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7/27/2019 Mechanics of Machines Cleghorn Answers
1/23
P1.1
(a)n = 6; (b) n = 6
turning pair, 5; turning pair, 5;
sliding pair, 1; sliding pair, 2;
rolling pair, 1; rolling pair, 0;
2 dof pair,0; 2 dof pair,0;
m = 1 m = 1
P1.2
(a)m = 1
(b)m = 1
P1.3(a)n = 6; (b) n = 8
turning pair, 7; turning pair, 10;
sliding pair, 0; sliding pair, 0;
rolling pair, 0; rolling pair, 0;
2 dof pair,0; 2 dof pair,0;
m = 1 m = 1
P1.4(a)n = 4; (b) n = 4
turning pair, 0; turning pair, 0;
sliding pair, 5; sliding pair, 3;
rolling pair, 0; rolling pair, 0;
2 dof pair,0; 2 dof pair,2;
m = -1 (by inspection, m = 1) m = 1
P1.5
(a)n = 4; (b) n = 8
turning pair, 3; turning pair, 9;
sliding pair, 0; sliding pair, 0;
rolling pair, 2; rolling pair, 2;
2 dof pair,0; 2 dof pair,0;
m = -1 (by inspection, m = 1) m = -1 (by inspection, m = 1)
7/27/2019 Mechanics of Machines Cleghorn Answers
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P1.6
(a)
(i)
l < s + p + q, link can form a mechanism
(ii)
s + l < p + q, class I four-bar kinematic chain, drag link mechanism.
(b)
(i)
l < s + p + q, link can form a mechanism
(ii)
s + l = p + q, change point mechanism
(c)
(i)
l < s + p + q, link can form a mechanism
(ii)
s + l < p + q, class I four-bar kinematic chain, crank-rocker four-bar mechanism
(d)
(i)
l < s + p + q, link can form a mechanism
(ii)
s + l < p + q, class I four-bar kinematic chain, rocker-rocker four-bar mechanism(e)
(i)
l > s + p + q, link cannot form a mechanism
(f)
(i)
l < s + p + q, link can form a mechanism
(ii)
s + l > p + q, rocker-rocker four-bar mechanism
P1.7
(a)0 < r2 < 0.5 (cm)
(b)1.5 < r2 < 3.5 (cm)
(c) r2 = 0.5 or r2 = 1.5 or r2 = 3.5 (cm)
(d)0.5 < r2 < 1.5 or 3.5 < r2 < 5.5 (cm)
7/27/2019 Mechanics of Machines Cleghorn Answers
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P1.8
(a)0 < r4 < 0.5 (cm)
(b)1.5 < r4 < 4.5 (cm)
(c) r4 = 0.5 or r4 = 1.5 or r4 = 4.5 (cm)
(d)0.5 < r4 < 1.5 or 4.5 < r4 < 6.5 (cm)
P1.9
(a)0 < r2 < 1.5 (cm)
(b) r2 > 0 (cm)
P1.10
(a) r3 3.5 (cm)
(b) r3 > 0
P1.11
7/27/2019 Mechanics of Machines Cleghorn Answers
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P2.1
(a)2 = 200 or 20
3 max = 50 rpm
(b)2 = 110 or 290
P2.2
(a)2= 90 or 270
v4 max =v2 = 52.36 (cm/s)
(b)2 = 0 or 180
vs max =v2 = 52.36 (cm/s)
(c)2 = 0 or 180
a4 max =a2N
= 548.1 (cm/s2)
P2.3
(a) 2 = 53.1 or 306.9
(b)v6 = 150 (cm/s), 2 = 0
P2.4
(a)(
4,avg )cw = 7.22 (rad/s)
(b)(
4,avg )ccw = 7.58 (rad/s)
(c)TR= 0.953
P2.5
(a) (v4,avg )right = 124.18 (cm/s)
(b) (v4,avg )left = 85.56 (cm/s)
(c) TR= 1.45
P2.6
(a)2 = 306.9
(b)2 = 0 or 360
(c) 2 = 306.9 or 0 or 360
P2.7
w3 ,max = 72.4 rad/s CW ; 2 =180
7/27/2019 Mechanics of Machines Cleghorn Answers
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P2.10
(a) 2.79 r2 3.1 (cm)
(b)The four-bar mechanism is a drag link.
P2.11
(a) 1.82 r4 2.71 (cm)
(b)The four-bar mechanism is a drag link.
P2.12
0 r2 0.76 (cm)
P2.13
r2 4.95 (cm)
P2.14
(a)m = 1
7/27/2019 Mechanics of Machines Cleghorn Answers
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Cleghorn Ch3 answers
3.1
60=Bv in/sec, 5.39=
Cv in/sec, 15=
Dv in/sec,
3.2
1.473 =v in/sec
3.3
28.64 =v cm/sec
3.4
(a) 1.214 =& rad/sec CW
(b) 52.20=Dv in/sec
3.5
84=Dv cm/sec
3.6
7.36 =& rad/sec CW
3.7
89.2=Cv in/sec
3.8
2.7=Dv in/sec
3.9
(a) 28.63 =& rad/sec CCW
(b) 84.404 =v cm/sec to the left
3.10
(a) 6.52 =Pv cm/sec
(b) 6.133 =Pv cm/sec
(c) 6.923 =PPv cm/sec
(d) 8.12
3=
&
&
3.11
(a) 4.182 =Pv cm/sec
(b) 183 =Pv cm/sec
(c) 2.1123 =PPv cm/sec
(d) 7.12
3=
&
&
3.12 (b)
7/27/2019 Mechanics of Machines Cleghorn Answers
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(i) 13.25=Bv in/sec@ 45
(ii) 4.24=Cv in/sec@ 29
(iii) 5.27=Dv in/sec@ 0
(iv) 0.20=BDv in/sec@ 1.118
(v) 7.63 =& rad/sec CCW
3.13(b)
(i) 0.22=Bv in/sec@ 30 (ii) 1.8=
Cv in/sec@ 18
(iii) 1.13=Dv in/sec@ 10 (iv) 53 =
& rad/sec CCW
(v) 24.54 =& rad/sec CCW
3.14(b)
(i) 45=Dv cm/sec@ 98 (ii) 24=
Ev cm/sec@ 180
(iii) 1.63 =& rad/sec CW (iv) 165 =
& rad/sec CCW
3.15(c)
(i) 12032 =s
BBa cm/sec2 (ii) 1.273 =
&& rad/sec2 CW, 9.135 =&& rad/sec2 CCW
3.16(c)
1547 =&& rad/sec2 CCW
3.17(c)
805 =&&
rad/sec2
CW
3.18(c)
876 =&& rad/sec2 CCW
3.19(c)
336 =&& rad/sec2 CCW
3.20
(c)
7/27/2019 Mechanics of Machines Cleghorn Answers
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(i) 12806 =Ba cm/sec2 (ii) 483 =
&& rad/sec2 CW, 1606 =&& rad/sec2 CW
(d) 6.606 =&& rad/sec2 CW
3.21(c)
(i) 6344 =&& rad/sec2 CW
(ii) 650=Ca cm/sec
2
3.22(c)
(i) 188042 =s
BBa in/sec2
(ii) 4882 =&& rad/sec2 CW
3.23(d)
1773 =&& rad/sec2 CCW, 3534 =
&& rad/sec2 CCW
3.26
(b)
(i) 9.23 =& rad/sec CW, 8.104 =
& rad/sec CW
(ii) 59=Dv cm/sec, 102=
Ev cm/sec, 42=
Fv cm/sec
(d)
(i) 2863 =&& rad/sec2 CCW, 4904 =
&& rad/sec2 CCW
(ii) 1780=Da cm/sec2 , 3240=
Ea cm/sec2 , 2780=
Fa cm/sec2
3.27(c)
413 =&& rad/sec2 CCW
3.28(c)
(i) 253 =&& rad/sec2 CW, 2205 =
&& rad/sec2 CCW
(ii) 30=Da cm/sec2
3.29(b)
(i) 43=Gv cm/sec
(ii) 62715 =c
EEa cm/sec
2
7/27/2019 Mechanics of Machines Cleghorn Answers
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p4.1 w)rad/sec(cc212.02 =& o
v70@sec)/(53.132 = inv
s
p4.2 3&& =42.3 (ccw)
sec2rad
p4.3
sin
)sin(212
+
=
rr
sin
sin213
rr = =2r&
sin
)cos(221
+&r
sin
)sin()cos( 22
212212
++=
&&&
&&rr
r
sin
sin 213
rr =
sin
cos 2213
&
&r
r =
sin
sincos 22
212213
&&&
&&rr
r
=
p4.4 ov
4.119@sec)/(3.102 cmvc =
p4.5 (a) (cw)sec/39.23 rad=
&
(b) 37.4 cm/sec @71.9 o
(c) 17932=
c
AAav
cm/sec 2 @-18.1 o
p4.6 (a) (cw)rad/sce29.73 =&
(b) 71843=
c
BBa cm/sec2 @-101.8 o
p4.7 (a)ov
0@sec/2690 2cmaB =
(b) ov
225@sec/2990 242
cma s AA =
p4.8 (ccw)sec/69.13 rad=&
p4.9 (a) sec/)7.10671.4( cmjivc
vvv=
(b) sec/)8.241( cmjvBvv
=
p4.10 (a) (cw)sec/39.73 rad=&
(b) sec/7.2913
cmv BB =v
@-16.8 o
(c) o8.106@sec/439 213
= cmac BB
p4.11 (a) )sec(/8.283 cwrad=&
(b) ov
8.215sec@/14032
cmvPP=
7/27/2019 Mechanics of Machines Cleghorn Answers
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p4.12 sec/)41.1198.10( injivcvvv
+=
p4.13 (a) sec/cos577.0sin78.11)cos3
1(sin 222222 inrvc +=+=
&v
(b) sec/cos6.13 224 inv BB =v
(c) 223 sec/sin8.106 inr =&&
(d) sec/6.13max24
inv BB =v
p4.14 (a) )sec(/5.83 ccwrad=&
(b) sec/)192.177( cmjivcvvv
=
p4.15 (a) sec/1.4413 cmv DD=
v
@16.3
o
(b) )sec(/13.23 cwrad=&
7/27/2019 Mechanics of Machines Cleghorn Answers
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5.1 5.97d in=
5.2 450min
rev =
5.3 66.3 rpm =
5.4 (a) velocity ratio=1.333
(b)3
30N =
5.53 2
100 20N N= =
5.63 2
25 15 2356sec
mmN N pitch line velocity= = =
5.7 (a) 64B
N =
(b) c=13 in
(c) 0.785c
p in=
(d) 16.5in
(e) 15.49 in
5.8 (a)2 3
23.3 11.7d in d in= =
(b)3
5.48b
r in=
(c)3
cp in
=
(d) c=17.5 in
(e) 600rpm CCW
(f) m=1.755.9
5.10
5.113
117N =
(a)2
12.56c
p mm=
(b)3
219.9br mm=
(c)4
192d mm=
(d) c=282 mm
5.12 (a)3
2.69b
r in=
(b) c=4.65 in
(c ) 0.395c
p in=
(d) 4 5 18.4 = =D
(e)4
3.52d in=
(f) 3.76 in
5.13 (a) 2.17 in
(b) 60BN =
(c) 2.27Bb
r in=
7/27/2019 Mechanics of Machines Cleghorn Answers
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(d) 0.131 in
5.14 (a) Z=0.546 in
(b) 1.48c
m =
5.15 (a) 4 1.496br in=
(b) c=3.501 in
(c) 1.93cm =
(d) 0.55 in
(e) 3 24.6 =D
(f)3
10.85d in=
(g) 10.53 in
7/27/2019 Mechanics of Machines Cleghorn Answers
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P6.1
(a). 0.75 (b). ( )56.25 rpm CCW P6.2
(a).2 3 4 5
15 120 150 75N N N N= = = =
(b).2 3 4 5
39 120 150 115N N N N= = = =
P6.3
Choose1 8
150 250N N= =
5 6 1
3 4 2
2 7 3
229 171 0.8035
182 218 0.5009
267 133 0.2989
N N e
N N e
N N e
= = =
= = =
= = =
P6.4
Select 6 732 64N N= =
4 5 8 969 48 27 48N N N N = = = =
Select6
60N =
1124N =
10 16N =
7/27/2019 Mechanics of Machines Cleghorn Answers
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P6.5
1 2 3 4 5
All locked
at x rpmx x x x x
Fix 2,
move 1 at
y rpm
y 01
3
Ny
N 1
4
Ny
N 1
5
Ny
N
Absolute
rpmx y+ x 1
3
Nx y
N 1
4
Nx y
N+ 1
5
Nx y
N+
P6.6
1 2 3 4 5 6 7
All locked
& move at
x rpm
x x x x x x x
Fix 2,
move 1 at
y rpm
y 01
3
Ny
N 1
3
Ny
N 1 4
3 5
N Ny
N N 1 4
3 5
N Ny
N N 1 4 6
3 5 7
N N Ny
N N N
Absolute
rpm x y+ x 1
3
N
x yN1
3
N
x yN1 4
3 5
N N
x yN N+1 4
3 5
N N
x yN N+ 1 4 6
3 5 7
N N N
x yN N N+
P6.7
1 2 3 4 5 6
All
locked,
move at
x rpm
x x x x x x
Fix 2,
move 1
at y rpm
y 01
3
Ny
N 1
3
Ny
N 1 4
3 5
N Ny
N N 1 4
3 6
N Ny
N N
Absolute
rpmx y+ x 1
3
Nx y
N 1
3
Nx y
N 1 4
3 5
N Nx y
N N+ 1 4
3 6
N Nx y
N N+
7/27/2019 Mechanics of Machines Cleghorn Answers
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P6.8
10 4
198
25 =
10 264 rpm =
P6.9
(a). 1 4 63 5 7
1N N N
eN N N
= +
(b).1
3 5 7
1 4 6
1N N N
eN N N
=
(c).1
3 5 7
1 4 6
1N N N
eN N N
=
(d). 1 54 6
1N N
eN N
=
P6.1056 rpm
P6.11
(a).(i) 1412.5 rpm (ii) 2087.5 rpm
(b). 3n = P6.12
(a). 405 turns (opposite to indicated direction)1
108
turns (in the indicated direction)
(b). 2214 turns (opposite to indicated direction)
7/27/2019 Mechanics of Machines Cleghorn Answers
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P6.13
(a). ( )9442 rpm CW
(b).
( )6328 rpm CW
P6.14
3105N =
0.429e =
P6.15
19.61 rads
P6.16
(a)2
1
1
7e = (b)
21
1
7e =
P6.17 P6-18 P6-19
(a). 1 4 48
11 3 7 7
1N N N
eN N N N
=
+
(b).4
7
81
4
7
1
NN
eN
N
=
P6-181
17542 rpm(opposite to output)c
x = =
P6-192
5198.7 rpm(CW) =
7/27/2019 Mechanics of Machines Cleghorn Answers
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P6-20
1 1
1 2 2
1
1 1 2
4
2 6
2 2
5 7
0
x y
x x y
Nx y x
NN N
x yN N
+ = = +
=
+ =
,
condition:1 4
2 6 5 7
1 1 0 0
1 0 1 1det 0
1 / 1 0
0 0 1 /
N N
N N N N
P6-21
1
6
1
7
2
2
1 1 1 00
31 0 1 0
220
1 51/ 20 0 00
0.48 0 1 1
xN
yN
x
y
=
required relation:
6
7
1 1 1 0
31 0 122det 0
1 51/ 20 0 0
0.48 0 1 1
NN
=
P6-22 (a)2 1
28.571 rpm(CCW)x = =
(b) 2 1 28.571 rpm(CCW)x = =
P6-23 output3 5
750 rpm, 700 rpm = =
P6-24 (a) Carrier makes 12 rotation clockwise.
Angular displacement is 30rortations.
(b)
4 3
7cf g
+
=
7/27/2019 Mechanics of Machines Cleghorn Answers
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P6-25 1.4 rpm(CCW)f
=
P6-26 (a) 0.4985, 249.25 rpm(same direction as the input)ce = =
(b) 564.71 Nm(opposite to input ratation)f
T =
P6-277
19.9 rad/s =
P6-28 316.67 Nmf
T =
P6-29 (a)3
/ 5 inp =
(b) 190 rpm(same direction as the input)c
=
(c) 324.3 Nmf
T =
P6-302
7312.5 rpm(same direction as output) =
546.84 Nm(same direction as output)f
T =
P6-31 (a) 34.9 rpm(CCW)c
=
(b) 2/ 258 rpm(CCW)c =
(c) max. number: 4
P6-32 (a) 7 / 4 rev(CCW)x =
(b) 7.15 rev(CCW)
(c) 2.75 rev(CW)
7/27/2019 Mechanics of Machines Cleghorn Answers
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(d)
'C
270
'D
270
'A
'E
1
2
34
5
P6-33 (a)2
1.496 inb
r =
(b) 4.77 inc =
(c) 1.64(d)
4 5 6 739, 19, 33, 27N N N N= = = =
(e) 0.517 in
(f)4
14.8 =
(g)4
0 =
(h) 5 3.127 ind =
(i) 3.455 in
P6-34 (a) 159.3 rpm(CW)
(b) 1614 rpm(CC W)
(c) max. number: 8
7/27/2019 Mechanics of Machines Cleghorn Answers
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P7.2
=
cos12
L
s = 120L = 35
P7.4
Required length of follower face:
= max.length min.length + 2 clearance
= 13.1 - 9.6 + 2 0.5 = 4.5cm
P7.5
(b) @= 30 = 23
@= 60 = 13
@= 90 = 3
P7.6
Cam Rotation (deg)Duration
(sec)
Motion
(cm)Start of
IntervalEnd of Interval
(deg)
1 0.5 +2.0 0 1203601.5
0.5= 120
2 0.3 0 120 1923601.5
0.3120 =+ 72
3 0.7 -2.0 192 360 168
(deg)
cos s = (deg)
0 1 0
20 2/3 2.4
40 1/2 8.8
60 0 17.5
80 -1/2 26.3
100 - 2/3 32.7
120 -1 35.0
7/27/2019 Mechanics of Machines Cleghorn Answers
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(a)one rotation occurs in 1.5secsec
188.440sec
666.0sec5.1
1 radrpm
revrev==== ccw* ( *direcrtion selected )
(b)maximum velocity and acceleration of follower occur during interval I( L = 2.0cm ; = 120 = 2.094rad ; = 60)
Form Figure 7.13 :
2
2
=
Ls ;
7/27/2019 Mechanics of Machines Cleghorn Answers
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Chapter 8 Graphical Force Analysis of Planar Mechanics
Ans:
P8.1 NQNF 1800,150012
P8.2 NQblbPa 129)(,14)(
P8.3 NFNFbccwmNmNMa 24,28)(),(.090.0032.028)( 131212
P8.4 lbFlbP 5.17,5.157 41
P8.5 lbF 20P8.6 )(1.68.14.33212 cwlbininlbdFM
P8.7 Mpa2.8
P8.8 )(90012 cwcmNM
P8.9 Mpa796.0
P8.10 )(49.2 cwcmNM
P8.11 )(5.012 cwmNM
P8.12 lbF 5.10
P8.13
)(9155)(),(7.89)(),(5.55)(),(9300)( cwlbindccwlbincccwlbinbcwlbina
P8.14 )(32.0 cwmN
P8.15 )(16.4 cwcmN
P8.16 )(875)(),(710)(),(165)( ccwlbincccwlbinbccwlbina
P8.17 )(04.1)(),(71.0)(),(33.0)( ccwmNcccwmNbccwmNa
P8.18 )(7.2 ccwmN
P8.19 )(011.0 ccwmN
P8.20 )(1.8 ccwcmN
P8.21 )(85.1)(),(21.1)(),(64.0)( ccwcmNcccwcmNbccwcmNa
7/27/2019 Mechanics of Machines Cleghorn Answers
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11.1
cm786.02 r ; cm364.13 r ; cm384.14 r
11.2
cm700.11 r ; cm326.22 r ; cm047.43 r
11.3
Employing: os
100)( 2 ; cm2ss
cm881.51 r ; cm3216.52 r ; cm082.33 r
11.4
Employing: os
120)( 2 ;o
s70)( 4
cm000.11 r ; cm759.12 r ; cm724.03 r ; cm066.24 r
11.6
cm099.51 r ; cm177.12 r ; cm898.23 r ; cm225.64 r