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Unformatted text preview: ENGINEERING MECHANICS ~ Dynamics W.F. Riley & L.D. Sturges 16—5 A 450—lb crate 18 in. wide by 45 in. tall is sitting on the back of a truck as shown. The truck is traveling at 65 mi/h when the driver suddenly applies the brakes. Determine the minimum distance required to
stop the truck (without causing the crate to tip over) and the minimum static coefficient of friction required between the crate and the truck to prevent the crate from slipping. Solution The crate is in translation with ay = a = 0. The equations of motion are +_) 217 _F __..4_§_0_ a
32.2 X N450 o
N(9a) — 22.517 0 At the point of impending tip, the back corner of the crate begins to lift
away from the bed of the truck and the normal force resultant acts at the
front corner (a = 0). At this point N = 450 lb = constant
F = 9N/22.5 = 180 1b = constant
ax = 0.40g = 12.88 ft/s2 = constant
and the minimum required coefficient of friction is
u F/N = 9/22.5 = 0.4 Since a relationship between velocity and position is desired, rewrite
the acceleration using the chain rule of differentiation and integrate I v dv I a dx = 12.88 I dx 2
V /2 = 12.88x + Cll = 4544.222  12.88x where the constant of integration is has been chosen such that V = 65 mi/h
= 95.33 ft/s when x = 0. Then the truck will come to a stop (v = 0) when V = O ; 4544.222 — 12.88x
when ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1627* The 201b slender bar AB shown is supported by a frictionless
pin at A and a cord at B. If the cord at B
breaks, determine the acceleration of the mass center of the bar and the reaction at support A at the instant that motion begins. Solution
The bar rotates about a fixed
axis through A. The equations of motion are 20
+9 2 = = ——————
Fx Ax 32.2 aGX
20
+¢ = _ = ______
_ Egy 20 Ay 32.2 aGy
= . = a
C+ EMA _1 5(20) 1; where w = 0 (starts from rest) 2 2
a = —1.5w = O ft/s
Gx
— 1 5a ft/s2
aqy — .
20 2
I = —l—(—————— (3) 1 A 3 32.2
2
1.86335 slugft Therefore
2
a = 16.100 rad/s D 2
24.150 ft/s l $1)
I! Gy
— 2
ac = 24.2 ft/s l . . . . . . . . . . ... . . . . . . . . . . . ........... Ans.
A = 0 lb
X 3
II
U1
5
_) ENGINEERING MECHANICS  Dynamics 16—35 A 300lb force is applied
to the end of a rope which is
wrapped around the outside of
a 6ft diameter cable drum as shown in Fig. a. a. If the 600—lb cable drum can
be treated as a uniform
cylinder,determine the
angular acceleration of the
cable drum and the time
required to rotate the cable
drum 3 revolutions. b. Repeat part a if the 300—lb
force is replaced with a
300—lb block as in Fig. b. Solution The wheel rotates about a fixed
axis through its mass center G while
the block moves in a straight vertical
line. Therefore, separate free—body
diagrams must be drawn of the two
separate objects. The equations of
motion of the drum and block are C+ ZMb 3T 16g 300
32.2 900 2
[ Jm +¢ SF 300 — T = a Y and l 2 32.2 2
125.7764 slugft a. If the to the drum, then T = 300 lb and Eq. a 2
a 7.16 rad/s D b. If the block is attached to the
then solving Eqs.
245 lb 2
5.85 rad/s D T a W.F. Riley & L.D. Sturges 3UHb (M 300—lb force is applied directly gives drum, a and b together gives ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. sturges 1644 A SO—kg rectangular plate of uniform thickness is supported by a
smooth pin and a cable as shown. If the cable at B breaks, determine the angular velocity of the plate and the magnitude of the force exerted on the plate by the pin at support A a. When the mass center of the plate is
directly below A. b. When side AB of the plate is vertical
(corner B is directly below A).
Solution
The mass center of the plate travels
a circular path about the fixed axis of rotation A. Therefore ﬂ=raA+UA
aG AG et en where rAG = 279.509 mm, ¢ = 26.5650, and the equations of motion are +F ZFt = 50(9.81) cos 6 — At = 50(o.27951 or) 50638
2
+5. 217'“ = An  50(9.81) sin 9 = 50(o.27951 (a)
+1) 214 = 50(9.81)(0.27951 cos 6) = I or
A A 'QV‘
where r
2 2 2 2
IA = I%(50)[0.250 + 0.500 ] + 50[0.125 + 0.250 ]
2
= 5.20833 kg°m
The moment equation can be solved for the A\
angular acceleration )6
a = 26.32313 cos 9 0L f‘
G .—
Rewriting the angular acceleration using ‘R<\
the chain rule of differentiation Y;GPL Y. Us;
Ac.
a_ do _ dw d6 _w do)
' dt ' d6 dt ‘ ale and integrating gives (Problem 16—44 continues ...) ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges (Problem 1644 — cont.) I a d9 = I 26.32313 cos 6 d9 I... w2/2 = 26.32313 sin 9 — 11.77204 where the constant of integration has been 0
chosen so that w = 0 when 9 = ¢ = 26.565 ° 2
a. When 9 = 90 , a = O rad/s D m = 5.39 rad/s D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans.
A = 897 N
n
A =
t 0 N
2 2
A = A + A = 897 N . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans.
n t O 0
b. When 9 = 90 + ¢ = 116.565 , a = —11.77204 rad/S2 w 4.85 rad/s D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. 767.759 N In
ll At = 54.838 N A = 770 N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1654* Two 400mm diameter disks and a 240mm diameter disk are joined
to form a spool as shown. The 125 kg spool has a radius of gyration of
125 mm with respect to the axis of the spool. If a force p of 500 N is
applied to the spool through a cord that is wrapped around the cylinder, determine the acceleration 56 of the mass
center of the spool and the angular
acceleration a of the spool if a. The horizontal surface is
smooth (u = O). b. The horizontal surface is
rough (u = 0.25).
Solution The equations of motion for the
spool are +—) 2F = 500 — c =
X f 125aG
+T 2F = c  125(9.81) = o
y 12
(1+ ZMG = 0.120(500) + 0.2oocf = 160: K.
where a = 0 since the mass center of ‘ Gn (if the spool has no motion normal to the surface, c 2 2 'n IG = 125(O.125) = 1.95313 kg°m a. If the surface is smooth (Cf = 0) then
2 a6 = 4 m/s % .. . . . . .................. . . . . . . . . . ....... Ans.
2 a = 30.7 rad/s D ..... . . . . . ............... . . . . . . . . .. Ans. b. If the surface is rough (u = 0.25) then the spool may roll without f
Guessing that the wheel will roll without slipping slipping (a6 = 0.2a; C S uCn) or it may slip (Cf = uch; aG ¢ 0.2a). C = 500  125(O.2a) = 9.76563a  300 f 2 a = 23.0 rad/s D ................................... Ans.
2 a6 = 4.60 m/s % ..... . . . . . . . . . . .................... Ans. an = 1226.25 N 1‘ cf = 75.28 N = 75.28 N —) Note that the amount of friction required to prevent slip (75.28 N) is less than the amount of friction available (0.25Cn = 306.6 N). Therefore, the assumption of rolling without slip is valid. ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 1673 A 16lb ladder 12 ft long slides in a smooth corner. The ladder is initially at rest 0
with 9 = 0 when the lower end is disturbed slightly. Determine the angle 9, the angular
velocity 0, and the velocity 93 of end B when end A loses contact with the vertical wall. Solution The equations of motion of the ladder are +9 2F = A = ma x Gx
+T 2F = B  m = ma Y 9 Gy
(2+ ZIMG = B(L/2) sin 6  A(L/2) cos 6 = I60:
where 2
IG = mL /12 To get the relationship between the accelerations a , a , and a, must do
Gx Gy the kinematics of the bar. First, using the known information about the Es
motion of the ends of the bar I\
a = a + a
as 3!! aB/A I“
A _ A 6 A _ e A x
a3 1 — aA J + La(cos 1 + sm J) a
2 A A
+Lw(—sin91+cosej) 8
get 2
aA = La sin 9 — Lw cos 9 t:\
Then,
a =  + a
ac; aA aG/A = (La sin 6 + sz cos 9) j + (L/2)a(cos 9 i + sin 9 j)
+ (L/2)w2(—sin 9 i + cos 6 3) (Problem 16—73 continues ...) ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges (Problem 16—73 — cont.) Therefore,
2 .
aGX  (L/2)(a cos 6  w Sin 9)
aGy = (L/2)(a sin 9 + w2 cos 6) A = (mL/2)(a cos 6 — wz sin 6)
B = mg  (mL/2)(a sin 6 + oz cos 6) and the moment equation can be simplified to 2Lu = 39 sin 6 Rewriting the angular acceleration using the Chain Rule of Differentiation
do do d9 do
a=—=——=w—
dt d9 dt d9 and integrating to get the angular velocity gives I 2L0 dw I 2La d9 = I 39 sin 9 d9 2
L0 39(1  cos 6) where the constant of integration has been chosen so
that w = 0 when 6 = 0. Then, A will become zero when
A = m(L/2)(a cos 9 —‘w2 sin 6) = O
2La cos 9 = 2sz sin 9 3g sin 9 cos 9 = 69(1  cos 9) sin 6 II
N 3 cos 9
9 = 48.190 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans.
(independent of m and L!) and at this angle
a = 3.048 rad/s2 C
m = 1.651 rad/s c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans.
B = 3.81 lb T and the relative velocity equation gives _. —. —. = +
VB VA VB/A
A = A + 6 A + ' 6 A
VB 1 VA J Lw(cos 1 Sin J)
v = 13.21 ft/s 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans. ...
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