PROBLEM 16.1 Two identical 0.4-kg slender rods AB and BC are welded together to form an L-shaped assembly. The assembly is guided by two small wheels that roll freely in inclined parallel slots cut in a vertical plate. Knowing that G = 30°, determine (a) the acceleration of the assembly, (b) the reactions at A and C.
SOLUTION ' f
T
ma
0-8?-
OC. _______ A-. _______ R* / h . — us- n
/
^ iiSi» m ZFt = ma
LFX = (0.8 kg)(9.81 m/s2)(cos60°) = (0.8 kg)a
(a)
a
=
* 2 a = 4. 4.91 m /s2 /s2 ^
30° A
+ ) ZM C = (-0.8 kg )(g)(4 5 mm) + /? /?/)(0.866)(l80 /)(0.866)(l80 mm) - /?^(0.5)(l80 /?^(0.5)(l80 mm)
(b)
= (-0.8 kg )(a)(-0.866)(l35 mm) - (0.8 (0.8 kg)(a)(0.5)(45 mm) /?,, (65.88 (65.88 mm) = -(39.52 8 kg m m )a -(39.528 kg-mm)[ kg-mm)[
M
m/s2 j
R.4 R.4 = ------------ 7 (65.88 mm)
•m/s" = -2- 943 kg •m/s" or R . = 2.94 2.94 N 7
60° A
or R f = 9.74 N
60° A
IF,. = R a + Rc - (0.8g)cos30° = 0 or
RA + Rc = 6.7964 Re = 6.7965 + 2.943 = 9.7395 N
PROBLEM 16.2 Two identical 0.4-kg slender rods AB and BC are welded together to form an L-shaped assembly. The assembly is guided by two small wheels that roll freely in inclined parallel slots cut in a vertical plate. Determine at A is zero, (b) the (a) the angle of inclination 0 for which the reaction at corresponding acceleration of the assembly.
SOLUTION See free body diagram for PI6.1 £F r = 0.8 # sin # = 0.8r/ 0.8r/
a = g s i n #
+) IA/(; = 0 = R( cos#(45 mm) - Rc sin#(135 mm)
(a)
tan # = -
(b)
a = 9.81 m/s" (sin 18.43°) = 3.10 m/s2
or
# = 18.4 18.43° 3° A
or a = 3.1 3.10 0 m/s2 m/s2 ^
18.43° <
PROBLEM 16.2 Two identical 0.4-kg slender rods AB and BC are welded together to form an L-shaped assembly. The assembly is guided by two small wheels that roll freely in inclined parallel slots cut in a vertical plate. Determine at A is zero, (b) the (a) the angle of inclination 0 for which the reaction at corresponding acceleration of the assembly.
SOLUTION See free body diagram for PI6.1 £F r = 0.8 # sin # = 0.8r/ 0.8r/
a = g s i n #
+) IA/(; = 0 = R( cos#(45 mm) - Rc sin#(135 mm)
(a)
tan # = -
(b)
a = 9.81 m/s" (sin 18.43°) = 3.10 m/s2
or
# = 18.4 18.43° 3° A
or a = 3.1 3.10 0 m/s2 m/s2 ^
18.43° <
PROBLEM 16.3 A 60-lb uniform thin panel is placed in a truck with end A resting on a rough horizontal surface and end B supported by a smooth vertical surface. surface. Knowing that the deceleration deceleration of o f the truck is 12 ft/s2, determine (a) the reactions at ends A and B, ( b) the minimum required coefficient of static friction at end A.
SOLUTION S
NS
m a
f e d
ss
(u )
3
t N f y -
(lm
+ ) I M a = N b ( 5 ft)(0.866) ft)(0.8 66) - (60 lb)(2. lb )(2.5 5 ft)(0.5 ft)( 0.5))
(12 ft/s2 )(0.866) -(. -( .60 60 lb) l b) (12 ' 32.2 ft/ ft/'s 's2 2' N b = 28.501 lb
J i - ZFX ft/s2) ZFX = N B - F = 60 lb, (l2 ft/s2) B ’ 32 ft/s2 ft/ s2 V F = 6.1404 lb 0)
R a = N / n 2 a+ F 2 = 60.3133 lb
or R a = 60.31 lb a = tan~' 603133 = 6.1404
84.2° <
34. 18° and N„ = 28.5 28.5 lb — A
(b)
6AAM
N a
10234
60 or n = 0.1023 <
PROBLEM 16.4 A 60-lb uniform thin panel is placed in a truck with end A resting on a rough horizontal surface and end B supported by a smooth vertical surface. Knowing that the panel remains in the position shown, determine (a) the maximum allowable acceleration of the truck, (b) the corresponding minimum required coefficient of static friction at end A.
SOLUTION
yr) 0 . ^
—>
—
3 2-2.
'
l< \)
1
a = ( 6 0 l b )( 0 Z M a )( 2 . 5 f t ) ( 0 . 5 ) = ^
^
(2.5 f t ) (s i n 60 60 °) °)
a = 18.5907 fUs‘
or a = 18.59 18.59 ft/s2 — <
(a)
(b)
A
a i . i - . J 2 a 7 (l8.5912 ft/s2) = 34.6419 34.6419 lb lb 32.2 ft/s F _ 34.64191b ~ Na ~
60 lb
= 0.5773 or n = 0.577 <
PROBLEM 16.5 Knowing that the coefficient of static friction between the tires and the road is 0.80 for the automobile shown, determine the maximum possible acceleration on a level road, assuming (a) four-wheel drive, ( b ) rearwheel drive, (c) front-wheel drive. I-
1.5 hi
-k
-I
I in
SOLUTION (a) Four-wh eel dr ive : vv = w 3 i Vv -1
sTyiOL
£ F - #
\ / / a 1
n -f b
+ 1 IF,, = 0: N a + N B - W = 0 Thus:
N a + N b = W = mg
F a + F b = /.ikN A + n kN B = n k { N A + N B) = n kW = 0.80mg
I F r = l ( F j c : FA + FB = ma 0.80mg = ma a = 0.80g = 0.80(9.81 m/s2) = 7.848 m/s2 d = 7.85 m/s2 < (b)
Rear-whe el drive:
L g y ± L ( ? y > if A +) I M b = l( A /fl)e(r: (1 m ) W - (1.5 m)/V4 = -(0 .5 m )m d N a = 0.41V + 0.2 md
Thus:
F a = n kN B = 0.80(0.40'' + 0.2 ma) = 0.32 mg + 0.16 ma
- ± .I F t = l( F r)en.: F a = ma 0.32 mg + 0.16 ma = ma 0 3 2 g = 0.84a
_ 0.32 a = — (9.81 m/s2) = 3.7371 m/s2 0.84' ' or a = 3.74 m/s 2 A
PRO BLEM 16.5 CONTINUED (c) Front-wheel drive : •
f
e
. *
I g
p
< L 0 _ f —
@
3 _ ^ » r
I •S *»i
'1 *»
+) I M a = Z ( M A)e(r: (2.5 m ) N B - (1.5 m ) ^ = -(0.5 m)ma N„ = 0.61V - 0 .2md
Thus:
Fh = it kN B = 0.80(0.6^ - 0.2md) = 0.48m# - 0.16md
I F t = £ (F v)cfr: FB = m d 0.48m# - 0.16md = md 0.48g = 1.16a a = —
'
(9.81 m/s2) = 4.0593 m/s2
1.16'
or a = 4.06 m/s2 —* A
PROBLEM 16.6 For the truck of Sample Prob. 16.1, determine the distance through which the truck will skid if (a) the rear-wheel brakes fail to operate, ( b ) the front-wheel brakes fail to operate.
SOLUTION (a) I f rear-wheel brakes fail to operate :
/mg. ]
(2
f a ); { h I
(\
M 7f t
■<*S
4 f t >
Ft t/a
S f £ N „(\2 ft) - W (5 ft) = md(4 ft)
+) Z.M , =
N„ = L W + - — a 3 X * 12 = ^(^)cr
tf _ FH = ma' t‘kN H = — a
( 5 W | W ^ 0.699 \ — W + - — a = — a 1 12 3 g ) g
a =
° '6" ( ^ ) ( 32'2 1 - 0.233
a = 12.227 ft/s- -
Uniformly accelerated motion
v2 = v02 + la x
0 = (30 ft/s2) - 2(12.227 ft/s2).t x = 36.8 ft <
(b)
If fro nt- wh eel brakes fail to operate:
H - f t
u(/ct t~A &
I s-rt
1 f t
+) IM h = S (M fi)eff:
W (1 ft) - N A(12 ft) = m d ( 4 ft)
N a = — W - - — d 12 3 g
PROBLEM 16.6 CONTINUED w . * F X =
s ( F * ) e ff :
f a
=
m a ,
f ik N A
=
—
a
g
0.699
7 07 i.f/ 12
0.699 0 2 . 2 a -
1+ 0.233
1W~ a 3 g
-----------
W . i
---
g
» s!) 0 = 10.648 ft/s2
Uniformly accelerated motion v2=vl+lax
0 = (30 ft/s2 ) - 2 ( l0.648 ft/s2)x x = 42.3 ft <
PROBLEM 16.7 L
A 50-lb cabinet is mounted on casters that allow it to move freely (/ / = 0) on the floor. I f a 25-lb force is applied as shown, determine (a) the acceleration of the cabinet, (b) the range of values of h for which the cabinet will not tip.
G *
Il>
|
»
e
m i*
1,
|
:
■
LJ
.36 in.
i
SOLUTION
(a)
Acceleration - i . I F , = I ( f T)c„ :
25 lb = ma 25 lb =
501b _ r-o 32 ft/s a = 16.10 ft/s" <
(b)
For tipping to impend J);
A = 0 <
Z M B = l { M B)c n :
(25 lb)/; - (50 lb)( 12 in.) = m a(36 in.) 25/j = 600 • (2 5)( 36 ) For tipping to impend *);
h = 60 in.
5 = 0 + C I A / , = I ( M , ) c(T:
(25 lb)/; + (50 lb)(l2 in.) = ma(36)
or
h = 12 in.
cabinet will not tip for 12 in. < h < 60 in. -4
PROBLEM 16.8 Solve Prob. 16.7 assuming that the casters are locked and slide on the rough floor {/ js = 0.25).
SOLUTION Acceleration
(a)
+1
= 0
N ., + N „ - W = 0 N a + N h = 50 lb
w /. S o | b
h 4_
TT, ^ But F = / iN , Thus
G
\ G
z s lk
Fa + F* = A (50 lb), e 1 * © F»-
6
Fs - 0.25
a
— 2/rv = S (F v)cn.:
a j
k
l a i n .
25 lb - (F , + Fh ) = ma
12 i n
25 lb - [(0.25)(50 lb)] =
50 lb 32.2 ft/s-
U a
a = 8.05 ft/s(b) Tipping
For tipping to impend J : N , = 0 + C
c •
A
Soli G
u
&
t" r
hl | I n ■h **t 4 li in ^ 12 in.
■ma
r a
a
= I(M b)m :
(25 lb)/; - (50 Ib)(l2 in.) = —‘—
■, (8 05 ft/s: )(36 in.)
/; = 42 in. For tipping to impend ]): N H = 0 + ( SA/, = S( A /, )c1T: (25 lb) + (50 lb)(l2 in.) = - J i ^ j g . O S ft/s: )(36 in.) /; = -6 in.
impossible cabinet will not tip if /; < 42 in. A
PROBLEM 16.9 The support bracket shown is used to transport a cylindrical can from one elevation to another. Knowing that p y = 0.30 between the can and the bracket, determine («) the magnitu de of the upward acceleration a for which the can will slide on the bracket, (b) the smallest ratio hid for which the can will tip before it slides.
SOLUTION (a)
Sliding impends
O-
IffG
r
&
—
!
I H ±_ IF,. = l( F r )cfr: F = ww cos 30° + | IF,. = l ( F , )
: N - mg = ///usin30° N = m {g + «sin30°)
F P. = — ; N
0.25 =
h/£/cos 30 °
. , g + a sin 30° = 3.33 33« cos3 0° m [g + a sin 30°) a = 0. 4 19g
g = 2.3867u (b)
30° <
Tipping impends
'ma-
^ IA /c = l(A#c )eff: F
' hy
a ,
- N
' d \ _
a )~
F __ d_
N ~ h p = — ; N
0.3 =
h
- = 3.33 < d
PROBLEM 16.10 Solve Prob. 16.9 assuming that the acceleration a o f the bracket is downward.
SOLUTION (a) Sliding impends’ . -— — -j
(T
- i - E Fx =Z{F x)e S: F = ma cos30°
+ } E Fy = E(/^)eff: N - m g = -masin30° TV= m (g - a sin 30°)
F a, = — ; N
maeos30° m „ 0.3 = —7---------------- m [ g - asm30°)
g - asin30° = 3.3333acos30°
— = --------------------- g 3.3333cos30° + sin30°
=
0.29527
a = 0.295 P7 <
(b) Tipping impends W O-
' h N
f d ' = W ,2, ’ ,2 ,
F _ d N ~ h p = — ; N
0.30 = - ; h
- = 3.33 < d
PROBLEM 16.11 An 8-lb uniform slender rod AB is held in position by two ropes and the link CA which has a negligible weight. After rope BD is cut the assembly rotates in a vertical plane under the combined effect of gravity and a 4 lb ft couple M applied to link CA as shown. Determine, immediately after rope BD has been cut, (a) the acceleration of rod AB, (b) the tension in rope EB.
PROBLEM 16.12 (T(yf
O'
1.5ft
60^
SO/
A -2 ft-
,
g
An 8-lb uniform slender rod AB is held in position by two ropes and the link CA which has a negligible weight. After rope BD is cut the assembly rotates in a vertical plane under the combined effect of gravity and a couple M applied to link CA as shown. Knowing that the magnitude of ”1 the acceleration of rod AB is 12 ft/s immediately after rope BD has been cut, determine (a) the magnitude of the couple M, ( b) the tension in rope EB.
PROBLEM 16.13 A uniform circular plate of mass 6 kg is attached to two links AC and BD of the same length. Knowing that the plate is released from rest in the positi on shown, in which lines joini ng G to A and B are, respectively, horizontal and vertical, determine (a) the acceleration of the plate, (b ) the tension in each link.
SOLUTION
+/ Z F = ZFC|1: /wg cos 75° = mi7 a = gcos75° = 2.5390 m/s*
a = 2.54 m/s2 7 ^ 15° A (b ) Tension in each link +*) ZAf/, = Z
(F ,c o s7 5 ° )r + (F , sin 75 °)r = (/mm sin 75° )r Fi(cos75° + sin75°) = (6 kg)(2.539 m/s2)sin75° F a = 12.0146
F , = 12.01 N Tension A
m : F, + Ftt - M/g sin 75° = 0 75° Z F = Z F
6.009 lb + F„ -(6kg)(9.81 m/s2)sin75° = t Fl{ = 44.839 N
Fm - ^4.8 N Tension <
PROBLEM 16.14 Bars AB and BE, each of mass 4 kg, are welded together and are pinconnected to two links AC and BD. Knowing that the assembly is released from rest in the position shown and neglecting the masses of the links, determine (a) the acceleration of the assembly, ( b ) the forces in the links.
SOLUTION A-
IA/(; = 0 => Fa = 0
/
/
F a /
/
V 1
\ + IF iflO = (78.48N)(0.5) = (8k g)(fl)
/ ? NN r n a - 3 ^
a = 4.905 m/s2
or
I ' A
+/
a - — 2
= f b ~ ( 78-48 N) (0 .866 ) = 0 F„ = 67.966 N
(«) (b)
a = 4 .9 1m/s2
30° <
F a = 0 , FB = 68.0 N compression A
7
_g'V
I—inM-4———
PROBLEM 16.15
"JJflP [7
Cranks BE and CF, each of length 15 in., are made to rotate at a constant speed of 90 rpm counterclockwise. For the position shown, and knowing that P = 0, determine the vertical components of the forces exerted on the 15-lb uniform rod ABC D by pins B and C.
•j- )■>■»— 1
*iw—
SOLUTION a> = 90 rpm
60
| = 3x rad/s
Bar AD in translation a = a B = a c = rco2 = (15 in.) — (3* rad/s)2 = 111.03 ft/s2
\ 12 J
A
60°
J yyfl CX.
LtJ
G
£
8 & A & ° — *------------- r
C.
o .
3^
0 , 2 ^
+) Z Ma = Z(A/c )e|T: C,( 1.5 in.) - B,(1.5 in.): By = C ,
+ | Z Fv = £(/\.)
Bv + Cy - mg = ma sin 60°
2 B v - (l 15 lb x 32.2 ft/s2) = (l 11.03 ft/s2) sin 60° B v = +29.897 lb
or By = Cy = 29.9 lb A
PROBLEM 16.16 nSHp ' 7
At the instant shown the angular velocity of links BE and CF is 6 rad/s i counterclockwise and is decreasing at the rate of 12 rad/s2. Knowing that the length o f each link is 15 in. and neglecting the weight o f the links, determine (a) the force P, ( b) the corresponding force in each link. The weight of rod AD is 15 lb.
1—■»!».— I
SOLUTION Links: an = rco2 = (1.25 ft)(6rad/s)‘ = 45 rad/s2
eg- 6
at = ra = (1.25 ft)(l2 rad/s2 j = 15 rad/s2
Bar AD is in translation
a - aB - a c
f & £
f c .
\ 1<~r
3
- IS——in
£ 6
I* 15 —
+ Q EMG = E (M G)eff: FCFsin60 °(l5 in.) - T’B£sin6 0°(l5 in.) be F ce = F
F cf sin60°(30 in.) - (15 lb)(l5 in.) = mat sin 30°(15 in.) + man sin60° (l5 in.)
F cf (25.981 in.) - 225 lb in. =
(
15 lb
^ j (15 in.)(l5 rad/s2 sin30° + 45 rad/s2 sin 60°) 32.2 ft/s
F cf = 21.15861b be = 21.2 lb T < or F cf = F
PROBLEM 16.16 CONTINUED be + F cf ) cos 60° + P = -ma, cos 30° + man cos 60° +( IF . = l ( F x)ef| : {F
21.15861b + P
151b 32.2
ft/s2
^ 4 — cos 30° + 45 rad/s2 cos 60°
21.15861b + P = 4.42991b P = -16.72871b
or P = 16.731b-— A
PROBLEM 16.17 Draw the shear and bending-moment diagrams for each of the bars AB and BE of Prob. 16.14.
SOLUTION
1
a
V n
T ma„ =
'A
y 0.5 y
= ijigu
= 0; - V = man - V = 33.983 u,
at
u = 0.5 m V = - 1 6 . 9 9 N ([+ SM - M = man
M = 3 3 . 9 8 3 — = 1 6. 99 1m 2 2
at
u = 0.5 m, M = 4.25 N m SbeAY
rn
eht
PROBLEM 16.17 CONTINUED
/ *
ma„
0.5
A IF„ -
- 2yf3gu A
2 .
-V =
(2j 3g)u
V = 2>/3 gu
+ ( EM - M + ( 8 « g x ) f = 2A ? « 2/2
m = -V 3gw 2 /Y fo m n tr
shear
V A'
(V'B -*<•25 Of-n;
PROBLEM 16.18 Draw the shear and bending-moment diagrams for the horizontal rod ABCD of Prob. 16 .15.
SOLUTION a = 111.03 ft/s'
.r
/ J S i t —
2? 9 £*>
2
Bx = Cx = 29.9 lb r ) ll II --3
S / i 30 /5 -0
jc ---- 3j>----- 1 = 0 S f S S Z U/iH
'" i
v
3
15 1 ^ / s C ivi - O h U/iv) 29-9 /0 t-H
, 29-9 1
2,0 i.M
----------
■4___ 1___ it____it___ ’1
.1
Y
V
S . M r
i— ZD r 1/
0 ' i 9 S 3 2 /b/'-" 7
/ “f
PROBLEM 16.19 For a rigid slab in translation, show that the system of the effective forces consists of vectors ( Am, ) a attached to the various particles of the slab, where a is the acceleration of the mass center G of the slab. Further show, by computing their sum and the sum of their moments about G, that the effective forces reduce to a single vector ma attached at G.
SOLUTION Since slab is in translation, each particle has same acceleration as G, namely a. The effective forces consist of (Am/) a.
The sum of these vectors is:
I (Am,) a = ( I Am,) a
or since I Aw, = w, l(Am,)a = ma The sum of the moments about G is:
I rj x (Am, )a = ( I Am,/;') x a
(I )
But, IAm,/jf = nir = 0, because G is the mass center. It follows that the right-hand me mber of Eq. (1) is zero. Thus, the moment about G of ma must also be zero, which means that its line of action passes through G and that it may be attach ed at G.
PROBLEM 16.20
( A m M a x r ',)
For a rigid slab in centroidal rotation, show that the system of the effective forces consists of vectors -(A m ,)m V , and (Am,) ( a x r ', ) attached to the various particles P, of the slab, where co and a are the angular velocity and angular acceleration of the slab, and where r', denotes the position of the particle Pt relative to the mass center G o f the slab. Further show, by computing their sum and the sum of their moments about G, that the effective forces reduce to a couple l a .
SOLUTION For centroidal rotation: a, = (a,-) + (a y) = a x r/ - co2v\ Effective forces are:(Am ,)a, = ( Am ,)(a x r,) - (Am,)
1<*
£(Am,)a,- = £ (A m ,)(a x r/) - l(A m ,)m 2^' = a x l( A m , )r/ -
E(A m, )r/ = 0
effective forces reduce to a couple, Summiml moments about G l( r, ' x Am,a,) = E r/ x (Am, ) |a x r /j
- Er/ x (Ami)o)2r'i
But,
r/ x (Am,.)m2r/ = co2(Am,)(r; x r') = 0
and,
r' x (Am,)(a x rfj = (Am,-)r,,2a
Thus,
E(r,' x Am,a,) = E (Am,) r2a = [^E(Am,.)r'2J a
Since the moment of the couple is / a
E(Am )rJ2= / ,
PRO BLEM 16.21 It takes 10 min for a 2.4-Mg flywheel to coast to rest from an angular velocity o f 300 rpm. K nowing that the radius o f gyration o f the flywheel is 1 m, determine the average magnitude of the couple due to kinetic friction in the bearing.
SOLUTION / = m k 2 = 2400 kg(l m)2 = 2400 kg m2 (2n\
<3)n = 300 rpm — v 60
J
= 10/r rad/s
co - g)0 + at;
0 = 10^ rad/s + a ( 60 0 s) a = -0.05236 rad/s2 M = 7a = (2400 kg m2)(-0.05236 rad/s2) = 125.66 N m
or M = 125.7 N-m A
PROBLEM 16.22 The rotor of an electric motor has an angular velocity of 3600 rpm when the load and power are cut off. The 120-lb rotor, which has a centroidal radius of gyration of 9 in., then coasts to rest. Knowing that kinetic friction results in a couple o f magnitude 2.5 lb •ft exerted on the rotor, determine the number of revolutions that the rotor executes before coming to rest.
SOLUTION
V32.2 ft/s2/v «. , M = la ;
2.5 lb ft = (2.0963 lb ft-s2)a
a = 1.1926 rad/s" (deceleration ) (2n\
a>0 = 3600 rpm — 2 j O
o> q
= 120;r rad/s
+ 2a 0;
6 = 59,585.07 radf 1 r6V 1 = 9,483.26 rev
or
9 = 9480 rev A
PROBLEM 16.23 A 20-lb uniform disk is placed in contact with an inclined surface and constant 7.5 lb ft couple M is app lied as shown. The w eigh t o f the lin] AB is negligible. Knowing that the coefficient of kinetic friction at D i 0.4, determine (a) the angular acceleration o f the disk, (b) the force in th link AB.
SOLUTION W = 2<0 ft 7 - S
U
f t
0,
+)
201b ,32 .2 ft/s2
9_ 12
ft j a = (0.17469 lb ft-s2) a
N = 7.5 lb ft - (0.4 N lb) A f t 12
1
201b
2 32.2 ft/s2
f 9 f t x2 12
a
- j T,Fy = 0.4 N cos30° + Aco s60° - 20 lb = 0; N = 23.629 lb
7.5 lb-ft - (0.4)(23.629 lb)| ^
ft
a =
0.4112 lb-ft 0.17469 lb-ft-s2
2.3539 rad/s2
a = 2.35 rad/s2 J) A
(a)
1FX = C - (0.866)A + 0.4A(0.5) - 0 C = 0.6660 N = (0.6660)(23.629 lb)
= 15.7369 lb (b) or
C - 15.74 lb (compression)
^
PROBLEM 16.24 A y r
n ■
,
9 hi.* “
A 20-lb uniform disk is placed in contact with an inclined surface and a constant couple M is applied as shown. The weight of the uniform link AB is 10 lb and the coefficient of kinetic friction at D is 0.25. Knowing that the angular acceleration o f the disk is 16 rad/s* clockwise, determine (a) the magnitude o f the couple M, (b) the force exerted on the disk at D.
/ O f
:«) y
SOLUTION
IC/L B
C
Ik
/a -If 20 lb 2132.2 ft/s+) ZA/„ = M - (0.25ATlb) ^
ft j = 2.7950 lb-ft
+ | ZF V = (0.25 N)(0 .866 ) + 7V(0.5) - 20 lb - 5 lb = 0 IV = 34.8915 lb
(a) Now
M = 2.7950 lb-ft + (34.8915 l b ) ^ ft j
= 9.3372 lb ft or M = 9.34 lb-ft A P = tan -i
0.25
\ 1.0
= 14.04°
F = J n 2 + (0.25 N ) 2 = 35.965 lb
or F = 36.0 lb ^
44° <
PROBLEM 16.25 The 160-mm-radius brake drum is attached to a larger flywheel that is not shown. The total mass moment of inertia of the drum and the flywheel is 18 kg n r and the coefficient of kinetic friction between the drum and the brake shoe is 0.35. Knowing that the angular velocity of the flywheel is 360 rpm counterclockwise when a force P of magnitude 300 N is applied to the pedal C, determine the number of revolutions executed by the flywheel before it comes to rest.
SOLUTION Static Static Equilibrium (Friction Force | )
Lever Lev er ABO.
F = n kN = 0.35JV a = 0: + ) I M a
N (0.20 (0.20 m) - F(0.0 4 m) - 300 300 N(0.18 m) = 0 0.20N - 0.04(0.35 0.04(0.35 N) = 54 N N = 290.32 N
N
u r
F = n kN = 0.35(290.32 N) = 101.61 N
O.o
Dru m : r = 0.160 m
(2 n
//
360 rpm rpm — co0 = 360 v60 (o0 (o0 = 12/r rad/s
F - / oI. UN
C “ Fr = Ta-.
o =
(101.61 N)(0.16 m) = (l8kg m2)a
(deceleration ation)) a = 0.9032 rad/s2 (deceler o r = col + 2a d \
0 = (l2 ;r)2 + 2(-0.9 2(-0.9032) 032)<9 <9 6 = 786.77094 rad 6 = 786.77094
= 125.22 rev 9 = 125.2 rev <
PROBLEM 16.26 1211nun nun
Solve Prob. 16.25 16.25 assuming that the initial angular velocity of the flywheel is 360 rpm clockwise.
SOLUTION Static Equilibrium Equilibrium (Friction Force j )
Lever Leve r ABC:
F = n kN = 0.35W
+ ) I M a = 0:
■/ &
300 N
0.10 m
A
JV(0 JV(0.2 .20 0 m) + F(0 .04 m) - 300 N(0.18 m) = 0 JV[0.20 JV[0.20 m + 0.35( 0.04 m )] = 54 N m
I
N = 252.3364 N
% 0 . 0 ‘I *
F = fikN fi kN = 0.35(252.34 N) = 88.319 N
Dru m: r = 0.16 m
a f a = 360 rpm —
Uo
o)Q = 12/r rad/s
+) ZA/d A/d = l(A /D)e /D)e(T = 0
Fr = l a :
88.318 N(0.16 m) = (l8k m2)a a = 0.78505 0.78505 rad/s2 (decelera (deceleration tion)) o r = o> o>l + 2aO:
0 = (12* )2 + 2(-0.78505)<9 2(-0.78505)<9 6 = = 905.1799 rad
/ 9 = 905.18
1 x
2*
= 144.0638 rev or 9 = 144.1 rev A
PROBLEM 16.27 The flywheel shown has a radius of 600 mm, a mass of 144 kg, and a radius of gyration of 450 450 mm. An 18-kg block A is attached to a wire that is wrapped around the flywheel, and the system is released from rest. Negle cting the effect o f friction, friction , determine deter mine (a) the acceleration of blo b lo c k s, (b) the speed of block A after it has moved l.8 m.
PROBLEM 16.28 In order to determine the mass moment of inertia of a flywheel of radius 1.5 ft, a 20-lb block is attached to a wire that is wrapped around the flywheel. flywheel . The block is release d and is observe d to fall 12 ft in 4.5 s. To eliminate bearing friction form the computations, a second block of weight weig ht 40 lb is used and is observed observe d to fall 12 ft in 2.8 s. s. Assum ing that the moment of the couple due to friction remains constant, determine the mass moment of inertia of the flywheel.
SOLUTION Kinetics Kineti cs
Kinem atics
* ) Z M ,= ! ( * , ) * : imA g)r ~
= T a
+ ( m Aa ) r
m 4gr - M , = / —+ m ,ar r
Case 1: y - 12 12 ft,
( 1)
/ = 4.5 s; y - —at 2 => 12 ft = -^0(4.5 s)‘ a = 1.1852 ft/s-
(20 lb)(l.5 lb)(l.5 ft) - M , = I
From (1)
^ 1.1852 1.1852 ft/s2N
20 lb
T )(l.l85 2 ft7 ft7s2)(l.5 ft) 32.2 ft/s-
1.5 ft
(2 )
30 - M , = 0.790 13/ + 1.1042 1.1042 Case 2: y = 12 ft,
t - 2.8 s; y = —at 2 => 12 ft = —a ( 2.8 s)‘ a = 3.0612 ft/s
From (1)
(40 lb)(l.5 lb)(l.5 ft) - M t = I
f 3.0612 3.06 12 ft/s2 ft/ s2 "l 1.5 ft
/
40 lb
Tj(3.0612 ft/s2j(l.5 ft) 32.2 ft/s
60 - M , = 2.0408 / + 5.70 5.7041 41 Subtract (2) from (3)
(3)
30 = 1.25067/ +4.5999 / = 20.309 20.309 lb-ft lb-ft s2
or
/ = 20.3 lb-ft-s2 <
PROBLEM 16.29 f»
<»
I ( H
Each of the double pulleys shown has a mass moment o f inertia o f 20 k g-m 2 and is initially at rest. The outside radius is 400 mm, and the inner radius is 200 mm. Determine (a) the angular acceleration of each pulley, (b) the angular velocity of each pulley after point A on the cord has moved 3 m.
SOLUTION (a)
Case 1:
- 0 SAf0 = E(A/0)e(r: (785 N)(0.2 m) = (20 k g m 2)a a = 7.85 rad/s2 ') A
( b)
I S S N
G=
0.2 m
= 15 rad
o r = 2 a 9 = 2(7.85 rad/s2)(l5 rad)
to = 15.35 rad/s2 ) A Case 2:
(«)
+) T.M q = l (A /0 )e(T:
(80 kg)(9.81 m/s2)(0.2 m) = 20a + ma(0.2 m) 157.0 = 20a + (80)(0.2a)(0.2) a = 6.767 rad/s2
0 2>*)
a = 6.77 rad/s2
3 0 kcj
/yno~- on( 0-2
( b)
0 =
0.2 m
A
= 15 rad
o r = l a d = 2(6.767 rad/s2)(l5 rad)
to = 14.25 rad/s2 ") ^ Case 3 :
(a)
+ ) ZA/0 = l( A /0)efr: (230 kg)(9.81 m/s2)(0.2 m) - (150 kg)(9.81 m/s2)(0.2 m) = 20a + 230a(0.2 m) + 150a(0.2 m) 157.0 = 20a + 230(0.2)2a + 150(0.2)2a a = 4.460 rad/s2
23o
0 . 2**J
o.2 >•i
2$0q
I^OQ
or a = 4.46 rad/s2 ") A
PROBLEM 16.29 CONTINUED (b)
0 =
0.2 m
= 15 rad
a? = l a d = 2(4.46 rad/s2)(l5 rad)
a = 11.57 rad/s ^ 4 Case 4\
(a)
+ ) Z M 0 = l ( M
0 )eff:
(40 kg)(9.81 m/s2j(0.4 m) = 20a + 40a(0.4 m) 157.0 = 20 a + 40(0 .4)2 a ,
a = 5.947 rad/s2 a = 5.95 rad/s2 ") A
... (2>)
_ 3m _, 0 = ------- = 7.5 rad 0.4 m co2 = 2a 9 = 2(5.947 rad/s2)(7.5 rad)
<0 = 9.44 rad/s
*) M
PROBLEM 16.30 The weight and radius of friction of disk A are WA = 12 lb and rA = 6 in.; the the weight and radius of friction of disk B are WB = 6 lb and rB = 4 in. in. The disks are at at rest when a couple M o f moment 7.5 lb in. in. is applied to disk A. Assuming that no slipping occurs between the disks, determine (a) the angular acceleration of each disk, ( b) the friction force that disk A exerts on disk B.
PROBLEM 16.30 CONTINUED Substitute for F from from Eq. (3) into Eq. (2): M -
rA = JA JAa A
Substitute for F from Eq. (3), and for a B from Eq. (1).
r ^
- ~ m BrarA
\ ~<*A
\ fB
)
1
2
= ~ mArA«A rA«A
M = j ( m A + mB) r Ac A 2 cxA
Data:
WA = 12 lb,
2 M
2Mg
(mA + m B)rJ
(WA (WA + WB)r \
WB = 6 lb rA = 6 in. = 0.5 ft;
rB = 4 in. = 0.33333 ft
M = 7.5 lb in. = 0.625 lb-ft
(a)
2(0.625 lb-ft)(32.2 ft/s2) ^ = 8.94 8.9444 44 rad/ rad/s2 s2 (12 lb + 6 lb)(0.5 ft2)
a. = —
a , = 8.94 rad/s 2 ) < a B = — a , = ^ - ^ ( 8 .9 .9 4 4 4 rad/s2) = 13.417 rad/s2 r„ 4 in .v '
13.42 rad/s2') 4 (b)
r, 1 If 6 lb F - —mR —m RrRa R = — ---------- ^ (0.33333 ft)(l3.417 rad/s2) 2 2 1 32.2 32.2 ft/s2 ft/s2 F = 0.416671b
or F = 0 .4 .4 17 17 l b j ^
PROBLEM 16.31 Solve Prob. 16.30 assuming that the couple M is applied to disk B.
SOLUTION Kinem Kinematic atics'. s'. Since the tangential acceleration of the outside of the disks are equal.
a., a. , = — a R
(1)
Kinetics: Disk A:
T
*
2
l A = ~ Zm Zm A r A
F r A = J Aa A
1
2
F r A = ~Zm Ar A a A
r,
1
= ~ m A r A< A
(2)
M - FrB FrB = I Ha H
(3)
F
Disk B: B:
Substitute for F from from Eq. (2) into Eq. (3) M
PROBLEM 16.31 CONTINUED 1
,
—mBrB Substitute for a A from Eq. (1), and I B = —mBrB
1
M - - mA mArA rArrB\ — <*B VrA ) 2
M = aR
Data: WA = 12 lb,
2
mA + fnB)r fnB)r j a B
2M
2M g
(mA + m By B
(WA + WB)rl
WB = 6 lb, rA - 6 in. (= 0.5 ft),
rB = 4 in. (= 0.33333 ft)
m = 7.5 lb-in. = 0.625 lb-ft
2(0.625 lb-ft)(32.2 ft/s2) ft/s2)
(a)
aR=
20.125 rad/s
(12 lb)(0.33333 ft)2 a , = 20.1 rad/s2 J) M
rR a A = — a R -
4 in. 6 ^ j( 2 0 .1 rad/s2) rad/s2) = 13. 13.41 417 7 rad/s2
a. (b)
1 rAa A = — 1 r = —mArA F —mA /I /I /I ^ y
121b 32.2 ft/s2
13.42 rad/s2 ^ ^
(0.5 (0.5 ft)(l3.4 17 rad/s2) rad/s2)
= 1.25001b Friction force on disk B:
F = 1.250 lb | <
PROBLEM 16.32 Two identical 8-kg uniform cylinders are at rest when a constant couple M of magnitude 5 N m is applied to cylinder A. Knowing that the coefficient of kinetic friction between cylinder B and the horizontal surface is 0.2 and that no slipping occurs between the cylinders, determine (a) the angular acceleration of each cylinder, (b ) the minimum allowable value o f the coefficient o f static friction between the cylinders.
M
' 100 mm V
SOLUTION A:
/„ = 0.04 a /
+) I.V/<; = 5 - 10. If] = 0.0 4a
7S.«(8 ^
__
+ | IF , = NX- 78.48 = 0
))
i f
5 N-w
B : +) IA/c; = 0.1^ - O .l^ /V ;,) = 0.04a
\ + 1 I F , = N 2 - 78.48 - 78.48
c n f(4(
Now
N, = IS ‘iS
5 - 0.1/*] = 0. IF, - 0.0 2W , Fx = 40.696 N
or
a A = 23.3 rad/s: x 4
(a) J 18
(
I(
««•
1
A \
a„ = 23.3 rad/s:
A ( b) 0-0 4**
1 N/z «. I!T6 .1C
4 _ 40.696 N
F,
AT,
V j /m,n
78.48 N
= 0.51855
or /js =0.519 4
PROBLEM 16.33 Two identical 8-kg uniform cylinders are at rest when a constant couple M of magnitude 5 N m is applied to cylinder A. Knowing that the coefficient of kinetic friction between the cylinders is 0.45 and the coefficient of kinetic friction between cylinder B and the horizontal surface is 0.15, determine the angular acceleration of each cylinder.
SOLUTION A:
+)
= 5 N m - ( 0 .1 m ) ( 3 5 . 3 1 6 N ) = 0.04 a A a A = 36.71 rad/s2
or a A = 36.7 rad/s2 J) A
= is.3 lt B:
+) 1 M h = (0.1 m)(35.316N)
-(0.1 m)(23.544N) = 0.04aB a 8 = 29.43 rad/s-
or a B = 29.4 rad/s2 \ A
PROBLEM 16.34 Disk A has a mass of 8 kg and an initial angular velocity of 480 rpm clockwise; disk B has a mass o f 4 kg and is initially at rest. The disks are brought toge ther by apply ing a horizontal force o f magnitude 30 N to the axle of disk A. Knowing that fjk = 0. 15 between the disks and neglecting bearing friction, determine (a) the angular acceleration o f each disk, (b) the final angular velocity of each disk.
SOLUTION While slipping occurs, a friction force F j is applied to disk A , and F \ to disk B. Disk A : i
^
2
1A = r mArA
= 1(8 kg)(0.11m)‘
= 0.0484 kg -n r ZF :
N = P = 30 N
F = fi N = 0.15(30) = 4.5 N
+ ) ZM a = Z(A ^) efr: FrA = l Aa A (4.5 N )(0.1 1m) = (0.0484 kg-m2) ^ a A = 10.23 rad/s2 ") A
a A = 10.227 Disk B: v - S W
I G ° (B
1
T
2
h = ~ mBrB
= 1 (4 kg)(0.08 m)‘
= 0.0128 kg m+)
= Z { M H)cn : FrB = I Ba B
(4.5 N)(0.08 m) = (0.0128 kg-m2) ^ a B = 28.125 rad/s'
a B = 28.1 rad/s2 ^ ^
PROBLEM 16.34 CONTINUED (<»a )b = 480 rprnj ~
= 16;r rad/s J
Sliding stops when Vc = \ c or a>ArA = coBrB rA[{
(0.11 m i 16/r rad/s - (10.227 rad/s* jf j = (0.08 m)(28.125 rad/s 2\t t = 1.6383 s +'C
03 ~Mo a
~
a At
~16^ rad/s -
(l0.227 rad/s* ((1.6383 s)
6», = (33.511 r a d /s ) ^
= 33.511 rad/s
= 320.00 rpm or at. = 320 rpm ) A
+ Q coB = a Bt = 28.125 rad/s2 (1.6383 s) = 46.077 rad/s ' 60 ' coB = (46.077 rad/s) — = 440.226 rpm OX (S)B = 440 rpm ”) 4
PROBLEM 16.35 Solve Prob. 16.34 assuming that initially disk A is at rest and disk B has an angular velocity o f 480 rpm clockwise.
SOLUTION (a)
From P 16.34: or a A - 10.23 rad/s2 x: 4
a A = 10.227 rad/s
and or a B - 28.1 rad/s'1^ A
a B = 28.125 rad/s2 f 2
cb)
( coa ) b = 0, {&B\ = (480 rpm)
60
1bn rad/s
Sliding starts when Vc - Vc . That is when <°ArA = 0JBrB
= [( ® b ) o - a B{ ] rB
(10.227 rad/s2)/ ](0.11 m) = 16;r rad/s - (28.125 rad/s2){/) (0.08 m) 3.37497/ = 4.02124 => t = 1.1915 s £ mA a a At = (l0.227 rad/s2)(l.l915 s) = 12.18535 rad/s
+(
( 60 A 0 )A = (12.18535 rad/s) — = 116.3615 rpm ') \2n)
or (oA = 116.4 rpm *) A +(k coB = (®b)0 - ocBt = 16^r rad/s - (28.125 rad/s 2j ( l . 1915 s)
= 16.755 rad/s') 60 a>B - 116.755 rad/s) — | = 159.9985 rpm 2/r
or (aB = 160 rpm J) A
PROBLEM 16.36 /Of*— / --------------------------
r
£
s f ' - 4 R
r >
Disk B has an angular velocity eo0 when it is brought into contact with disk A, which is at rest. Show that (a) the final angular velocities of the disks are independent of the coefficient of friction n k between the disks as long as n k * 0, (b) the final angular velocity of disk B depends only upon g>{) and the ratio o f the masses m A and m B of the two disks.
PROBLEM 16.36 CONTINUED 2 PM kr „ l p M k r t =
Substitute from Eqs (1) + (2):
2 PMt
' 1
1 ' 1---- t - co0rB‘ , t
\ mA
mBj
conr, WB
2 pMk _L + J_ m,
t _ = “>0rB 2PUk
Eq (3):
aoA = a A
m Am B
mA + mB
_ 2 Pf ik m ArA
00 a _= ' B rA
»is
®0 rB
m AmB
2 P p k
mA + mB
Wfl
-COa
mA + mB
(a>A is independent of fik , QED)
Eq (4):
(0B = o)o - a Bt = 0 )n -
g>b
_ <*>0
2 P Mk m B'B ar.
gVl 2 Pjuk
r»AmB mA + mB
/n. = ay (1 -■ mA + m B
m A + mB - mA _ — /n* +
(On =
+ mB
fi>n
+1 (ys dep end s only upon
0 and
(QED)
PROBLEM 16.37 The 12-lb disk A has a radius rA = 6 in. and an initial angular velocity
PROBLEM 16.37 CONTINUED f 2n ' M o = 750 fP” 1| ^ J = 78.5 398 rad/ s ) ;
(a>„)0 = 0
When disks stop sliding vP = vP
coArA = (0BrB
[ M o “ a A r* =
[78.5398 rad/s - (l3.417 rad/s2)(t )] (6 in.) = [(3.22 rad/s2)(t ) (10 in.) => t = 4.1813 s ooA = 78.5398 rad/s - (l3.417 rad/s2)(4.1813 s) = 22.439 rad/s coA = 22.439 rad/s
2n
214.2767 rpm or toA = 214 rpm ) 4
oB =
= 0.6(214.2767) = 128.5660 rpm
or a \B = 128.6 rpm ^ 4
PROBLEM 16.38 Solve Prob. 16.37 assuming that disk A is initially at rest and that disk B has an angular velocity of 750 rpm clockwise.
SOLUTION Based on the solution for PI6.37 M o = °; M o = 750115111 Eq. (1):
a>ArA = coBrB\
= 1SrT 115111 )
a AtrA = [(«„ )Q- a Bt ] rB
(13.417r)(6 in.) = [25* - 3.2 2/](l 0 in.) t = 6.9688 s
( f*C\\ (O a = a At = (l3.417)(6.9688) = 93.5 rad/sf — = 892.86 rpm \ 2 tt or atA = 893 rpm ') A coB = — coB = 0.6(892.86) = 535.716 rpm
or (oB = 536 rpm J A
PROBLEM 16.39 A cylinder of radius r and mass m rests on two small casters A and B as shown. Initially, the cylinder is at rest and is set in motion by rotating caster B clockwise at high speed so that slipping occurs between the cylinder and caster B. Denoting by /.ik the coefficient of kinetic friction and neglecting the moment of inertia of the ffee caster A, derive an expression for the angular acceleration of the cylinder.
PROBLEM 16.40 In Prob. 16.39 assume that no slipping can occur between caster B and the cylinder (such a case would exist if the cylinder and caster had gear teeth along their rims). Derive an expression for the m aximum allowable counterclockwise acceleration a o f the cylinder if it is not to lose contact with the caster at A.
PROBLEM 16.41 Show that the system of the effective forces for a rigid slab in plane motion reduces to a single vector, and express the distance from the mass center G of the slab to the line^of action of this vector in terms of the centroidal radius of gyration k o f the slab, the magnitude a of the acceleration of G, and the angular acceleration a.
SOLUTION We know that the system of effective forces can be reduced to the vector ma at G and the couple la. We further know from Chapter 3 of statics that a force-couple system in a plane can be further reduced to a single force.
The perpendicular distance d from G to the line o f action o f the single vector ma is expressed by writing
+) ™ G = I ( M G)eff :
l a = (ma)d
_ l a _ m k2a ma
ma
a
PROBLEM 16.42 Am, ;i
For a rigid slab in plane motion, show that the system of the effective forces consists o f vectors (Am ,)a, -(A m ,)© 2i^, and (Aw, )(a x r/) attached to the various particles Pt o f the slab, where a is the acceleration o f the mass center G o f the slab, co is the angular velocity of the slab, a is its angular acceleration, and r d e n o te s the position vector of the particle P, relative to G. Further show, by computing their sum and the sum of their moments about G, that the effective forces reduce to a vector ma attached at G and a couple I a .
SOLUTION The acceleration of PL is
Kinematics
a, = a + a^/c a, = a + a x i;' + to x( to xi ;') = a + a x i ^ -
Thus, the effective forces are as shown in Fig PI6.47 (also shown above). We write (Am,)a, = (Am,)a + (Am ,)(a x <) - (Am,)© 2/)' The sum of the effective forces is E(Am ,)a, = E(Am/)a + E(Am ,)(a x r,') - E(Am,)
l(Am ,)a, = m a
The sum of the moments about G of the effective forces is: E(i;' x Am,a,) = Zr- x Am,a + Zr;' x (Am;) (a x r/) - Er/ x (Am, )©2i;' Z(i;' x Am,a,) = (Zr;Am ,)a + z [i ;' x (a x i;')Am,-J - ©2Z (r / x r/)Am,
(1)
PROBLEM 16.42 CONTINUED Since G is the mass center,
I,r'Aml = 0
Also, for each particle,
r/ x r/ = 0
Thus, l(if x Aw,a,) = S[i ■ ;x (a,, x Since a 1 r/, we have i - x ( a x r ;•) = r fa and E( r/ x A/w,a; ) = 5>/2(Aiwf) a = ( & / 2Afft,)a Since Er/ 2Ami = I E (r' x A
= la
From Eqs. (1) and (2) we conclude that system of effective forces reduce to ma attached at G and a couple la.
PROBLEM 16.43 A uniform slender rod AB rests on a ffictionless horizontal surface, and a force P of magnitude 2 N is applied at A in a direction perpendicular to the rod. Knowing that the rod has a mass of 3 kg, determine the acceleration of (a) point A, (b) point B.
SOLUTION ft
-P T m—
m (3 k6)( l m) : - 0 25 kg m;
12
12
+ . Z Fx = Z ( F x )eff: P = ma cPa
I N = (3 kg)o
LJ%
a = 0.66667 m/s'
3 <
'’(f) ( 2 N ) f t
m j
-
= (0.25 kg-nr
)a
_ U N)(0 .5 m ) _ 4 raj/s; (o.25 kg-m:)
(«)
= a + i a = 0.66667 m/s2 + ^
m j(4 rad/s2)
= 2.66667 m/sor (b)
aH - a + -^a = 0.66667 m/s2 -
m
j (4r a d / s 2 )
a 4 = 2.67 m/s2-
= -1.33333 m/s2 or
a B = 1.333 m/ s'
PROBLEM 16.44 (a) In Prob. 16.43, determine the point o f the rod AB at which the force P should be applied if the acceleration of point B is to be zero. ( b ) Knowing that P = 2 N, determine the corresponding acceleration of point A.
SOLUTION A —
- i . w . - I ( F ,L
V
L
P = ma,
a
2
p = — m
+ ( XA/C = X(A/c ) cn
L
D U Ph = I7 a =
Z
73
a =
_
L
a« = a
(a)
2
"
0
P
m L r
12
a
12 Ph mLt )
a: Z. 12^/7
= — 11 ml
m 2\ ml? ,
L
h = — = 0.16667 m 6
Thus, P is located 0.333 m from A A (b)
.
L _ _ a,, = a + —a: 2
m
a , = -^-^-(l + l)
'
3kgV
’
2
mL j - -mI '
i )
= —m/s2 = 1.3333 m/s2 3 or
a .
= 1.333 m/s2
y
w 7
i n
j
I
p
PROBLEM 16.45 A force P of magnitude 0.75 lb is applied to a tape wrapped around the body indicated. Kno wing that the body rests on a fricti onless horizontal surface, determine the acceleration of (a) point A, (b) point B. A thin hoop o f weight 6 lb.
PROBLEM 16.46 ¥
A force P of magnitude 0.75 lb is applied to a tape wrapped around the body indicated. Knowing that the body rests on a ffictionless horizontal surface, determine the acceleration o f (a) point A, ( b) point B.
-•
A uniform disk o f weight 6 lb.
p
PROBLEM 16.47 A force P is applied to a tape wrapped around a uniform disk that rests on a frictionless horizontal surface. Show that for each 360° rotation o f the disk the center of the disk will move a distance nr.
SOLUTION Let r, = tim e required for 360° rotation
1 2 „ 0 = -air, 2
^ l f 2 P "1 2 n = - — g tr 2\w r )
=
2nw r Pg
jc, = distance G moves during 360° rotation U p 1-2 = T « 'i = d — g 2 21 w
2nwr\
= n r
~~Pg~)
x, = n r Q.E.D. A
PROBLEM 16.48 A uniform semicircular plate of mass 6 kg is suspended from three vertical wires at points A, B, and C, and a force P of magnitude 5 N is applied to point B. Immediately after P is applied, determine the acceleration of (a) the mass center of the plate, ( b) point C.
'J
SOLUTION
s
= C 't y r
la =
1
m
2
—mr | - m 2
f 4 Ar ' 2
3n
= (6 kg)(0.3 m )2 0.5 (a)
a - mr
2 ( 1
^
16
|a
1.7777 ^ •> a = ^0.17273 k g m 2 j a
XFx = 5 N = 6 kg ac ,
aG = 0.8333 m/s
or
aG = 0.8331 m/s A
ZM g = 5 N — m j = ^0.17273 k g m 2) a \ n a = 3.6856 rad/s2 ar = a(; + aC76- =
d-2333 *■ +
a,- = 0.8333 — + — 0.63641 = 0.1969 — m/s2
or ar = 0.1969im/s2 A
PROBLEM 16.49 Immediately after the force P is applied to the plate o f Prob. 16.48, determine the acceleration of (a) point A, ( b ) point B.
' H
PROBLEM 16.50 A drum of 10-in. radius is attached to a disk of radius rA = 7.5 in. The disk and drum have a combined weight of 12 lb and a combined radius of gyration of 6 in. and are suspended by two cords. Knowing that TA = 9 lb and TB = 6 lb, determine the accelerations of points A and B on the cords.
SOLUTION
s. f
is"
\
1°
..
4
^
I I I
=
(0-3)
*
' t
(SO 12
i l —
\ 3Z-2
a
1 >11 lb • t i p ,.
12 lb
T a = 15 lb - 12 lb; 32.2 ft/s
+)ZAf,T= 9 lb y
a = 8.05 ft/s 2 |
7.5 in .) . (1 0 in.'I 3 lb ft 2 , - 6 lb ------ = ------------ a ( 12 ; ( 12 j 32.2 ft/s
a = 6.70833 rad/s2 ) aA = | 8.05 ft/s 2 + | 6.70833 rad/s2( ^
ft ] = 12.24269 ft/s2 or
aB = | 8.05 ft/s 2 + j 6.70833 rad/s2^
a A = 12.24 ft/s2 | A
ftJ = 24597 ^ s 2 | or
a B = 2.46 ft/s2 | A
PROBLEM 16.51 A drum of 10-in. radius is attached to a disk of radius rA = 6.92 in. The disk and drum have a combined weight o f 12 lb and are suspended by two cords. Knowing that the acceleration of point B on the cord is zero. a = 10 lb, and TB - 5 lb, determine the combined radius of gyration of T the disk and drum.
\«
\
f X {
y: <-rA— -—UMn.—*
SOLUTION \0 ih
-t IF.. = 12 lb (a) = 15 lb - 12 lb v 32.2 ft/s 1 a = 8.05 ft/s2 |
+ ) ZM a = 10 lb
6.92 12
a = 9.66 rad/s2 J)
12 1b , 7 ft 1-5 lbl — ft I = A:2 (9.66 rad/s2) 12 ) 32.2 ft/s
k- = 0.44444 ft'
k = 0.66666 ft
or k - 8 in. A
PROBLEM 16.52 A 5-m beam of mass 200 kg is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground, the crane operators apply brakes to slow the unwinding motion. Knowing that the deceleration o f cable A is 5 m/s2 and the deceleration of cable B is 0.5 m /s2, determ ine the tension in each cable.
SOLUTION Kinematics:
0 . 5-
■
^r=
S
3
Kinetics : nwo-
7 b
t
a iS
F S 2 Soo K
w*fzoa -Jz c o {:jj aB
=
a A +
5a
| 0.5 m/ s2 = | 5 m/s 2 + (5 m ) a => a = 0.9 rad/s 2 J) a
-
aA
| + 2.5« | = 5 m/s 2 - (2.5 m)(o.9 rad/s2) = 2.75 m/ s2 *)
I -
=
^g)(5 m)2 - 416.67 kg m 2 ( Z M B = E(MB)eff:
T a (5 m) - (200 kg)(# )(2.5 m ) = (200 kg )(a)(2 .5 m) + (416.67 kg-m2) a
Substituting g = 9.81 m/s2, a = 2.75 m/s2, a = 0.9 rad/s 2 T a (5 m) = (4905 + 1375 + 375)kg-m 2/s2, TA = 1 3 3 1 N a = 1331N < or T
ZFy = E(Fy)eff:
a +T b - (200 kg)(g) = (200 kg)a T
b - (200 kg)(9.81 m/s2) = (200 kg)(2.75 m/s2) 1331 N + T b = 1181 N T b = 1181 N < or T
PROBLEM 16.53 A 5-m beam of mass 200 kg is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground, the crane operators apply brakes to slow the unwinding motion. Knowing that the deceleration o f cable A is 5 m/s 2 and the deceleration o f cable B is 0.5 m/s2, determine the tension in each cable.
SOLUTION Kinetics :
t |—
*
r r*
J
6
_
2
Kinematics:
y ^
A
g
f 8
Of
a a = a B + 4a
5 m /s 2 | = 0.5 m/s 2 j + (4 m )a
a = 1.125 rad/s2 J
a = afl + 1.5a = 0.5 m/s2 + (1.5 m )(l.l25 rad/s2 j
a = 2.1875 m/s2
/ = ^ m L 2 = ^ ( 2 0 0 k g )( 5 m )2 = 416.67 kg m 2 + (
= S(A /„)cfr: 7^(4 m) - (200 kg )(g ) = m a (1.5 m) + 7a
T a (4 m) - (200 kg)(9.81 m/s2)(l.5 m) = (200 kg)(2.1875 m/s2)(l.5 m) + (416.67 kg m2)(l.l25 rad/s2) a = 1017 N T
a = 1017 N A or T
PROBLEM 16.53 CONTINUED , - (200 kg )(g) = j +T j ) IF , = z (F ,) =ir: T
1017 N + Tb = (200 kg)(9.81 m/s2) + (200 kg)(2.1875 m/s2)
PROBLEM 16.54 The 400-lb crate shown is lowered by means of two overhead cranes. Knowing that at the instant shown the deceleration of cable A is 3 ft/s 2 and that o f cable B is 1 ft/s2, determine the tension in each cable. cable.
SOLUTION Kinetic Kin etic s:
*
e\
W = mg
f 0tlS~ I = -± -m (b 2 + c 2)
£
1 400'b ( 6.6 ft )2 + (3.6 ft)2] 12 32.2 ft/s58.5093 lb ft s 2 + |£ /V = E (f;.)ejr: (f;.)ejr: TA + T B - 400 =
( 1 )
f ~\ a + T b = 400 1 + 2 . T + ( Z MC M C = l(M c )efr: (T a - r fl) (l ( l . 8) = l a (T a - Tb)( 1.8 1.8 ft) = (58.5093 lb- ft-s 2) a
(2)
Kinematics'. Kinematics'. a = a + ra (aA)v - a + 1.8a; 3 - a + 1.8a
(3)
(as )v = a - 1 .8a ; 1 = a - 1 .8a
(4)
(■Sft
Solving (3); (4);
a = 2 2 ft/s2 | ,
a = 0.55556 rad/s2
PROBLEM 16.54 CONTINUED From (1): From (2):
r = 400 f 1 + 2 1 = 424.84 lb T, + T A B \ 32.2 J ta
~
tb
= (32.505 lb-s2)(0.55556 rad/s2)
a = 221.4492 lb T
a = 221 lb A T
b = 203.3907 lb T
b - 203 T 203 lb A
PROBLEM 16.55 The 400-lb crate shown is lowered by means of two overhead cranes. Knowing that at the instant shown the deceleration of cable A is 3 ft/s 2 and that of cable B is is 1 ft/s2, determine the tension in each cable.
b -
r
3.6 ft
! i .hrt
:.
t
■3.3 It 6.6 ft
SOLUTION Kinetics:
J f i
j /Tf) /Tf) &&*
"
f?
s
p &
*
* \
• L lA lt 1 = — m ( a 2 + b 2) = — 400 lb T(6.6 ft )2 + (3.6 ft)2] = 58.509 Ib-ft> 12 32.2 ft/s2 LV ’ ' -I 12 v LV V TA + TB - m
=^ a
a + T b = 400 \ + ± T g
( 1)
( ZM Ma a = Z(Mc )eff )eff:: (T a - TB){3.3 ft) = l a = (58.509 lb-ft-s2)a a ~ T b) {T T b) = ( 17.7 17.73 3 l b s 2)a
(2 )
Kinematics: (fejtj
(9, (9, ft f t )*} R
fS-
6
-*■ ( a s ) y
( aA)y = a + 3.3a 3 = o + 3.3a
(3)
M y - a - 3-3ar
l = o - 3 .3 a Solving (3); (4)
a = 2 ft/s2, ft/s2,
a = 0.30303 rad/s2
(4)
PROBLEM 16.55 CONTINUED From (1); (2)
2 ft/s 2 %
(400 lb) r.1 + TA + TB = (400 -2 o 424.84 lb ^ 32.2 ft/sz / T T b = (l7.73 lb-s2)(0.30303 rad/s2) = 5.3727 lb a - T b
Solve
T a = 215.10636 lb
or TA = 215 lb <
b = 209.73366 lb T
or T Bt = 21 0 lb <
PROBLEM 16.56 The uniform disk shown, of mass m and radius r, rotates counterclockwise. Its center C is constrained to move in a slot cut in the vertical member AB and a horizontal force P is applied at B to maintain contact at D between the disk and the vertical wall. The disk moves downward under the influence of gravity and the friction at D. Denoting by /Jk the coefficient of kinetic friction between the disk and the wall and neglecting friction in the vertical slot, determine (a) the angular acceleration of the disk, (b ) the acceleration of the center C of the disk.
SOLUTION
P
*■ ZP
I P
s - L w r 2- ^ \ _ i i
■m o
. 4 fikP a = - L-::— mr
(«) (b)
+J IF , =
mg +(2
=ma m
A
PROBLEM 16.57
n /
The 6-kg uniform disk shown, of radius r = 80 mm, rotates counterclockwise. Its center C is constrained to move in a slot cut in the vertical member AB and a 50-N horizontal force P is applied at B to maintain contact at D between the disk and the vertical wall. The disk moves downward under the influence of gravity and the friction at D. Knowing that the coefficient of kinetic friction between the disk and the wall is 0.12 and neglecting friction in the vertical slot, determine (a) the angular acceleration of the disk, ( b) the acceleration o f the cen ter C o f the disk.
SOLUTION From PI6.56: (a)
a =
4ftkP _ 4(0.12)(50 N) mr
(6 kg)(0.08 m) or a = 50 rad/s2 ) A
m
6 kg or a = 11.81 m/s 2 j A
PROBLEM 16.58 A beam AB of mass m and of uniform cross section is suspended from two springs as shown. If spring 2 breaks, determine at that instant (a) the angular acceleration of the beam, (b) the acceleration of point A, (c) the acceleration o f point B.
SOLUTION 7] = r 2 = i ^ = i m g
Statics: (a)
Z
A
«?
A Lf a
" ]/V - on g \ = r ( M 0 )c„ :
*)
= la
- m g f —1 = — ml} a
2
{2J a =
12
3 g
ma mg - —mg = ma
1 _ a = - g
1 | _ a = - g J
(b) Acceleration o f A : a A ~ a C + *A/G
a
t P\-
_
J - z L - J l
%
4.1
1
L
1
L
+ l *a = 2 g ~ 2 a
- _ / t a .- - 3
V
= 2g " 2 U , ° a =
= g \<
PROBLEM 16.58 CONTINUED (c) Acceleration o f B:
PROBLEM 16.59 A beam AB of mass m and of uniform cross section is suspended from two springs as shown. If spring 2 breaks, determine at that instant (a) the angular acceleration of the beam, (b) the acceleration o f point A, (c) the acceleration of point B.
u
-I
PROBLEM 16.60 — T T
------------
1 3;
2 H
Hi •t
.
4
i H
A beam AB of mass m and of uniform cross section is suspended from two springs as shown. If spring 2 breaks, determine at that instant (a) the angular acceleration of the beam, (b) the acceleration of point A, (c) the acceleration of point B.
PROBLEM 16.61 A thin plate of the shape indicated and of mass m is suspended from two springs as shown. If spring 2 breaks, determine the acceleration at that instant (a) o f point A, (b) of point B. A circular plate o f radius b.
SOLUTION 1 m = —m\ — 2 {2, Statics:
= —mb 2 8
^ = Tl = \ W = \ m g
+ ) ZM g = l ( A f c )eft
Kinet ics :
t b 71— = l a
1 2 mg
E>
1 .2 — | = —mb a
8
a = 2 — ) b v + lZ 7v = z (F v)eff: W - T , = ma m g
1
_
mg = ma
a
Kinematics:
Translation b
(a)
( b)
= ac
+ a fi/G = a
\ +
\
*
a \ =
+
1
i
i +
Rotation b (
2^ IT;
PROBLEM 16.62
1
---- —
h — — i f
springs as shown. If spring 2 breaks, determine the acceleration at that instant (a) o f point A, (b) o f point B. A thin hoop of radius b.
v
•
(
>
PROBLEM 16.63 31$ • £g M
"
'!§< ;=5 m
springs as shown. If spring 2 breaks, determine the acceleration at that instant (a) o f point A, (b) o f point B.
\
—
A square plate o f side *-
r
_
b.
B
SOLUTION I = — n ti b2 + ft2) ' ’ 12 I = - mb 2 6 Statics:
T \ = h = \ w = img
Kinetics:
J -
+) ZMa = S(A/6 )ic(T
A T — = / a 2
1 \ 6‘ W='W 3
— m g f — 1 = — m b 2a
2
6
_ 3S
+| IF , = Z (F,)e(r: IT- 7 ; = ma 1
m g - - m g = ma
1 , _ a = -g J
Kinematics:
a n
\a
Rotati on
PROBLEM 16.63 CONTINUED
PROBLEM 16.64
* P
A bowler projects a 200-mm-diameter ball of mass 5 kg along an alley with a forward velocity v 0 of 5 m/s and a backspin
