o;234) TA + UA_B. =1}; 0 + 15495 91251;; (154.951) I 2 — 5,
VB " (125 kg). v12; = 124.0 1112/52; Newtons Law at B +/ N — Wcos40° = _R ; v; =124.0m2/s2 1' 250 1:: 124.0 2/ 2
N = (250 kg X 9.81 nl/sz')(cos40°) — £___g_)(—mg_)
‘ ' ' r 4 7 27m N = 1879 —1148 = 731N ::; ' A small block slides at a ‘speed v = 3 m/s on a horizontal surface at a
height L‘h"'=‘1f'i/n abové the groimd. Determine (a) the angle 0 at which it
will leave the cylindrical surface BCD, (b) the distance x at which it will
hit the grownNeglectfrictionand resistance. 7 7 SOLUTION
(a) Block leaves surface at C when the, normal force, N = 0 i V ' mgcosH = man 2 v
cosB = —C—
g h v3 = ghcosB = gy Workenergy principle
1 2 TB =§mvc UB—C =W(h‘y)=mg(h—yc) TB + UB—C = Tc Use Equation (1) 4.5m + mg(h  y) = émvé 4.5 + g(h — y) gyC l
2
3 4.5 + gh = Egyc
(4.5 + gh)
J’C = "'—
89 A
(4.5 + (9.81)(1)2) y =
3
E(9.81) y = 0.97248 m PROBLEM 1%;453ﬁiJEE " 17c ’='hc‘o‘sl9' “hose: 19': 0'37248‘2' 0.97248
0 = cos10.97248 =13.473° , a =,1'3.47°< ‘ (b) From Equations (1) and (3), , ,
vci = J3 = 9.8:1ﬁ({0.97248) = 3.0887 m/s .f' At C; (vc)x = vC c056 = 3.0887 cos1’3y.47° = 3.0037 m/s (vC)y 7': —vcsin_0 = 3.0887sin13.47° = ,—0.71947 m/s y = yC + (vC)yt — 1;.th = 0.97248 — 0.71947r — —:—(9.81)t2 At E: yE = 0: 4.905t2 + 0.7194t — 0.97248 = 0 ,
t = 0.37793 5 At E: V ‘ ' x1 “="hcos0' +‘(isc')x t' “=’ 1(sin13.47°) + 3.0037(0.37793)
= 0.33894 51.3519; 1.3681 m
x = 1.368 m 4 PRLEM 13.55 . A force P is slowly applied to a plate that is attached to two springs and
causes a deﬂectiOn x0. In, each of the two cases shown, derive an
expression for the constant ke, in terms of k1 and k2, of the single
spring equivalent to the given system, that is, of the single spring which
will undergo the same deﬂection x0 when subjected to the same force P. SOLUTION System is in equilibrium in deﬂected x0 position. Case (a) Force both springs is the same = P Case (b) Deﬂection in both springs is the same = x0
W104 P = k‘xo + kzxo P = kexo
Equating the two expressions for ke =kl+k2‘ PROBLEM 5 113.68
A l.2—kg collarcan slide along the rod shown. It is attached to an elastic
cord anchored at F, which has an undeformed length of 300 mm and a spring constant of 70 N/m. Knowing that the collar is released from rest at A and neglecting friction, determine the speed of the collar (a) at B,
(b) at E. LAF = (0.5)2’+ (0.4)2 + (0.3)2
LAF = 0.707 m
LBF = (0.4)2 + (0.3)?
LB'F = 0.5 m
LFE = (0.5)2 + (0.3)2
LFE = 0.5831m
V=n+g
(a) Speed atB vA = 0, TA = 0 at pointA
(me = %k(ALAF)2 ALAF = LAF — L0 = 0.707 — 0.3
ALAF = 0.407m
(VAL = (me = 5.7977 J _1_ 2 (70 N/m)(0.407 111)2 (mg = (W)(0.4 m) = (1.2 kg)(9.81m/sz)(0.4 m) = 4.71 J VA = (me + (mg = 5.80 + 4.71 = 10.51 J .1. 1 2
T =mv =
.B 2 B 2 (1.2 kg)v§ = 0.6v12, = §k(ALBF)2 ALBF = LBF — L0 = 0.5 m — 0.3m ALBF = 0.2m (VB)e = %(70 N/m)(0.2 m)2 = 1.4 J PROBLEM 13.68, CONTINUED
(VB)g = (W)(0.4) = (1.2 kg)(9.81 n1/52)(0.4 m) = 4.71 J VB =(VB)'e+(I_/B)g>='1.4J + 4.71J = 6.11J, TA + VA = TB + VB: 0 +1051 = 0.6v}3 + 6.11 VB =_2.71m/s”4
(b) Speed atE ‘
PointA _ TA = 0, VA =10.51J (from Part (a)) Point E
1 TE = gmva = 3(12 kg)v§ = 0.611%; (VE)e = %k(ALFE)2 ALFE = LFE — L0 = 0.5831m — 0.3m (VE)e = %(70 N/m)(0.2831m)2 = 2.805 J (VE)g = 0 VE = 2.805] TA + VA = TE + VE; 0 + 10.51 = 0m}; + 2.805 .796
v}; = (7 ) = 12.9933 m2/52 (0.6) VE = m/S ‘ ...
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 Summer '06
 
 Force, Friction, Trigraph, kg, Belt, Conveyor belt

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