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PROBLEM 1650
A force P of magnitude 3 N is applied to a tape wrapped around the body indicated Knowing that the body rests on a frictionless horizontal surface determine the acceleration of (a) PointA (b) PointB
A thin hoop of mass 24 kg
SOLUTION
Hoop I o
p a
m - 2Ia=mr a
a- p ) mr
(a) Acceleration ofPointA
- p P P a +ra=-+r 2-aA m mr) m
3N 2 aA =2--=25ms aA 250 ms2 - 24 kg
(b) Acceleration of Point B
- paB =a -ra=-- o m
PROPRiETARY JlJATERTAL copy 2009 The McGraw-Hill Companies Inc All reserved No part of this Manual may be displayed reproduced or distribllled in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribUlion to teachers and educators permilled by McGraw-Hillfor their individual course preparation Jfyou are a student using this Manual you are using it without permission
1401
11 PROBLEM 1655
B A 3-kg sprocket wheel has a centroidal radius of gyration of70 mm and is suspended from a chain as shown Determine the acceleration of Points A and B of the chain knowing that T j == 14 Nand == 18 N
SOLUTION
+h=lS + TB
I==mP (3 kg)(O07 m)2
147xl0-3 kgmiddotm2
r=O08 m W=mg
(3 kg)(981 rnIs2 )
W 2943 N
W == ma TA + TB 2943 N (3 kg)a
1-(T +3 A -2943) (1)
+IMG I(Mdelf TB(OOSm)-T4(OOSm) fa
(TB m)==(147xlO-3 kgmiddotm2 )a
+a=5442 (TB -TA ) (2)
== 14 N TB == 18 N
Eg (1) 18 3
2943) 08567 rnIs2
Eg (2) +a== 5442(18-14) 21769 rads2
PROPRIETARY MATERIAL (9 2009 The McGraw-Hill Companies Inc All rights reserved No part 0 this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited disiriblllion to teachers and educators permilled by AlcGraw-HilIor their individual course preparatioll lfyou are a stlldem lIsing this Manual you are using it withollt permission
1407
fWOf4M[t-J OOttampL_ -bullAilS maampIIamp-aHF WW JiWJiZWLiHk
PROBLEM 1655 (Continued)
-+ta A (OA)t
a+ra 08567 - (008)(21769)
=-0885 ms2 0885 ms2
+taB =(oB)t =o+ra
08567 + (008)(21769)
=+260 ms2 260 ms 2 t
PROPRlETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrilten permission of the publisher or used beyond the limited distribution to teachers and educators permilted by McGrav-Hilifor their individual course preparation Ifyou are a student using this 1v1anuol YOlt are using it without permission
1408
=ze-
PROBLEM 1697
A homogeneous sphere S a uniform cylinder C and a thin pipe P are in contact when they are released from rest on the incline shown Knowing that all three objects roll without slipping determine after 4 s of motion the clear distance between (a) the pipe and the cylinder (b) the cylinder and the sphere
SOLUTION
General case I mP a=ra
-2 mgsmf3r=mk a+mr-a c IJiIf-
a
a =ra sin f3 (1)
For pipe k r
r -=----==-
S1l1 =-g5mf3 1 f3 2
-2 1 - f3 2 f3For cylinder k =- 5111 = - g S1l1 2 3
2
For sphere =-2 ----g smf3 -5 g 5111f3 5 ) 2 2 7 r- + r
5
(a) Between pipe and cylinder
21 ( 1 6gS1I1 f3) 1 r
Sl units ms2 jsin100 (45)2 =227m laquo1
US units ftJ5 2)sin 10deg(4 S)2 746 ft
PROPRJETARY jfATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Afmlllal may be displayed reproduced or distributed ill any form 01 by allY means without the prior wrillen permission of the publisher or used beyond the limited distributionta teachers and educators permitted by lvfcGraw-HiIlfor their individual cOme preparation lfyoll are a student using this Aianual you are using it without permission
1464
7
PROBLEM 1697 (Continued)
Between sphere and cylinder
SID3 = 1 gS111321
gSin3)t2 2 21
0SI units Xsc =21(121981mls2j511110 (4s)2 0649 m
US units 213 ft
PROPRIETARY MATERL4L 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators pemlitled by lifcGraw-Hillfor their individual course preparatiol1fyou are a siudent using lhis Afanual Y01l are using if withoul permission
1465
Ii
B
PROBLEM 16117
The ends of the lO-kg uniform rod AB are attached to collars of negligible mass that slide without friction along fixed rods If the rod is released from rest when f) = 25deg determine immediately after release (a) the angular acceleration of the rod (b) the reaction at A (c) the reaction at B
SOLUTION
Kinematics Assume a) (j)=O
e lzso
l-8A-r(middot1)rJ
aA+aB1A =[aA --- ]+[12a125deg]
-- aB (l2a)cos 25deg =10876a
(12a) sin 25deg =O5071aaA
aG =aA +aGIA =[aA ]+[O6a 125deg]
=[O5071a - ] + [O6a1 25deg]aG
ax =(aG)x =[O5071a ---+ ]+[O2536a --]
ax = O2535a -
ay [O6acos 25deg-] =O5438a
aWe have found for a ) ax O2535a
y 05438a
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc AU rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
]494
7
PROBLEM 16117 (Continued)
1 2-m(l2 m)Kinetics 12
(a) Angular acceleration
+) fME f(ME)eff mg(05438 m) Ia+max(02535 m)+may(05438 m)
m(12)2 +m(02535)2 a + m(05438)2 a 12
g(05438) 048a a 1l33g a 1111 rads2 )
(b) + t rPy = A - mg = -may = m(05438a)
A 10(981) -(10)(05438)(11115)
A = 981 - 6044 3766 N A 377 Nt
B = max m(02535a)
B = 10(02535)(11115) B =2818 N B
(c)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribution to teachers and educators permilled by Jv1cGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using itlvithout permission
1495
c
PROBLEM 1638
Two unifonn disks and two cylinders are assembled as indicated Disk A weighs 20 lb and disk B weighs 12 lb Knowing that the system is released from rest detennine the acceleration (a) of cylinder (b) of cylinder D
The cylinders are attached to a single cord that passes over the disks Assume that no slipping occurs between the cord and the disks
D
SOLUTION
Moments of inertia
Kinematics
-Kinetics
DiskA
1 2 1 20lb (8 ftfA mArA 013803 lbmiddot ftmiddot 52 2 2322 ftls 2 12 )
1 2 1 12 Ib (6 f IB -mBrB 004658 lbmiddot ft S2
2 2 322 ftls 2 12 t
a a ==l5a
a a = =2aaB rB 1
FAB == Tension in cord between disks
+)LMA =LMA
+mca ftJ 2 (2)15lb - FAB --10 == (013803)(150) + al-3 322 fls L 3
2 -10 =020705a + 031 056a 3 2 -10 =051760a 3
FAB = 15+077641a (1)
PROPRIETARY lIJATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No parI of this Manual may be displayed reproduced or distributed in any form or by means WilhoUl the prior wrillen permission of the publisher or used beyond the limiled distribution to teachers McGraw-Hill for their individual course preparation fYOII are a studem using this ivfanual you are Ising it wilholll permission
1380
7 s
PROBLEM 1538 (Continued)
DiskB ) 2MB = 2(A1B)eff
(6 (6 ft = 1flaB +llID( --ft)
12 12 9 =(004658)(2a)
1 322 ftls- 2)
9 05F4B = 009316a + 02795a
Substitute for FAB from Eq (l)
9 05(15 +077641a) =037266a 9 75 03882a = 037266a
15 076086a
a =1971 ftls2
Both ae and aD have the same magnitude
(a) Acceleration of C ae =1971 ftls2t
(b) Acceleration ofD aD = 1971 ftls2
PROPRIETARY MATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrillen permission of the publisher or lIsed beyond the limied distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyou are a student using lhis Manual you are using il withoul permission
1381
r
PROBLEM 1678
A unifonn slender rod of length L 900 mm and mass m 4 kg is suspended from a hinge at C A horizontal force P of magnitude 75 N is applied at end B Knowing that r 225 mm determine (a) the angular acceleration of the rod (b) the components of the reaction at C
SOLUTION It (a) Angular acceleration tf
a=ra
p[r + +la
- 1
( -)- 1 L7= mra r + m-a B 1 (-2
12 I L 1 1p(-r +- =m r + 2 J 12
Substitute data m + 09 m =(4 kg)[(0225 m)2 m)2la 15 N 2 12
50125 = 04725a
a = 10714 rads2 (l =1071 rads2
)
(b) Components of reaction at C
+ t = l(Fy)eff
Cy W == mg (4 kg)(981 mls2 ) C 392 Nt y
P=-ma r ex P-ma= P-m(fa)
0714 rads2=75 N-
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be (ffsJlalverJ repoduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the distribution to teachers and educators permilled by lvfcGraw-Hillfor their individual course preparation Ifyou are a student IISlI1g this Afanlal you are using il without permission
1439
PROBLEM 16109
Two unifonll disks A and B each of weight 4 lb are connected by a 3-lb rod CD as shown A counterclockwise couple M of moment 15 lbmiddot ft is applied to disk A Knowing that the disks roll without sliding determine (a) the acceleration of the center of each disk (b) the horizontal component of the force exerted on disk B D
SOLUTION
-
DiskA
+)LMs=L(M)eff
W=41b
- 1 2 =-mr 2
M-c(ftl (ma)r+a12
15lbmiddotft C 12
1 2(mra)r+-mr a 2
3=-mr21f 2
l51bmiddotft C = 3 4lb 12 2322
6 ft)2 a 12
15 O046584a 3
(1)
DiskB I
(ma)r+ a 2 1 0 mr 2
Rod CD
D =O046684a 8
(2)
6 2 8
6 2ftia 12 )
=-a+-a=-a 12 12 12
PROPRIE14RY IUATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or dislributed in any form or by any means without the prior written permission of the publisher or lIsed beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyoll are a student using this Manual Y01l are using it without permission
1480
PROBLEM 16109 (Continued)
Multiply
2 0041408a (3)By 3 3 3
Add (1) (2) (3) 15-1e +1D 1e-1D 0134576a
II =1 1146 rads2 )
6 7 ft(l1l46 Tads) =5573 ftls-(a)
12 )
(b) Substitute for a in (2)
D 0046584(11146) D 07791b -- 3
PROPRIETAllJMATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrilten permission athe or used he yond the limited dislrihllliol1ta teachers and educators permilled by McGraw-Hilior their individual course preparaliol1 are a stdl11I using Ihis Manual you are using it withoul permission
1481
__4mQmQbullbull a za I
I
7 7
I((r PROBLEM 16145
A unifonn slender bar AB of mass In is suspended as shown from a unifonn disk of the same mass m Neglecting the effect of friction detennine the accelerations of Points A and B immediately after a horizontal force P has been applied at B
Tiii
SOLUTION
Kinematics
Cylinder Rolling without sliding
RodAB )r
Yz L -GAB0ts Go
42poundIM f3
Cylinder
regtllt tl =-r C
C (1) t c
RodAB
_ L -PL = mGAB 2+ I a AB
PL (2)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No parI of this Manual may be dispJayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1553
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554
11 PROBLEM 1655
B A 3-kg sprocket wheel has a centroidal radius of gyration of70 mm and is suspended from a chain as shown Determine the acceleration of Points A and B of the chain knowing that T j == 14 Nand == 18 N
SOLUTION
+h=lS + TB
I==mP (3 kg)(O07 m)2
147xl0-3 kgmiddotm2
r=O08 m W=mg
(3 kg)(981 rnIs2 )
W 2943 N
W == ma TA + TB 2943 N (3 kg)a
1-(T +3 A -2943) (1)
+IMG I(Mdelf TB(OOSm)-T4(OOSm) fa
(TB m)==(147xlO-3 kgmiddotm2 )a
+a=5442 (TB -TA ) (2)
== 14 N TB == 18 N
Eg (1) 18 3
2943) 08567 rnIs2
Eg (2) +a== 5442(18-14) 21769 rads2
PROPRIETARY MATERIAL (9 2009 The McGraw-Hill Companies Inc All rights reserved No part 0 this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited disiriblllion to teachers and educators permilled by AlcGraw-HilIor their individual course preparatioll lfyou are a stlldem lIsing this Manual you are using it withollt permission
1407
fWOf4M[t-J OOttampL_ -bullAilS maampIIamp-aHF WW JiWJiZWLiHk
PROBLEM 1655 (Continued)
-+ta A (OA)t
a+ra 08567 - (008)(21769)
=-0885 ms2 0885 ms2
+taB =(oB)t =o+ra
08567 + (008)(21769)
=+260 ms2 260 ms 2 t
PROPRlETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrilten permission of the publisher or used beyond the limited distribution to teachers and educators permilted by McGrav-Hilifor their individual course preparation Ifyou are a student using this 1v1anuol YOlt are using it without permission
1408
=ze-
PROBLEM 1697
A homogeneous sphere S a uniform cylinder C and a thin pipe P are in contact when they are released from rest on the incline shown Knowing that all three objects roll without slipping determine after 4 s of motion the clear distance between (a) the pipe and the cylinder (b) the cylinder and the sphere
SOLUTION
General case I mP a=ra
-2 mgsmf3r=mk a+mr-a c IJiIf-
a
a =ra sin f3 (1)
For pipe k r
r -=----==-
S1l1 =-g5mf3 1 f3 2
-2 1 - f3 2 f3For cylinder k =- 5111 = - g S1l1 2 3
2
For sphere =-2 ----g smf3 -5 g 5111f3 5 ) 2 2 7 r- + r
5
(a) Between pipe and cylinder
21 ( 1 6gS1I1 f3) 1 r
Sl units ms2 jsin100 (45)2 =227m laquo1
US units ftJ5 2)sin 10deg(4 S)2 746 ft
PROPRJETARY jfATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Afmlllal may be displayed reproduced or distributed ill any form 01 by allY means without the prior wrillen permission of the publisher or used beyond the limited distributionta teachers and educators permitted by lvfcGraw-HiIlfor their individual cOme preparation lfyoll are a student using this Aianual you are using it without permission
1464
7
PROBLEM 1697 (Continued)
Between sphere and cylinder
SID3 = 1 gS111321
gSin3)t2 2 21
0SI units Xsc =21(121981mls2j511110 (4s)2 0649 m
US units 213 ft
PROPRIETARY MATERL4L 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators pemlitled by lifcGraw-Hillfor their individual course preparatiol1fyou are a siudent using lhis Afanual Y01l are using if withoul permission
1465
Ii
B
PROBLEM 16117
The ends of the lO-kg uniform rod AB are attached to collars of negligible mass that slide without friction along fixed rods If the rod is released from rest when f) = 25deg determine immediately after release (a) the angular acceleration of the rod (b) the reaction at A (c) the reaction at B
SOLUTION
Kinematics Assume a) (j)=O
e lzso
l-8A-r(middot1)rJ
aA+aB1A =[aA --- ]+[12a125deg]
-- aB (l2a)cos 25deg =10876a
(12a) sin 25deg =O5071aaA
aG =aA +aGIA =[aA ]+[O6a 125deg]
=[O5071a - ] + [O6a1 25deg]aG
ax =(aG)x =[O5071a ---+ ]+[O2536a --]
ax = O2535a -
ay [O6acos 25deg-] =O5438a
aWe have found for a ) ax O2535a
y 05438a
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc AU rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
]494
7
PROBLEM 16117 (Continued)
1 2-m(l2 m)Kinetics 12
(a) Angular acceleration
+) fME f(ME)eff mg(05438 m) Ia+max(02535 m)+may(05438 m)
m(12)2 +m(02535)2 a + m(05438)2 a 12
g(05438) 048a a 1l33g a 1111 rads2 )
(b) + t rPy = A - mg = -may = m(05438a)
A 10(981) -(10)(05438)(11115)
A = 981 - 6044 3766 N A 377 Nt
B = max m(02535a)
B = 10(02535)(11115) B =2818 N B
(c)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribution to teachers and educators permilled by Jv1cGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using itlvithout permission
1495
c
PROBLEM 1638
Two unifonn disks and two cylinders are assembled as indicated Disk A weighs 20 lb and disk B weighs 12 lb Knowing that the system is released from rest detennine the acceleration (a) of cylinder (b) of cylinder D
The cylinders are attached to a single cord that passes over the disks Assume that no slipping occurs between the cord and the disks
D
SOLUTION
Moments of inertia
Kinematics
-Kinetics
DiskA
1 2 1 20lb (8 ftfA mArA 013803 lbmiddot ftmiddot 52 2 2322 ftls 2 12 )
1 2 1 12 Ib (6 f IB -mBrB 004658 lbmiddot ft S2
2 2 322 ftls 2 12 t
a a ==l5a
a a = =2aaB rB 1
FAB == Tension in cord between disks
+)LMA =LMA
+mca ftJ 2 (2)15lb - FAB --10 == (013803)(150) + al-3 322 fls L 3
2 -10 =020705a + 031 056a 3 2 -10 =051760a 3
FAB = 15+077641a (1)
PROPRIETARY lIJATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No parI of this Manual may be displayed reproduced or distributed in any form or by means WilhoUl the prior wrillen permission of the publisher or used beyond the limiled distribution to teachers McGraw-Hill for their individual course preparation fYOII are a studem using this ivfanual you are Ising it wilholll permission
1380
7 s
PROBLEM 1538 (Continued)
DiskB ) 2MB = 2(A1B)eff
(6 (6 ft = 1flaB +llID( --ft)
12 12 9 =(004658)(2a)
1 322 ftls- 2)
9 05F4B = 009316a + 02795a
Substitute for FAB from Eq (l)
9 05(15 +077641a) =037266a 9 75 03882a = 037266a
15 076086a
a =1971 ftls2
Both ae and aD have the same magnitude
(a) Acceleration of C ae =1971 ftls2t
(b) Acceleration ofD aD = 1971 ftls2
PROPRIETARY MATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrillen permission of the publisher or lIsed beyond the limied distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyou are a student using lhis Manual you are using il withoul permission
1381
r
PROBLEM 1678
A unifonn slender rod of length L 900 mm and mass m 4 kg is suspended from a hinge at C A horizontal force P of magnitude 75 N is applied at end B Knowing that r 225 mm determine (a) the angular acceleration of the rod (b) the components of the reaction at C
SOLUTION It (a) Angular acceleration tf
a=ra
p[r + +la
- 1
( -)- 1 L7= mra r + m-a B 1 (-2
12 I L 1 1p(-r +- =m r + 2 J 12
Substitute data m + 09 m =(4 kg)[(0225 m)2 m)2la 15 N 2 12
50125 = 04725a
a = 10714 rads2 (l =1071 rads2
)
(b) Components of reaction at C
+ t = l(Fy)eff
Cy W == mg (4 kg)(981 mls2 ) C 392 Nt y
P=-ma r ex P-ma= P-m(fa)
0714 rads2=75 N-
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be (ffsJlalverJ repoduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the distribution to teachers and educators permilled by lvfcGraw-Hillfor their individual course preparation Ifyou are a student IISlI1g this Afanlal you are using il without permission
1439
PROBLEM 16109
Two unifonll disks A and B each of weight 4 lb are connected by a 3-lb rod CD as shown A counterclockwise couple M of moment 15 lbmiddot ft is applied to disk A Knowing that the disks roll without sliding determine (a) the acceleration of the center of each disk (b) the horizontal component of the force exerted on disk B D
SOLUTION
-
DiskA
+)LMs=L(M)eff
W=41b
- 1 2 =-mr 2
M-c(ftl (ma)r+a12
15lbmiddotft C 12
1 2(mra)r+-mr a 2
3=-mr21f 2
l51bmiddotft C = 3 4lb 12 2322
6 ft)2 a 12
15 O046584a 3
(1)
DiskB I
(ma)r+ a 2 1 0 mr 2
Rod CD
D =O046684a 8
(2)
6 2 8
6 2ftia 12 )
=-a+-a=-a 12 12 12
PROPRIE14RY IUATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or dislributed in any form or by any means without the prior written permission of the publisher or lIsed beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyoll are a student using this Manual Y01l are using it without permission
1480
PROBLEM 16109 (Continued)
Multiply
2 0041408a (3)By 3 3 3
Add (1) (2) (3) 15-1e +1D 1e-1D 0134576a
II =1 1146 rads2 )
6 7 ft(l1l46 Tads) =5573 ftls-(a)
12 )
(b) Substitute for a in (2)
D 0046584(11146) D 07791b -- 3
PROPRIETAllJMATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrilten permission athe or used he yond the limited dislrihllliol1ta teachers and educators permilled by McGraw-Hilior their individual course preparaliol1 are a stdl11I using Ihis Manual you are using it withoul permission
1481
__4mQmQbullbull a za I
I
7 7
I((r PROBLEM 16145
A unifonn slender bar AB of mass In is suspended as shown from a unifonn disk of the same mass m Neglecting the effect of friction detennine the accelerations of Points A and B immediately after a horizontal force P has been applied at B
Tiii
SOLUTION
Kinematics
Cylinder Rolling without sliding
RodAB )r
Yz L -GAB0ts Go
42poundIM f3
Cylinder
regtllt tl =-r C
C (1) t c
RodAB
_ L -PL = mGAB 2+ I a AB
PL (2)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No parI of this Manual may be dispJayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1553
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554
PROBLEM 1655 (Continued)
-+ta A (OA)t
a+ra 08567 - (008)(21769)
=-0885 ms2 0885 ms2
+taB =(oB)t =o+ra
08567 + (008)(21769)
=+260 ms2 260 ms 2 t
PROPRlETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrilten permission of the publisher or used beyond the limited distribution to teachers and educators permilted by McGrav-Hilifor their individual course preparation Ifyou are a student using this 1v1anuol YOlt are using it without permission
1408
=ze-
PROBLEM 1697
A homogeneous sphere S a uniform cylinder C and a thin pipe P are in contact when they are released from rest on the incline shown Knowing that all three objects roll without slipping determine after 4 s of motion the clear distance between (a) the pipe and the cylinder (b) the cylinder and the sphere
SOLUTION
General case I mP a=ra
-2 mgsmf3r=mk a+mr-a c IJiIf-
a
a =ra sin f3 (1)
For pipe k r
r -=----==-
S1l1 =-g5mf3 1 f3 2
-2 1 - f3 2 f3For cylinder k =- 5111 = - g S1l1 2 3
2
For sphere =-2 ----g smf3 -5 g 5111f3 5 ) 2 2 7 r- + r
5
(a) Between pipe and cylinder
21 ( 1 6gS1I1 f3) 1 r
Sl units ms2 jsin100 (45)2 =227m laquo1
US units ftJ5 2)sin 10deg(4 S)2 746 ft
PROPRJETARY jfATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Afmlllal may be displayed reproduced or distributed ill any form 01 by allY means without the prior wrillen permission of the publisher or used beyond the limited distributionta teachers and educators permitted by lvfcGraw-HiIlfor their individual cOme preparation lfyoll are a student using this Aianual you are using it without permission
1464
7
PROBLEM 1697 (Continued)
Between sphere and cylinder
SID3 = 1 gS111321
gSin3)t2 2 21
0SI units Xsc =21(121981mls2j511110 (4s)2 0649 m
US units 213 ft
PROPRIETARY MATERL4L 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators pemlitled by lifcGraw-Hillfor their individual course preparatiol1fyou are a siudent using lhis Afanual Y01l are using if withoul permission
1465
Ii
B
PROBLEM 16117
The ends of the lO-kg uniform rod AB are attached to collars of negligible mass that slide without friction along fixed rods If the rod is released from rest when f) = 25deg determine immediately after release (a) the angular acceleration of the rod (b) the reaction at A (c) the reaction at B
SOLUTION
Kinematics Assume a) (j)=O
e lzso
l-8A-r(middot1)rJ
aA+aB1A =[aA --- ]+[12a125deg]
-- aB (l2a)cos 25deg =10876a
(12a) sin 25deg =O5071aaA
aG =aA +aGIA =[aA ]+[O6a 125deg]
=[O5071a - ] + [O6a1 25deg]aG
ax =(aG)x =[O5071a ---+ ]+[O2536a --]
ax = O2535a -
ay [O6acos 25deg-] =O5438a
aWe have found for a ) ax O2535a
y 05438a
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc AU rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
]494
7
PROBLEM 16117 (Continued)
1 2-m(l2 m)Kinetics 12
(a) Angular acceleration
+) fME f(ME)eff mg(05438 m) Ia+max(02535 m)+may(05438 m)
m(12)2 +m(02535)2 a + m(05438)2 a 12
g(05438) 048a a 1l33g a 1111 rads2 )
(b) + t rPy = A - mg = -may = m(05438a)
A 10(981) -(10)(05438)(11115)
A = 981 - 6044 3766 N A 377 Nt
B = max m(02535a)
B = 10(02535)(11115) B =2818 N B
(c)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribution to teachers and educators permilled by Jv1cGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using itlvithout permission
1495
c
PROBLEM 1638
Two unifonn disks and two cylinders are assembled as indicated Disk A weighs 20 lb and disk B weighs 12 lb Knowing that the system is released from rest detennine the acceleration (a) of cylinder (b) of cylinder D
The cylinders are attached to a single cord that passes over the disks Assume that no slipping occurs between the cord and the disks
D
SOLUTION
Moments of inertia
Kinematics
-Kinetics
DiskA
1 2 1 20lb (8 ftfA mArA 013803 lbmiddot ftmiddot 52 2 2322 ftls 2 12 )
1 2 1 12 Ib (6 f IB -mBrB 004658 lbmiddot ft S2
2 2 322 ftls 2 12 t
a a ==l5a
a a = =2aaB rB 1
FAB == Tension in cord between disks
+)LMA =LMA
+mca ftJ 2 (2)15lb - FAB --10 == (013803)(150) + al-3 322 fls L 3
2 -10 =020705a + 031 056a 3 2 -10 =051760a 3
FAB = 15+077641a (1)
PROPRIETARY lIJATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No parI of this Manual may be displayed reproduced or distributed in any form or by means WilhoUl the prior wrillen permission of the publisher or used beyond the limiled distribution to teachers McGraw-Hill for their individual course preparation fYOII are a studem using this ivfanual you are Ising it wilholll permission
1380
7 s
PROBLEM 1538 (Continued)
DiskB ) 2MB = 2(A1B)eff
(6 (6 ft = 1flaB +llID( --ft)
12 12 9 =(004658)(2a)
1 322 ftls- 2)
9 05F4B = 009316a + 02795a
Substitute for FAB from Eq (l)
9 05(15 +077641a) =037266a 9 75 03882a = 037266a
15 076086a
a =1971 ftls2
Both ae and aD have the same magnitude
(a) Acceleration of C ae =1971 ftls2t
(b) Acceleration ofD aD = 1971 ftls2
PROPRIETARY MATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrillen permission of the publisher or lIsed beyond the limied distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyou are a student using lhis Manual you are using il withoul permission
1381
r
PROBLEM 1678
A unifonn slender rod of length L 900 mm and mass m 4 kg is suspended from a hinge at C A horizontal force P of magnitude 75 N is applied at end B Knowing that r 225 mm determine (a) the angular acceleration of the rod (b) the components of the reaction at C
SOLUTION It (a) Angular acceleration tf
a=ra
p[r + +la
- 1
( -)- 1 L7= mra r + m-a B 1 (-2
12 I L 1 1p(-r +- =m r + 2 J 12
Substitute data m + 09 m =(4 kg)[(0225 m)2 m)2la 15 N 2 12
50125 = 04725a
a = 10714 rads2 (l =1071 rads2
)
(b) Components of reaction at C
+ t = l(Fy)eff
Cy W == mg (4 kg)(981 mls2 ) C 392 Nt y
P=-ma r ex P-ma= P-m(fa)
0714 rads2=75 N-
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be (ffsJlalverJ repoduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the distribution to teachers and educators permilled by lvfcGraw-Hillfor their individual course preparation Ifyou are a student IISlI1g this Afanlal you are using il without permission
1439
PROBLEM 16109
Two unifonll disks A and B each of weight 4 lb are connected by a 3-lb rod CD as shown A counterclockwise couple M of moment 15 lbmiddot ft is applied to disk A Knowing that the disks roll without sliding determine (a) the acceleration of the center of each disk (b) the horizontal component of the force exerted on disk B D
SOLUTION
-
DiskA
+)LMs=L(M)eff
W=41b
- 1 2 =-mr 2
M-c(ftl (ma)r+a12
15lbmiddotft C 12
1 2(mra)r+-mr a 2
3=-mr21f 2
l51bmiddotft C = 3 4lb 12 2322
6 ft)2 a 12
15 O046584a 3
(1)
DiskB I
(ma)r+ a 2 1 0 mr 2
Rod CD
D =O046684a 8
(2)
6 2 8
6 2ftia 12 )
=-a+-a=-a 12 12 12
PROPRIE14RY IUATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or dislributed in any form or by any means without the prior written permission of the publisher or lIsed beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyoll are a student using this Manual Y01l are using it without permission
1480
PROBLEM 16109 (Continued)
Multiply
2 0041408a (3)By 3 3 3
Add (1) (2) (3) 15-1e +1D 1e-1D 0134576a
II =1 1146 rads2 )
6 7 ft(l1l46 Tads) =5573 ftls-(a)
12 )
(b) Substitute for a in (2)
D 0046584(11146) D 07791b -- 3
PROPRIETAllJMATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrilten permission athe or used he yond the limited dislrihllliol1ta teachers and educators permilled by McGraw-Hilior their individual course preparaliol1 are a stdl11I using Ihis Manual you are using it withoul permission
1481
__4mQmQbullbull a za I
I
7 7
I((r PROBLEM 16145
A unifonn slender bar AB of mass In is suspended as shown from a unifonn disk of the same mass m Neglecting the effect of friction detennine the accelerations of Points A and B immediately after a horizontal force P has been applied at B
Tiii
SOLUTION
Kinematics
Cylinder Rolling without sliding
RodAB )r
Yz L -GAB0ts Go
42poundIM f3
Cylinder
regtllt tl =-r C
C (1) t c
RodAB
_ L -PL = mGAB 2+ I a AB
PL (2)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No parI of this Manual may be dispJayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1553
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554
=ze-
PROBLEM 1697
A homogeneous sphere S a uniform cylinder C and a thin pipe P are in contact when they are released from rest on the incline shown Knowing that all three objects roll without slipping determine after 4 s of motion the clear distance between (a) the pipe and the cylinder (b) the cylinder and the sphere
SOLUTION
General case I mP a=ra
-2 mgsmf3r=mk a+mr-a c IJiIf-
a
a =ra sin f3 (1)
For pipe k r
r -=----==-
S1l1 =-g5mf3 1 f3 2
-2 1 - f3 2 f3For cylinder k =- 5111 = - g S1l1 2 3
2
For sphere =-2 ----g smf3 -5 g 5111f3 5 ) 2 2 7 r- + r
5
(a) Between pipe and cylinder
21 ( 1 6gS1I1 f3) 1 r
Sl units ms2 jsin100 (45)2 =227m laquo1
US units ftJ5 2)sin 10deg(4 S)2 746 ft
PROPRJETARY jfATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Afmlllal may be displayed reproduced or distributed ill any form 01 by allY means without the prior wrillen permission of the publisher or used beyond the limited distributionta teachers and educators permitted by lvfcGraw-HiIlfor their individual cOme preparation lfyoll are a student using this Aianual you are using it without permission
1464
7
PROBLEM 1697 (Continued)
Between sphere and cylinder
SID3 = 1 gS111321
gSin3)t2 2 21
0SI units Xsc =21(121981mls2j511110 (4s)2 0649 m
US units 213 ft
PROPRIETARY MATERL4L 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators pemlitled by lifcGraw-Hillfor their individual course preparatiol1fyou are a siudent using lhis Afanual Y01l are using if withoul permission
1465
Ii
B
PROBLEM 16117
The ends of the lO-kg uniform rod AB are attached to collars of negligible mass that slide without friction along fixed rods If the rod is released from rest when f) = 25deg determine immediately after release (a) the angular acceleration of the rod (b) the reaction at A (c) the reaction at B
SOLUTION
Kinematics Assume a) (j)=O
e lzso
l-8A-r(middot1)rJ
aA+aB1A =[aA --- ]+[12a125deg]
-- aB (l2a)cos 25deg =10876a
(12a) sin 25deg =O5071aaA
aG =aA +aGIA =[aA ]+[O6a 125deg]
=[O5071a - ] + [O6a1 25deg]aG
ax =(aG)x =[O5071a ---+ ]+[O2536a --]
ax = O2535a -
ay [O6acos 25deg-] =O5438a
aWe have found for a ) ax O2535a
y 05438a
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc AU rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
]494
7
PROBLEM 16117 (Continued)
1 2-m(l2 m)Kinetics 12
(a) Angular acceleration
+) fME f(ME)eff mg(05438 m) Ia+max(02535 m)+may(05438 m)
m(12)2 +m(02535)2 a + m(05438)2 a 12
g(05438) 048a a 1l33g a 1111 rads2 )
(b) + t rPy = A - mg = -may = m(05438a)
A 10(981) -(10)(05438)(11115)
A = 981 - 6044 3766 N A 377 Nt
B = max m(02535a)
B = 10(02535)(11115) B =2818 N B
(c)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribution to teachers and educators permilled by Jv1cGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using itlvithout permission
1495
c
PROBLEM 1638
Two unifonn disks and two cylinders are assembled as indicated Disk A weighs 20 lb and disk B weighs 12 lb Knowing that the system is released from rest detennine the acceleration (a) of cylinder (b) of cylinder D
The cylinders are attached to a single cord that passes over the disks Assume that no slipping occurs between the cord and the disks
D
SOLUTION
Moments of inertia
Kinematics
-Kinetics
DiskA
1 2 1 20lb (8 ftfA mArA 013803 lbmiddot ftmiddot 52 2 2322 ftls 2 12 )
1 2 1 12 Ib (6 f IB -mBrB 004658 lbmiddot ft S2
2 2 322 ftls 2 12 t
a a ==l5a
a a = =2aaB rB 1
FAB == Tension in cord between disks
+)LMA =LMA
+mca ftJ 2 (2)15lb - FAB --10 == (013803)(150) + al-3 322 fls L 3
2 -10 =020705a + 031 056a 3 2 -10 =051760a 3
FAB = 15+077641a (1)
PROPRIETARY lIJATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No parI of this Manual may be displayed reproduced or distributed in any form or by means WilhoUl the prior wrillen permission of the publisher or used beyond the limiled distribution to teachers McGraw-Hill for their individual course preparation fYOII are a studem using this ivfanual you are Ising it wilholll permission
1380
7 s
PROBLEM 1538 (Continued)
DiskB ) 2MB = 2(A1B)eff
(6 (6 ft = 1flaB +llID( --ft)
12 12 9 =(004658)(2a)
1 322 ftls- 2)
9 05F4B = 009316a + 02795a
Substitute for FAB from Eq (l)
9 05(15 +077641a) =037266a 9 75 03882a = 037266a
15 076086a
a =1971 ftls2
Both ae and aD have the same magnitude
(a) Acceleration of C ae =1971 ftls2t
(b) Acceleration ofD aD = 1971 ftls2
PROPRIETARY MATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrillen permission of the publisher or lIsed beyond the limied distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyou are a student using lhis Manual you are using il withoul permission
1381
r
PROBLEM 1678
A unifonn slender rod of length L 900 mm and mass m 4 kg is suspended from a hinge at C A horizontal force P of magnitude 75 N is applied at end B Knowing that r 225 mm determine (a) the angular acceleration of the rod (b) the components of the reaction at C
SOLUTION It (a) Angular acceleration tf
a=ra
p[r + +la
- 1
( -)- 1 L7= mra r + m-a B 1 (-2
12 I L 1 1p(-r +- =m r + 2 J 12
Substitute data m + 09 m =(4 kg)[(0225 m)2 m)2la 15 N 2 12
50125 = 04725a
a = 10714 rads2 (l =1071 rads2
)
(b) Components of reaction at C
+ t = l(Fy)eff
Cy W == mg (4 kg)(981 mls2 ) C 392 Nt y
P=-ma r ex P-ma= P-m(fa)
0714 rads2=75 N-
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be (ffsJlalverJ repoduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the distribution to teachers and educators permilled by lvfcGraw-Hillfor their individual course preparation Ifyou are a student IISlI1g this Afanlal you are using il without permission
1439
PROBLEM 16109
Two unifonll disks A and B each of weight 4 lb are connected by a 3-lb rod CD as shown A counterclockwise couple M of moment 15 lbmiddot ft is applied to disk A Knowing that the disks roll without sliding determine (a) the acceleration of the center of each disk (b) the horizontal component of the force exerted on disk B D
SOLUTION
-
DiskA
+)LMs=L(M)eff
W=41b
- 1 2 =-mr 2
M-c(ftl (ma)r+a12
15lbmiddotft C 12
1 2(mra)r+-mr a 2
3=-mr21f 2
l51bmiddotft C = 3 4lb 12 2322
6 ft)2 a 12
15 O046584a 3
(1)
DiskB I
(ma)r+ a 2 1 0 mr 2
Rod CD
D =O046684a 8
(2)
6 2 8
6 2ftia 12 )
=-a+-a=-a 12 12 12
PROPRIE14RY IUATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or dislributed in any form or by any means without the prior written permission of the publisher or lIsed beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyoll are a student using this Manual Y01l are using it without permission
1480
PROBLEM 16109 (Continued)
Multiply
2 0041408a (3)By 3 3 3
Add (1) (2) (3) 15-1e +1D 1e-1D 0134576a
II =1 1146 rads2 )
6 7 ft(l1l46 Tads) =5573 ftls-(a)
12 )
(b) Substitute for a in (2)
D 0046584(11146) D 07791b -- 3
PROPRIETAllJMATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrilten permission athe or used he yond the limited dislrihllliol1ta teachers and educators permilled by McGraw-Hilior their individual course preparaliol1 are a stdl11I using Ihis Manual you are using it withoul permission
1481
__4mQmQbullbull a za I
I
7 7
I((r PROBLEM 16145
A unifonn slender bar AB of mass In is suspended as shown from a unifonn disk of the same mass m Neglecting the effect of friction detennine the accelerations of Points A and B immediately after a horizontal force P has been applied at B
Tiii
SOLUTION
Kinematics
Cylinder Rolling without sliding
RodAB )r
Yz L -GAB0ts Go
42poundIM f3
Cylinder
regtllt tl =-r C
C (1) t c
RodAB
_ L -PL = mGAB 2+ I a AB
PL (2)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No parI of this Manual may be dispJayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1553
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554
7
PROBLEM 1697 (Continued)
Between sphere and cylinder
SID3 = 1 gS111321
gSin3)t2 2 21
0SI units Xsc =21(121981mls2j511110 (4s)2 0649 m
US units 213 ft
PROPRIETARY MATERL4L 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators pemlitled by lifcGraw-Hillfor their individual course preparatiol1fyou are a siudent using lhis Afanual Y01l are using if withoul permission
1465
Ii
B
PROBLEM 16117
The ends of the lO-kg uniform rod AB are attached to collars of negligible mass that slide without friction along fixed rods If the rod is released from rest when f) = 25deg determine immediately after release (a) the angular acceleration of the rod (b) the reaction at A (c) the reaction at B
SOLUTION
Kinematics Assume a) (j)=O
e lzso
l-8A-r(middot1)rJ
aA+aB1A =[aA --- ]+[12a125deg]
-- aB (l2a)cos 25deg =10876a
(12a) sin 25deg =O5071aaA
aG =aA +aGIA =[aA ]+[O6a 125deg]
=[O5071a - ] + [O6a1 25deg]aG
ax =(aG)x =[O5071a ---+ ]+[O2536a --]
ax = O2535a -
ay [O6acos 25deg-] =O5438a
aWe have found for a ) ax O2535a
y 05438a
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc AU rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
]494
7
PROBLEM 16117 (Continued)
1 2-m(l2 m)Kinetics 12
(a) Angular acceleration
+) fME f(ME)eff mg(05438 m) Ia+max(02535 m)+may(05438 m)
m(12)2 +m(02535)2 a + m(05438)2 a 12
g(05438) 048a a 1l33g a 1111 rads2 )
(b) + t rPy = A - mg = -may = m(05438a)
A 10(981) -(10)(05438)(11115)
A = 981 - 6044 3766 N A 377 Nt
B = max m(02535a)
B = 10(02535)(11115) B =2818 N B
(c)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribution to teachers and educators permilled by Jv1cGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using itlvithout permission
1495
c
PROBLEM 1638
Two unifonn disks and two cylinders are assembled as indicated Disk A weighs 20 lb and disk B weighs 12 lb Knowing that the system is released from rest detennine the acceleration (a) of cylinder (b) of cylinder D
The cylinders are attached to a single cord that passes over the disks Assume that no slipping occurs between the cord and the disks
D
SOLUTION
Moments of inertia
Kinematics
-Kinetics
DiskA
1 2 1 20lb (8 ftfA mArA 013803 lbmiddot ftmiddot 52 2 2322 ftls 2 12 )
1 2 1 12 Ib (6 f IB -mBrB 004658 lbmiddot ft S2
2 2 322 ftls 2 12 t
a a ==l5a
a a = =2aaB rB 1
FAB == Tension in cord between disks
+)LMA =LMA
+mca ftJ 2 (2)15lb - FAB --10 == (013803)(150) + al-3 322 fls L 3
2 -10 =020705a + 031 056a 3 2 -10 =051760a 3
FAB = 15+077641a (1)
PROPRIETARY lIJATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No parI of this Manual may be displayed reproduced or distributed in any form or by means WilhoUl the prior wrillen permission of the publisher or used beyond the limiled distribution to teachers McGraw-Hill for their individual course preparation fYOII are a studem using this ivfanual you are Ising it wilholll permission
1380
7 s
PROBLEM 1538 (Continued)
DiskB ) 2MB = 2(A1B)eff
(6 (6 ft = 1flaB +llID( --ft)
12 12 9 =(004658)(2a)
1 322 ftls- 2)
9 05F4B = 009316a + 02795a
Substitute for FAB from Eq (l)
9 05(15 +077641a) =037266a 9 75 03882a = 037266a
15 076086a
a =1971 ftls2
Both ae and aD have the same magnitude
(a) Acceleration of C ae =1971 ftls2t
(b) Acceleration ofD aD = 1971 ftls2
PROPRIETARY MATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrillen permission of the publisher or lIsed beyond the limied distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyou are a student using lhis Manual you are using il withoul permission
1381
r
PROBLEM 1678
A unifonn slender rod of length L 900 mm and mass m 4 kg is suspended from a hinge at C A horizontal force P of magnitude 75 N is applied at end B Knowing that r 225 mm determine (a) the angular acceleration of the rod (b) the components of the reaction at C
SOLUTION It (a) Angular acceleration tf
a=ra
p[r + +la
- 1
( -)- 1 L7= mra r + m-a B 1 (-2
12 I L 1 1p(-r +- =m r + 2 J 12
Substitute data m + 09 m =(4 kg)[(0225 m)2 m)2la 15 N 2 12
50125 = 04725a
a = 10714 rads2 (l =1071 rads2
)
(b) Components of reaction at C
+ t = l(Fy)eff
Cy W == mg (4 kg)(981 mls2 ) C 392 Nt y
P=-ma r ex P-ma= P-m(fa)
0714 rads2=75 N-
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be (ffsJlalverJ repoduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the distribution to teachers and educators permilled by lvfcGraw-Hillfor their individual course preparation Ifyou are a student IISlI1g this Afanlal you are using il without permission
1439
PROBLEM 16109
Two unifonll disks A and B each of weight 4 lb are connected by a 3-lb rod CD as shown A counterclockwise couple M of moment 15 lbmiddot ft is applied to disk A Knowing that the disks roll without sliding determine (a) the acceleration of the center of each disk (b) the horizontal component of the force exerted on disk B D
SOLUTION
-
DiskA
+)LMs=L(M)eff
W=41b
- 1 2 =-mr 2
M-c(ftl (ma)r+a12
15lbmiddotft C 12
1 2(mra)r+-mr a 2
3=-mr21f 2
l51bmiddotft C = 3 4lb 12 2322
6 ft)2 a 12
15 O046584a 3
(1)
DiskB I
(ma)r+ a 2 1 0 mr 2
Rod CD
D =O046684a 8
(2)
6 2 8
6 2ftia 12 )
=-a+-a=-a 12 12 12
PROPRIE14RY IUATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or dislributed in any form or by any means without the prior written permission of the publisher or lIsed beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyoll are a student using this Manual Y01l are using it without permission
1480
PROBLEM 16109 (Continued)
Multiply
2 0041408a (3)By 3 3 3
Add (1) (2) (3) 15-1e +1D 1e-1D 0134576a
II =1 1146 rads2 )
6 7 ft(l1l46 Tads) =5573 ftls-(a)
12 )
(b) Substitute for a in (2)
D 0046584(11146) D 07791b -- 3
PROPRIETAllJMATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrilten permission athe or used he yond the limited dislrihllliol1ta teachers and educators permilled by McGraw-Hilior their individual course preparaliol1 are a stdl11I using Ihis Manual you are using it withoul permission
1481
__4mQmQbullbull a za I
I
7 7
I((r PROBLEM 16145
A unifonn slender bar AB of mass In is suspended as shown from a unifonn disk of the same mass m Neglecting the effect of friction detennine the accelerations of Points A and B immediately after a horizontal force P has been applied at B
Tiii
SOLUTION
Kinematics
Cylinder Rolling without sliding
RodAB )r
Yz L -GAB0ts Go
42poundIM f3
Cylinder
regtllt tl =-r C
C (1) t c
RodAB
_ L -PL = mGAB 2+ I a AB
PL (2)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No parI of this Manual may be dispJayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1553
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554
B
PROBLEM 16117
The ends of the lO-kg uniform rod AB are attached to collars of negligible mass that slide without friction along fixed rods If the rod is released from rest when f) = 25deg determine immediately after release (a) the angular acceleration of the rod (b) the reaction at A (c) the reaction at B
SOLUTION
Kinematics Assume a) (j)=O
e lzso
l-8A-r(middot1)rJ
aA+aB1A =[aA --- ]+[12a125deg]
-- aB (l2a)cos 25deg =10876a
(12a) sin 25deg =O5071aaA
aG =aA +aGIA =[aA ]+[O6a 125deg]
=[O5071a - ] + [O6a1 25deg]aG
ax =(aG)x =[O5071a ---+ ]+[O2536a --]
ax = O2535a -
ay [O6acos 25deg-] =O5438a
aWe have found for a ) ax O2535a
y 05438a
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc AU rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
]494
7
PROBLEM 16117 (Continued)
1 2-m(l2 m)Kinetics 12
(a) Angular acceleration
+) fME f(ME)eff mg(05438 m) Ia+max(02535 m)+may(05438 m)
m(12)2 +m(02535)2 a + m(05438)2 a 12
g(05438) 048a a 1l33g a 1111 rads2 )
(b) + t rPy = A - mg = -may = m(05438a)
A 10(981) -(10)(05438)(11115)
A = 981 - 6044 3766 N A 377 Nt
B = max m(02535a)
B = 10(02535)(11115) B =2818 N B
(c)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribution to teachers and educators permilled by Jv1cGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using itlvithout permission
1495
c
PROBLEM 1638
Two unifonn disks and two cylinders are assembled as indicated Disk A weighs 20 lb and disk B weighs 12 lb Knowing that the system is released from rest detennine the acceleration (a) of cylinder (b) of cylinder D
The cylinders are attached to a single cord that passes over the disks Assume that no slipping occurs between the cord and the disks
D
SOLUTION
Moments of inertia
Kinematics
-Kinetics
DiskA
1 2 1 20lb (8 ftfA mArA 013803 lbmiddot ftmiddot 52 2 2322 ftls 2 12 )
1 2 1 12 Ib (6 f IB -mBrB 004658 lbmiddot ft S2
2 2 322 ftls 2 12 t
a a ==l5a
a a = =2aaB rB 1
FAB == Tension in cord between disks
+)LMA =LMA
+mca ftJ 2 (2)15lb - FAB --10 == (013803)(150) + al-3 322 fls L 3
2 -10 =020705a + 031 056a 3 2 -10 =051760a 3
FAB = 15+077641a (1)
PROPRIETARY lIJATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No parI of this Manual may be displayed reproduced or distributed in any form or by means WilhoUl the prior wrillen permission of the publisher or used beyond the limiled distribution to teachers McGraw-Hill for their individual course preparation fYOII are a studem using this ivfanual you are Ising it wilholll permission
1380
7 s
PROBLEM 1538 (Continued)
DiskB ) 2MB = 2(A1B)eff
(6 (6 ft = 1flaB +llID( --ft)
12 12 9 =(004658)(2a)
1 322 ftls- 2)
9 05F4B = 009316a + 02795a
Substitute for FAB from Eq (l)
9 05(15 +077641a) =037266a 9 75 03882a = 037266a
15 076086a
a =1971 ftls2
Both ae and aD have the same magnitude
(a) Acceleration of C ae =1971 ftls2t
(b) Acceleration ofD aD = 1971 ftls2
PROPRIETARY MATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrillen permission of the publisher or lIsed beyond the limied distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyou are a student using lhis Manual you are using il withoul permission
1381
r
PROBLEM 1678
A unifonn slender rod of length L 900 mm and mass m 4 kg is suspended from a hinge at C A horizontal force P of magnitude 75 N is applied at end B Knowing that r 225 mm determine (a) the angular acceleration of the rod (b) the components of the reaction at C
SOLUTION It (a) Angular acceleration tf
a=ra
p[r + +la
- 1
( -)- 1 L7= mra r + m-a B 1 (-2
12 I L 1 1p(-r +- =m r + 2 J 12
Substitute data m + 09 m =(4 kg)[(0225 m)2 m)2la 15 N 2 12
50125 = 04725a
a = 10714 rads2 (l =1071 rads2
)
(b) Components of reaction at C
+ t = l(Fy)eff
Cy W == mg (4 kg)(981 mls2 ) C 392 Nt y
P=-ma r ex P-ma= P-m(fa)
0714 rads2=75 N-
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be (ffsJlalverJ repoduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the distribution to teachers and educators permilled by lvfcGraw-Hillfor their individual course preparation Ifyou are a student IISlI1g this Afanlal you are using il without permission
1439
PROBLEM 16109
Two unifonll disks A and B each of weight 4 lb are connected by a 3-lb rod CD as shown A counterclockwise couple M of moment 15 lbmiddot ft is applied to disk A Knowing that the disks roll without sliding determine (a) the acceleration of the center of each disk (b) the horizontal component of the force exerted on disk B D
SOLUTION
-
DiskA
+)LMs=L(M)eff
W=41b
- 1 2 =-mr 2
M-c(ftl (ma)r+a12
15lbmiddotft C 12
1 2(mra)r+-mr a 2
3=-mr21f 2
l51bmiddotft C = 3 4lb 12 2322
6 ft)2 a 12
15 O046584a 3
(1)
DiskB I
(ma)r+ a 2 1 0 mr 2
Rod CD
D =O046684a 8
(2)
6 2 8
6 2ftia 12 )
=-a+-a=-a 12 12 12
PROPRIE14RY IUATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or dislributed in any form or by any means without the prior written permission of the publisher or lIsed beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyoll are a student using this Manual Y01l are using it without permission
1480
PROBLEM 16109 (Continued)
Multiply
2 0041408a (3)By 3 3 3
Add (1) (2) (3) 15-1e +1D 1e-1D 0134576a
II =1 1146 rads2 )
6 7 ft(l1l46 Tads) =5573 ftls-(a)
12 )
(b) Substitute for a in (2)
D 0046584(11146) D 07791b -- 3
PROPRIETAllJMATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrilten permission athe or used he yond the limited dislrihllliol1ta teachers and educators permilled by McGraw-Hilior their individual course preparaliol1 are a stdl11I using Ihis Manual you are using it withoul permission
1481
__4mQmQbullbull a za I
I
7 7
I((r PROBLEM 16145
A unifonn slender bar AB of mass In is suspended as shown from a unifonn disk of the same mass m Neglecting the effect of friction detennine the accelerations of Points A and B immediately after a horizontal force P has been applied at B
Tiii
SOLUTION
Kinematics
Cylinder Rolling without sliding
RodAB )r
Yz L -GAB0ts Go
42poundIM f3
Cylinder
regtllt tl =-r C
C (1) t c
RodAB
_ L -PL = mGAB 2+ I a AB
PL (2)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No parI of this Manual may be dispJayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1553
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554
7
PROBLEM 16117 (Continued)
1 2-m(l2 m)Kinetics 12
(a) Angular acceleration
+) fME f(ME)eff mg(05438 m) Ia+max(02535 m)+may(05438 m)
m(12)2 +m(02535)2 a + m(05438)2 a 12
g(05438) 048a a 1l33g a 1111 rads2 )
(b) + t rPy = A - mg = -may = m(05438a)
A 10(981) -(10)(05438)(11115)
A = 981 - 6044 3766 N A 377 Nt
B = max m(02535a)
B = 10(02535)(11115) B =2818 N B
(c)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means without the prior wrillen permission of the publisher or used beyond the limited distribution to teachers and educators permilled by Jv1cGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using itlvithout permission
1495
c
PROBLEM 1638
Two unifonn disks and two cylinders are assembled as indicated Disk A weighs 20 lb and disk B weighs 12 lb Knowing that the system is released from rest detennine the acceleration (a) of cylinder (b) of cylinder D
The cylinders are attached to a single cord that passes over the disks Assume that no slipping occurs between the cord and the disks
D
SOLUTION
Moments of inertia
Kinematics
-Kinetics
DiskA
1 2 1 20lb (8 ftfA mArA 013803 lbmiddot ftmiddot 52 2 2322 ftls 2 12 )
1 2 1 12 Ib (6 f IB -mBrB 004658 lbmiddot ft S2
2 2 322 ftls 2 12 t
a a ==l5a
a a = =2aaB rB 1
FAB == Tension in cord between disks
+)LMA =LMA
+mca ftJ 2 (2)15lb - FAB --10 == (013803)(150) + al-3 322 fls L 3
2 -10 =020705a + 031 056a 3 2 -10 =051760a 3
FAB = 15+077641a (1)
PROPRIETARY lIJATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No parI of this Manual may be displayed reproduced or distributed in any form or by means WilhoUl the prior wrillen permission of the publisher or used beyond the limiled distribution to teachers McGraw-Hill for their individual course preparation fYOII are a studem using this ivfanual you are Ising it wilholll permission
1380
7 s
PROBLEM 1538 (Continued)
DiskB ) 2MB = 2(A1B)eff
(6 (6 ft = 1flaB +llID( --ft)
12 12 9 =(004658)(2a)
1 322 ftls- 2)
9 05F4B = 009316a + 02795a
Substitute for FAB from Eq (l)
9 05(15 +077641a) =037266a 9 75 03882a = 037266a
15 076086a
a =1971 ftls2
Both ae and aD have the same magnitude
(a) Acceleration of C ae =1971 ftls2t
(b) Acceleration ofD aD = 1971 ftls2
PROPRIETARY MATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrillen permission of the publisher or lIsed beyond the limied distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyou are a student using lhis Manual you are using il withoul permission
1381
r
PROBLEM 1678
A unifonn slender rod of length L 900 mm and mass m 4 kg is suspended from a hinge at C A horizontal force P of magnitude 75 N is applied at end B Knowing that r 225 mm determine (a) the angular acceleration of the rod (b) the components of the reaction at C
SOLUTION It (a) Angular acceleration tf
a=ra
p[r + +la
- 1
( -)- 1 L7= mra r + m-a B 1 (-2
12 I L 1 1p(-r +- =m r + 2 J 12
Substitute data m + 09 m =(4 kg)[(0225 m)2 m)2la 15 N 2 12
50125 = 04725a
a = 10714 rads2 (l =1071 rads2
)
(b) Components of reaction at C
+ t = l(Fy)eff
Cy W == mg (4 kg)(981 mls2 ) C 392 Nt y
P=-ma r ex P-ma= P-m(fa)
0714 rads2=75 N-
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be (ffsJlalverJ repoduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the distribution to teachers and educators permilled by lvfcGraw-Hillfor their individual course preparation Ifyou are a student IISlI1g this Afanlal you are using il without permission
1439
PROBLEM 16109
Two unifonll disks A and B each of weight 4 lb are connected by a 3-lb rod CD as shown A counterclockwise couple M of moment 15 lbmiddot ft is applied to disk A Knowing that the disks roll without sliding determine (a) the acceleration of the center of each disk (b) the horizontal component of the force exerted on disk B D
SOLUTION
-
DiskA
+)LMs=L(M)eff
W=41b
- 1 2 =-mr 2
M-c(ftl (ma)r+a12
15lbmiddotft C 12
1 2(mra)r+-mr a 2
3=-mr21f 2
l51bmiddotft C = 3 4lb 12 2322
6 ft)2 a 12
15 O046584a 3
(1)
DiskB I
(ma)r+ a 2 1 0 mr 2
Rod CD
D =O046684a 8
(2)
6 2 8
6 2ftia 12 )
=-a+-a=-a 12 12 12
PROPRIE14RY IUATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or dislributed in any form or by any means without the prior written permission of the publisher or lIsed beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyoll are a student using this Manual Y01l are using it without permission
1480
PROBLEM 16109 (Continued)
Multiply
2 0041408a (3)By 3 3 3
Add (1) (2) (3) 15-1e +1D 1e-1D 0134576a
II =1 1146 rads2 )
6 7 ft(l1l46 Tads) =5573 ftls-(a)
12 )
(b) Substitute for a in (2)
D 0046584(11146) D 07791b -- 3
PROPRIETAllJMATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrilten permission athe or used he yond the limited dislrihllliol1ta teachers and educators permilled by McGraw-Hilior their individual course preparaliol1 are a stdl11I using Ihis Manual you are using it withoul permission
1481
__4mQmQbullbull a za I
I
7 7
I((r PROBLEM 16145
A unifonn slender bar AB of mass In is suspended as shown from a unifonn disk of the same mass m Neglecting the effect of friction detennine the accelerations of Points A and B immediately after a horizontal force P has been applied at B
Tiii
SOLUTION
Kinematics
Cylinder Rolling without sliding
RodAB )r
Yz L -GAB0ts Go
42poundIM f3
Cylinder
regtllt tl =-r C
C (1) t c
RodAB
_ L -PL = mGAB 2+ I a AB
PL (2)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No parI of this Manual may be dispJayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1553
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554
c
PROBLEM 1638
Two unifonn disks and two cylinders are assembled as indicated Disk A weighs 20 lb and disk B weighs 12 lb Knowing that the system is released from rest detennine the acceleration (a) of cylinder (b) of cylinder D
The cylinders are attached to a single cord that passes over the disks Assume that no slipping occurs between the cord and the disks
D
SOLUTION
Moments of inertia
Kinematics
-Kinetics
DiskA
1 2 1 20lb (8 ftfA mArA 013803 lbmiddot ftmiddot 52 2 2322 ftls 2 12 )
1 2 1 12 Ib (6 f IB -mBrB 004658 lbmiddot ft S2
2 2 322 ftls 2 12 t
a a ==l5a
a a = =2aaB rB 1
FAB == Tension in cord between disks
+)LMA =LMA
+mca ftJ 2 (2)15lb - FAB --10 == (013803)(150) + al-3 322 fls L 3
2 -10 =020705a + 031 056a 3 2 -10 =051760a 3
FAB = 15+077641a (1)
PROPRIETARY lIJATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No parI of this Manual may be displayed reproduced or distributed in any form or by means WilhoUl the prior wrillen permission of the publisher or used beyond the limiled distribution to teachers McGraw-Hill for their individual course preparation fYOII are a studem using this ivfanual you are Ising it wilholll permission
1380
7 s
PROBLEM 1538 (Continued)
DiskB ) 2MB = 2(A1B)eff
(6 (6 ft = 1flaB +llID( --ft)
12 12 9 =(004658)(2a)
1 322 ftls- 2)
9 05F4B = 009316a + 02795a
Substitute for FAB from Eq (l)
9 05(15 +077641a) =037266a 9 75 03882a = 037266a
15 076086a
a =1971 ftls2
Both ae and aD have the same magnitude
(a) Acceleration of C ae =1971 ftls2t
(b) Acceleration ofD aD = 1971 ftls2
PROPRIETARY MATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrillen permission of the publisher or lIsed beyond the limied distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyou are a student using lhis Manual you are using il withoul permission
1381
r
PROBLEM 1678
A unifonn slender rod of length L 900 mm and mass m 4 kg is suspended from a hinge at C A horizontal force P of magnitude 75 N is applied at end B Knowing that r 225 mm determine (a) the angular acceleration of the rod (b) the components of the reaction at C
SOLUTION It (a) Angular acceleration tf
a=ra
p[r + +la
- 1
( -)- 1 L7= mra r + m-a B 1 (-2
12 I L 1 1p(-r +- =m r + 2 J 12
Substitute data m + 09 m =(4 kg)[(0225 m)2 m)2la 15 N 2 12
50125 = 04725a
a = 10714 rads2 (l =1071 rads2
)
(b) Components of reaction at C
+ t = l(Fy)eff
Cy W == mg (4 kg)(981 mls2 ) C 392 Nt y
P=-ma r ex P-ma= P-m(fa)
0714 rads2=75 N-
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be (ffsJlalverJ repoduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the distribution to teachers and educators permilled by lvfcGraw-Hillfor their individual course preparation Ifyou are a student IISlI1g this Afanlal you are using il without permission
1439
PROBLEM 16109
Two unifonll disks A and B each of weight 4 lb are connected by a 3-lb rod CD as shown A counterclockwise couple M of moment 15 lbmiddot ft is applied to disk A Knowing that the disks roll without sliding determine (a) the acceleration of the center of each disk (b) the horizontal component of the force exerted on disk B D
SOLUTION
-
DiskA
+)LMs=L(M)eff
W=41b
- 1 2 =-mr 2
M-c(ftl (ma)r+a12
15lbmiddotft C 12
1 2(mra)r+-mr a 2
3=-mr21f 2
l51bmiddotft C = 3 4lb 12 2322
6 ft)2 a 12
15 O046584a 3
(1)
DiskB I
(ma)r+ a 2 1 0 mr 2
Rod CD
D =O046684a 8
(2)
6 2 8
6 2ftia 12 )
=-a+-a=-a 12 12 12
PROPRIE14RY IUATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or dislributed in any form or by any means without the prior written permission of the publisher or lIsed beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyoll are a student using this Manual Y01l are using it without permission
1480
PROBLEM 16109 (Continued)
Multiply
2 0041408a (3)By 3 3 3
Add (1) (2) (3) 15-1e +1D 1e-1D 0134576a
II =1 1146 rads2 )
6 7 ft(l1l46 Tads) =5573 ftls-(a)
12 )
(b) Substitute for a in (2)
D 0046584(11146) D 07791b -- 3
PROPRIETAllJMATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrilten permission athe or used he yond the limited dislrihllliol1ta teachers and educators permilled by McGraw-Hilior their individual course preparaliol1 are a stdl11I using Ihis Manual you are using it withoul permission
1481
__4mQmQbullbull a za I
I
7 7
I((r PROBLEM 16145
A unifonn slender bar AB of mass In is suspended as shown from a unifonn disk of the same mass m Neglecting the effect of friction detennine the accelerations of Points A and B immediately after a horizontal force P has been applied at B
Tiii
SOLUTION
Kinematics
Cylinder Rolling without sliding
RodAB )r
Yz L -GAB0ts Go
42poundIM f3
Cylinder
regtllt tl =-r C
C (1) t c
RodAB
_ L -PL = mGAB 2+ I a AB
PL (2)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No parI of this Manual may be dispJayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1553
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554
7 s
PROBLEM 1538 (Continued)
DiskB ) 2MB = 2(A1B)eff
(6 (6 ft = 1flaB +llID( --ft)
12 12 9 =(004658)(2a)
1 322 ftls- 2)
9 05F4B = 009316a + 02795a
Substitute for FAB from Eq (l)
9 05(15 +077641a) =037266a 9 75 03882a = 037266a
15 076086a
a =1971 ftls2
Both ae and aD have the same magnitude
(a) Acceleration of C ae =1971 ftls2t
(b) Acceleration ofD aD = 1971 ftls2
PROPRIETARY MATERIAL 2009 The McGraw-Hili Companies Inc All rights reserved No part of this Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrillen permission of the publisher or lIsed beyond the limied distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyou are a student using lhis Manual you are using il withoul permission
1381
r
PROBLEM 1678
A unifonn slender rod of length L 900 mm and mass m 4 kg is suspended from a hinge at C A horizontal force P of magnitude 75 N is applied at end B Knowing that r 225 mm determine (a) the angular acceleration of the rod (b) the components of the reaction at C
SOLUTION It (a) Angular acceleration tf
a=ra
p[r + +la
- 1
( -)- 1 L7= mra r + m-a B 1 (-2
12 I L 1 1p(-r +- =m r + 2 J 12
Substitute data m + 09 m =(4 kg)[(0225 m)2 m)2la 15 N 2 12
50125 = 04725a
a = 10714 rads2 (l =1071 rads2
)
(b) Components of reaction at C
+ t = l(Fy)eff
Cy W == mg (4 kg)(981 mls2 ) C 392 Nt y
P=-ma r ex P-ma= P-m(fa)
0714 rads2=75 N-
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be (ffsJlalverJ repoduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the distribution to teachers and educators permilled by lvfcGraw-Hillfor their individual course preparation Ifyou are a student IISlI1g this Afanlal you are using il without permission
1439
PROBLEM 16109
Two unifonll disks A and B each of weight 4 lb are connected by a 3-lb rod CD as shown A counterclockwise couple M of moment 15 lbmiddot ft is applied to disk A Knowing that the disks roll without sliding determine (a) the acceleration of the center of each disk (b) the horizontal component of the force exerted on disk B D
SOLUTION
-
DiskA
+)LMs=L(M)eff
W=41b
- 1 2 =-mr 2
M-c(ftl (ma)r+a12
15lbmiddotft C 12
1 2(mra)r+-mr a 2
3=-mr21f 2
l51bmiddotft C = 3 4lb 12 2322
6 ft)2 a 12
15 O046584a 3
(1)
DiskB I
(ma)r+ a 2 1 0 mr 2
Rod CD
D =O046684a 8
(2)
6 2 8
6 2ftia 12 )
=-a+-a=-a 12 12 12
PROPRIE14RY IUATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or dislributed in any form or by any means without the prior written permission of the publisher or lIsed beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyoll are a student using this Manual Y01l are using it without permission
1480
PROBLEM 16109 (Continued)
Multiply
2 0041408a (3)By 3 3 3
Add (1) (2) (3) 15-1e +1D 1e-1D 0134576a
II =1 1146 rads2 )
6 7 ft(l1l46 Tads) =5573 ftls-(a)
12 )
(b) Substitute for a in (2)
D 0046584(11146) D 07791b -- 3
PROPRIETAllJMATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrilten permission athe or used he yond the limited dislrihllliol1ta teachers and educators permilled by McGraw-Hilior their individual course preparaliol1 are a stdl11I using Ihis Manual you are using it withoul permission
1481
__4mQmQbullbull a za I
I
7 7
I((r PROBLEM 16145
A unifonn slender bar AB of mass In is suspended as shown from a unifonn disk of the same mass m Neglecting the effect of friction detennine the accelerations of Points A and B immediately after a horizontal force P has been applied at B
Tiii
SOLUTION
Kinematics
Cylinder Rolling without sliding
RodAB )r
Yz L -GAB0ts Go
42poundIM f3
Cylinder
regtllt tl =-r C
C (1) t c
RodAB
_ L -PL = mGAB 2+ I a AB
PL (2)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No parI of this Manual may be dispJayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1553
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554
r
PROBLEM 1678
A unifonn slender rod of length L 900 mm and mass m 4 kg is suspended from a hinge at C A horizontal force P of magnitude 75 N is applied at end B Knowing that r 225 mm determine (a) the angular acceleration of the rod (b) the components of the reaction at C
SOLUTION It (a) Angular acceleration tf
a=ra
p[r + +la
- 1
( -)- 1 L7= mra r + m-a B 1 (-2
12 I L 1 1p(-r +- =m r + 2 J 12
Substitute data m + 09 m =(4 kg)[(0225 m)2 m)2la 15 N 2 12
50125 = 04725a
a = 10714 rads2 (l =1071 rads2
)
(b) Components of reaction at C
+ t = l(Fy)eff
Cy W == mg (4 kg)(981 mls2 ) C 392 Nt y
P=-ma r ex P-ma= P-m(fa)
0714 rads2=75 N-
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No part of this Manual may be (ffsJlalverJ repoduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the distribution to teachers and educators permilled by lvfcGraw-Hillfor their individual course preparation Ifyou are a student IISlI1g this Afanlal you are using il without permission
1439
PROBLEM 16109
Two unifonll disks A and B each of weight 4 lb are connected by a 3-lb rod CD as shown A counterclockwise couple M of moment 15 lbmiddot ft is applied to disk A Knowing that the disks roll without sliding determine (a) the acceleration of the center of each disk (b) the horizontal component of the force exerted on disk B D
SOLUTION
-
DiskA
+)LMs=L(M)eff
W=41b
- 1 2 =-mr 2
M-c(ftl (ma)r+a12
15lbmiddotft C 12
1 2(mra)r+-mr a 2
3=-mr21f 2
l51bmiddotft C = 3 4lb 12 2322
6 ft)2 a 12
15 O046584a 3
(1)
DiskB I
(ma)r+ a 2 1 0 mr 2
Rod CD
D =O046684a 8
(2)
6 2 8
6 2ftia 12 )
=-a+-a=-a 12 12 12
PROPRIE14RY IUATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or dislributed in any form or by any means without the prior written permission of the publisher or lIsed beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyoll are a student using this Manual Y01l are using it without permission
1480
PROBLEM 16109 (Continued)
Multiply
2 0041408a (3)By 3 3 3
Add (1) (2) (3) 15-1e +1D 1e-1D 0134576a
II =1 1146 rads2 )
6 7 ft(l1l46 Tads) =5573 ftls-(a)
12 )
(b) Substitute for a in (2)
D 0046584(11146) D 07791b -- 3
PROPRIETAllJMATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrilten permission athe or used he yond the limited dislrihllliol1ta teachers and educators permilled by McGraw-Hilior their individual course preparaliol1 are a stdl11I using Ihis Manual you are using it withoul permission
1481
__4mQmQbullbull a za I
I
7 7
I((r PROBLEM 16145
A unifonn slender bar AB of mass In is suspended as shown from a unifonn disk of the same mass m Neglecting the effect of friction detennine the accelerations of Points A and B immediately after a horizontal force P has been applied at B
Tiii
SOLUTION
Kinematics
Cylinder Rolling without sliding
RodAB )r
Yz L -GAB0ts Go
42poundIM f3
Cylinder
regtllt tl =-r C
C (1) t c
RodAB
_ L -PL = mGAB 2+ I a AB
PL (2)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No parI of this Manual may be dispJayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1553
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554
PROBLEM 16109
Two unifonll disks A and B each of weight 4 lb are connected by a 3-lb rod CD as shown A counterclockwise couple M of moment 15 lbmiddot ft is applied to disk A Knowing that the disks roll without sliding determine (a) the acceleration of the center of each disk (b) the horizontal component of the force exerted on disk B D
SOLUTION
-
DiskA
+)LMs=L(M)eff
W=41b
- 1 2 =-mr 2
M-c(ftl (ma)r+a12
15lbmiddotft C 12
1 2(mra)r+-mr a 2
3=-mr21f 2
l51bmiddotft C = 3 4lb 12 2322
6 ft)2 a 12
15 O046584a 3
(1)
DiskB I
(ma)r+ a 2 1 0 mr 2
Rod CD
D =O046684a 8
(2)
6 2 8
6 2ftia 12 )
=-a+-a=-a 12 12 12
PROPRIE14RY IUATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or dislributed in any form or by any means without the prior written permission of the publisher or lIsed beyond the limited distribution to teachers and educators permitted by McGraw-Hili for their individual course preparation Ifyoll are a student using this Manual Y01l are using it without permission
1480
PROBLEM 16109 (Continued)
Multiply
2 0041408a (3)By 3 3 3
Add (1) (2) (3) 15-1e +1D 1e-1D 0134576a
II =1 1146 rads2 )
6 7 ft(l1l46 Tads) =5573 ftls-(a)
12 )
(b) Substitute for a in (2)
D 0046584(11146) D 07791b -- 3
PROPRIETAllJMATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrilten permission athe or used he yond the limited dislrihllliol1ta teachers and educators permilled by McGraw-Hilior their individual course preparaliol1 are a stdl11I using Ihis Manual you are using it withoul permission
1481
__4mQmQbullbull a za I
I
7 7
I((r PROBLEM 16145
A unifonn slender bar AB of mass In is suspended as shown from a unifonn disk of the same mass m Neglecting the effect of friction detennine the accelerations of Points A and B immediately after a horizontal force P has been applied at B
Tiii
SOLUTION
Kinematics
Cylinder Rolling without sliding
RodAB )r
Yz L -GAB0ts Go
42poundIM f3
Cylinder
regtllt tl =-r C
C (1) t c
RodAB
_ L -PL = mGAB 2+ I a AB
PL (2)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No parI of this Manual may be dispJayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1553
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554
PROBLEM 16109 (Continued)
Multiply
2 0041408a (3)By 3 3 3
Add (1) (2) (3) 15-1e +1D 1e-1D 0134576a
II =1 1146 rads2 )
6 7 ft(l1l46 Tads) =5573 ftls-(a)
12 )
(b) Substitute for a in (2)
D 0046584(11146) D 07791b -- 3
PROPRIETAllJMATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No pari of Ihis Manual may be displayed reproduced or distributed in any form or by any means withoul the prior wrilten permission athe or used he yond the limited dislrihllliol1ta teachers and educators permilled by McGraw-Hilior their individual course preparaliol1 are a stdl11I using Ihis Manual you are using it withoul permission
1481
__4mQmQbullbull a za I
I
7 7
I((r PROBLEM 16145
A unifonn slender bar AB of mass In is suspended as shown from a unifonn disk of the same mass m Neglecting the effect of friction detennine the accelerations of Points A and B immediately after a horizontal force P has been applied at B
Tiii
SOLUTION
Kinematics
Cylinder Rolling without sliding
RodAB )r
Yz L -GAB0ts Go
42poundIM f3
Cylinder
regtllt tl =-r C
C (1) t c
RodAB
_ L -PL = mGAB 2+ I a AB
PL (2)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No parI of this Manual may be dispJayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1553
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554
I
7 7
I((r PROBLEM 16145
A unifonn slender bar AB of mass In is suspended as shown from a unifonn disk of the same mass m Neglecting the effect of friction detennine the accelerations of Points A and B immediately after a horizontal force P has been applied at B
Tiii
SOLUTION
Kinematics
Cylinder Rolling without sliding
RodAB )r
Yz L -GAB0ts Go
42poundIM f3
Cylinder
regtllt tl =-r C
C (1) t c
RodAB
_ L -PL = mGAB 2+ I a AB
PL (2)
PROPRIETARY MATERIAL copy 2009 The McGraw-Hill Companies Inc All rights reserved No parI of this Manual may be dispJayed reproduced or distributed in any form or by any means without the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by McGraw-Hillfor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1553
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554
PROBLEM 16145 (Continued)
Substitute from (1)
3-mLaAB4 (3)
Multiply by L 9
(4) + (2)
1 2--mL a AB12
10 PL l( 1 5) L- 7 L-- = -+- m aAB =-m aAB9 2 18 9
(4)
10 P 7 m
(5)
(5) - (3) P)27m
3-mLaAB4
P =25p7
3 L4m aAB
3-mLaAB4 7 mL )
B
a P 2P a A =--
A 7 7 m 7m
12 10 P 22 P + aB = -lt411
7 7 m 7 m
PROPRIETARY MATERIAL 2009 The McGraw-Hill Companies Inc All rights reserved No part of this lvfanual may be displayed reprodlced or distributed in any form or by any means withour the prior written permission of the publisher or used beyond the limited distribution to teachers and educators permitted by lvfcGraw-Hilifor their individual course preparation Ifyou are a student using this Manual you are using it without permission
1554