PROBLEM 8.61 A 10° wedge is forced under an 80-kg cylinder as shown. Knowing that the coefficient of static friction between the cylinder and the vertical wall is 0.30, determine the smallest coefficient of static friction between the wedge and the cylinder for which shipping can be impending at B.
SOLUTION FBD Cylinder:
(
)
W = (80 kg) 9.81 m/s2 = 784.8 N For impending slip at B, ΣM A = 0:
FB = µ sB N B = 0.30N B
(r cos30°) NB − r (1 + sin30°) (0.30NB ) − r sin30°W = 0
N B = 1.20185W = 943.21 N FB = 0.30N B = 0.36055W ΣMG = 0: ΣFx = 0:
r ( FA − FB ) = 0, FA = FB = 0.36055W N A sin30° + FA cos30° − N B = 0 NA =
− (0.36055W ) cos30° + 1.20185W sin30°
N A = 1.77920W For minimum µ A, slip impends at A, so µ A min =
FA 0.36055W = = 0.2026 N A 1.77920W
µ A min = 0.203 t
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PROBLEM 8.62 Bags of grass seed are stored on a wooden plank as shown. To move the plank, a 9° wedge is driven under end A. Knowing that the weight of the grass seed can be represented by the distributed load shown and that the coefficient of static friction is 0.45 between all surfaces of contact, (a) determine the force P for which motion of the wedge is impending, (b) indicate whether the plank will slide on the floor. SOLUTION FBD plank + wedge:
( 2.8 m ) NB − ( 48 N/m)(.45 m )( .9 m)
ΣM A = 0: 0.9 m
− (.6 m)
1.5 m 96 N/m
48 N/m
1 (48 N/m )(.9 m) 2
1 − .9 + .5 m (96 N/m)( 1.5 m ) = 0 2
N B = 55.5 N 48 + 96 N/m (.9 m) NW + 55.5 N − 2
ΣFy = 0:
+
1 (96 N/m )( 1.5 m) = 0 2 NW = 62.7 N
Since NW > N B, and all µ s are equal, assume slip impends at B and between wedge and floor, and not at A. Then FW = µ s NW = 0.45( 62.7 N ) = 28.2 N FB = µ s N B = 0.45 ( 55.5 N ) = 25 N ΣFx = 0:
P − 28.2 N − 25 N = 0, P = 53.2 N
Check Wedge for assumption ΣFy = 0: 62.7 N − RA cosθ = 0 ΣFx = 0: 53.2 N − 28.2 N − RA sin θ = 0 53.2 N
28.2 N 62.7 N
so tanθ =
25 = .3987 < µ s + tan9° 62.7
so no slip here ∴ (a)
(b)
P = 53.2 N
t
impending slip at B t
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PROBLEM 8.63 Solve Prob. 8.62 assuming that the wedge is driven under the plank at B instead of at A. Problem 8.62: Bags of grass seed are stored on a wooden plank as shown. To move the plank, a 9° wedge is driven under end A. Knowing that the weight of the grass seed can be represented by the distributed load shown and that the coefficient of static friction is 0.45 between all surfaces of contact, (a) determine the force P for which motion of the wedge is impending, (b) indicate whether the plank will slide on the floor. SOLUTION FBD plank + wedge: ΣM A = 0:
− (.6 m )
1.5 m
0.9 m
(2.4 m ) NB − (.45 m)(48 N/m )( .9 m )
96 N/m
48 N/m
1 (48N/m)(.9 m ) 2
− [(.9 + .5)m] (96 N/m)(1.5 m) = 0
NW = 55.5 N 48 + 96 ΣFy = 0: N A + 55.5 N − N/m (.9 m ) 2 1 − (96 N/m)( 1.5 m) = 0 2 N A = 62.7 N Since N A > NW , and all µ s are equal, assume impending slip at top and bottom of wedge and not at A. Then FW = µ s NW = 0.45( 55.5 N ) FW = 25 N
FBD Wedge:
φs = tan−1 µ s = tan−1 (0.45) = 24.228°
55.5 N − RB cos (24.228° + 9°) = 0
ΣFy = 0:
RB = 66.3 N
25 N
ΣFx = 0:
(66.3 N ) sin33.228° + 25 N − P = 0 P = 61.3 N
55.5 N
Check assumption using plank/wedge FBD ΣFx = 0: FA + FW − P = 0, FA = 61.3 N − 25 N = 36.3 N FA max = µs N A = 0.45 (62.7 N ) = 28.2 N
FA < FA max, OK P = 61.3 N t ∴ (a) (b) no impending slip at A t PROPRIETARY MATERIAL. © 2007 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 written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1250
PROBLEM 8.64 The 10-kg block A is at rest against the 50-kg block B as shown. The coefficient of static friction µ s is the same between blocks A and B and between block B and the floor, while friction between block A and the wall can be neglected. Knowing that P = 150 N, determine the value of µ s for which motion is impending.
SOLUTION
(
)
(
)
WA = (10 kg) 9.81 m/s2 = 98.1 N, WB = (50 kg) 9.81 m/s2 = 490.5 N Slip must impend at all surfaces simultaneously, F = µ s N FBD I: A + B ΣFy = 0:
N B − 150 N − 98.1 N − 490.5 N = 0,
impending slip:
FBD II: A
N B = 738.6 N
FB = µ s N B = (738.6 N) µ s N A = (738.6 N) µs
ΣFx = 0:
N A − FB = 0,
ΣFy′ = 0:
FAB + (738.6µ s ) N sin20° − (150 N + 98.1 N) cos20° = 0 FAB = 233.14 − (252.62) µ s N
ΣFx′ = 0:
(738.6µ s ) N cos20° − (150 N + 98.1 N) sin20° − N AB = 0
N AB = 84.855 + (694.06) µs N µs =
FAB 233.14 − 252.62µ s = N AB 84.855 + 694.06µs
µ s2 = 0.48623µ s − 0.33591 = 0 µ s = − 0.24312 ± 0.62850
Positive root
µ s = 0.385 t
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PROBLEM 8.65 Solve Prob. 8.64 assuming that µ s is the coefficient of static friction between all surfaces of contact. Problem 8.64: The 10-kg block A is at rest against the 50-kg block B as shown. The coefficient of static friction µ s is the same between blocks A and B and between block B and the floor, while friction between block A and the wall can be neglected. Knowing that P = 150 N, determine the value of µ s for which motion is impending. SOLUTION
(
)
(
)
WA = (10 kg) 9.81 m/s2 = 98.1 N, WB = (50 kg) 9.81 m/s2 = 490.5 N Slip impends at all surfaces simultaneously FBD I: A + B ΣFx = 0: ΣFy = 0:
N A − FB = 0,
N A = FB = µ s N B
(1)
FA − (150 N + 98.1 N + 490.5 N) + N B = 0 µ s N A + N B = 738.6 N
Solving (1) and (2) N B = FBD II: B
ΣFx′ = 0:
FB =
738.6 µ s N 1 + µ s2
N AB + ( 490.5 N) cos70° − N B cos70° − FB sin70° = 0 N AB =
ΣFy′ = 0:
738.6 N , 1 + µ s2
(2)
738.6 N (cos70° + µs sin70°) − (490.5 N) cos70° (1) 1 + µs2
−FAB − ( 490.5 N) sin70° + N B sin70° − FB cos70° = 0
FAB =
738.6 N (sin70° − µs cos70°) − (490.5 N) sin70° = 0 1 + µs2
Setting FAB = µs N AB, µ s3 − 6.8847µ s2 − 2.0116µ s + 1.38970 = 0
Solving numerically, µ s = − 0.586, 0.332, 7.14 Physically meaningful solution:
µ s = 0.332 t
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PROBLEM 8.66 Derive the following formulas relating the load W and the force P exerted on the handle of the jack discussed in Section 8.6. (a) P = (Wr/a) tan (θ + φs ) , to raise the load; (b) P = (Wr/a) tan (φs − θ ) , to lower the load if the screw is self-locking; (c) P = (Wr/a) tan (θ − φ s ) , to hold the load if the screw is not self-locking.
SOLUTION FBD jack handle:
See Section 8.6 ΣMC = 0:
aP − rQ = 0 or P =
r Q a
FBD block on incline: (a) Raising load
Q = Wtan (θ + φs ) P=
r W tan (θ + φs ) t a continued
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PROBLEM 8.66 CONTINUED (b) Lowering load if screw is self-locking ( i.e.: if φs > θ )
Q = Wtan (φs − θ ) P=
r W tan (φs − θ ) t a
P=
r W tan (θ − φs ) t a
(c) Holding load is screw is not self-locking (i.e: if φs < θ )
Q = Wtan (θ − φs )
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PROBLEM 8.67 square-threadedworm wormgear gearshown shownhas hasa mean a mean radius The square-threaded radius of of 1.50.0381 in. andma lead 0.375 in. The gear gear is subjected to a toconstant clockwise and aoflead of 0.009 m. larger The larger is subjected a constant clockKnowing thatthatthethecoefficient couple of 7.2 Knowing coefficientofof static static friction wise couple of kip 7.2⋅ in. kN.m. between the two gears is 0.12, determine the couple that must be applied rotate the the large large gear gear counterclockwise. counterclockwise. Neglect friction in to shaft AB to rotate the bearings bearings at atA, A,B, B,and andC. C.
SOLUTION FBD large gear: kN.m
ΣMC = 0:
12 in. in. ==0,0, kN ⋅. m ((0.3 m)W − 7.2 kip
kN kips W = 24 0.600 = 600 lb
0.3 m
Block on incline: θ = tan−1 0.009 m
0.375 0.009in. m 2.153°° ==2.2785 2π (1.5 in.) ) 2p 0.0381
φs = tan−1 µ s = tan−1 0.12 = 6.8428°
Q = W tan (θ + φs )
2p (0.0381m)
= (600 lb) tan9.1213 ° == 96.333 24 kN tan 8.9958° 3.8 kNlb
FBD worm gear: 3.8 kN
r == 0.0381 1.5 in. m ΣMB = 0:
(0.0381 )(3.8 kN in.)(96.333 lb) − M (1.5
=0
M mt M==0.145 144.5kN lb ⋅.in.
PROPRIETARY MATERIAL. © 2007 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 written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1255
PROBLEM 8.68 In Prob. 8.67, determine the couple that must be applied to shaft AB to rotate the gear clockwise. Problem 8.67: 8.67: The square-threaded square-threaded worm wormgear gearshown shownhas hasaamean meanradius radius of 1.5 0.0381 m and a lead of 0.009 m. The larger gear is subjected to a in. and a lead of 0.375 in. The larger gear is subjected to a constant . constantclockwise couple of 7.2 kN m. Knowing that the coefficient clockwise couple of 7.2 kip ⋅ in. Knowing that the coefficient of staticof static friction between twoisgears 0.12, determine thethat couple friction between the twothe gears 0.12, is determine the couple must that be must be applied to shaft AB to rotate the large gear counterclockwise. applied to shaft AB to rotate the large gear counterclockwise. Neglect Neglectinfriction in the atbearings at C. A, B, and C. A, B, and friction the bearings
SOLUTION FBD large gear: ΣMC = 0:
kN.m
12 in. kip.⋅ m in.==0,0 ((0.3 m)W − 7.2 kN kN kips = 600 lb W = 24 0.600
0.3 m
Block on incline: θ = tan−1
0.375 0.009in. m ==2.2785 2.153°° π (1.5 22p in.) ) 0.0381
φs = tan−1 µ s = tan−1 0.12 = 6.8428°
Q = W tan (φs − θ ) 600kN lb) tan 4.5643 = (24 ° == 47.898 4.6898° 1.97 kNlb FBD worm gear: 1.97 kN
r = 1.5 in. ΣMB = 0:
M − (1.5 in.)(m 47.898 0.0381 )(1.97lbkN ) =) 0= 0
lb ⋅.in. = 71.8kN MM= 0.075 mt
PROPRIETARY MATERIAL. © 2007 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 written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1256
PROBLEM 8.69 High-strength bolts are used in the construction of many steel structures. For a 0.025 1-in.-nominal-diameter boltbolt thethe required minimum boltbolt tension is m-nominal-diameter required minimum tension 47.25 kips, coefficient of of friction to be determine the is 47.25 kN,Assuming Assumingthethe coefficient friction to 0.35, be 0.35, determine required couple thatthat should be be applied the required couple should appliedtotothe thebolt boltand and nut. nut. The mean diameter of the thread is 0.94 in., and the lead is 0.125 0.024m, 0.003 in. m. Neglect friction between the nut nut and and washer, washer, and andassume assumethe thebolt bolttotobebesquare-threaded. square-threaded.
SOLUTION Block/incline analysis: kN
θ = tan−1 0.003 m
0.003 in. m 0.125 = 2.2785° 2.4238° 2.9531 0.0754in. m
φs = tan−1 (0.35) = 19.2900°
0.0754 m
21.5685 ° 47.25 kN
18.68 kN Q = 47.25tan 47250tan((21.5685° 21.714°) )==18.816 lb
Couple =
d 0.94 lb)) == 8844 ⋅ in.. m Q = 0.024in. 18.68 kN 0.224lbkN m (18.516 2 2
. m⋅ ft t Couple N lb Couple==224 7.37
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PROBLEM 8.70 The position of the automobile jack shown is controlled by a screw ABC that is single-threaded at each end (right-handed thread at A, left-handed thread at C). Each thread has a pitch of 2 mm and a mean diameter of 7.5 mm. If the coefficient of static friction is 0.15, determine the magnitude of the couple M that must be applied to raise the automobile. SOLUTION FBD joint D:
FAD = FCD
By symmetry:
ΣFy = 0: 2FAD sin25° − 4 kN = 0
FAD = FCD = 4.7324 kN FBD joint A:
FAE = FAD
By symmetry:
ΣFx = 0: FAC − 2 ( 4.7324 kN) cos25° = 0
FAC = 8.5780 kN Block and incline A:
θ = tan−1
2 mm = 4.8518° π (7.5 mm)
φs = tan−1 µ s = tan−1 0.15 = 8.5308°
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PROBLEM 8.70 CONTINUED
Q = (8.578 kN) tan (13.3826°) = 2.0408 kN
Couple at A:
M A = rQ 7.5 mm (2.0408 kN) = 2 = 7.653 N ⋅ m
By symmetry: Couple at C:
MC = 7.653 N ⋅ m Total couple M = 2 (7.653 N ⋅ m)
M = 15.31 N ⋅ m t
PROPRIETARY MATERIAL. © 2007 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 written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1259
PROBLEM 8.71 For the jack of Prob. 8.70, determine the magnitude of the couple M that must be applied to lower the automobile. Problem 8.70: The position of the automobile jack shown is controlled by a screw ABC that is single-threaded at each end (right-handed thread at A, left-handed thread at C). Each thread has a pitch of 2 mm and a mean diameter of 7.5 mm. If the coefficient of static friction is 0.15, determine the magnitude of the couple M that must be applied to raise the automobile. SOLUTION FBD joint D:
By symmetry:
FAD = FCD ΣFy = 0: 2FAD sin25° − 4 kN = 0
FAD = FCD = 4.7324 kN FBD joint A:
By symmetry:
FAE = FAD ΣFx = 0: FAC − 2 ( 4.7324 kN) cos25° = 0
FAC = 8.5780 kN Block and incline at A:
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PROBLEM 8.71 CONTINUED θ = tan−1
2 mm = 4.8518° π (7.5 mm)
φs = tan−1 µ s = tan−1 0.15 φs = 8.5308°
φs − θ = 3.679°
Q = (8.5780 kN) tan3.679° Q = 0.55156 kN Couple at A: M A = Qr 7.5 mm = (0.55156 kN) 2 = 2.0683 N ⋅ m
By symmetry:
Couple at C : MC = 2.0683 N ⋅ m Total couple M = 2 (2.0683 N ⋅ m)
M = 4.14 N ⋅ m t
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PROBLEM 8.72 The vise shown consists of two members connected by two double-threaded screws of mean radius 6 mm and pitch 2 mm. The lower member is threaded at A and B ( µs = 0.35) , but the upper member is not threaded. It is desired to apply two equal and opposite forces of 540 N on the blocks held between the jaws. Determine (a) which screw should be adjusted first, (b) the maximum couple applied in tightening the second screw.
SOLUTION FBD lower jaw: By symmetry B = 540 N ΣFy = 0:
− 540 N + A − 540 N = 0,
A = 1080 N
(a) since A > B when finished, adjust A first when there will be no force t Block/incline at B: (b) θ = tan−1
4 mm = 6.0566° 12π mm
φs = tan−1 µ s = tan−1 (0.35) = 19.2900°
Q = (540 N) tan25.3466° = 255.80 N Couple = rQ = (6 mm)(255.80 N) = 1535 N ⋅ mm
M = 1.535 N ⋅ m t
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PROBLEM 8.73 Solve part b of Prob. 8.72 assuming that the wrong screw is adjusted first. Problem 8.72: The vise shown consists of two members connected by two double-threaded screws of mean radius 6 mm and pitch 2 mm. The lower member is threaded at A and B ( µs = 0.35) , but the upper member is not threaded. It is desired to apply two equal and opposite forces of 540 N on the blocks held between the jaws. Determine (a) which screw should be adjusted first, (b) the maximum couple applied in tightening the second screw. SOLUTION FBD lower jaw: By symmetry B = 540 N ΣFy = 0:
− 540 N + A − 540 N = 0,
A = 1080 N
since A > B, A should be adjusted first when no force is required. If instead, B is adjusted first, Block/incline at A:
θ = tan−1
4 mm = 6.0566° 12π mm
φs = tan−1 µ s = tan−1 (0.35) = 19.2900°
Q = (1080 N) tan25.3466° = 511.59 N Couple = rQ = (6 mm)(511.59 N) = 3069.5 N ⋅ mm
M = 3.07 N ⋅ m t
Note that this is twice that required if A is adjusted first.
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PROBLEM 8.74 In the gear-pulling assembly shown, the square-threaded screw AB has a 0.0234 in. m add m. Knowing Knowing that that the the mean radius of 0.9375 add aa lead lead ofof 0.006 0.25 in. coefficient of static friction is 0.10, determine the couple which must be N on the gear. Neglect applied to the screw to produce a force of 1000 lb friction at end A of the screw.
SOLUTION Block/incline: θ = tan−−11
0.25 in. 0.006 m = 2.4302 ° 2.3369° ππin. 1.875 m 0.0468
φss = tan−−11µ ss = tan−−11(0.10) = 5.7106°
8.0475° 1000 N
Q = (1000 lb °))== 143.048 N) tan (8.1408 8.0475° 141.4 Nlb Couple lb) == 33.09 134.108 Couple = rQ = ((00.9375 . 0 2 3 4in.m ) (141.4 N N .lb m⋅ in. )(143.048
M = 134.1 ⋅ in. t 33.1 Nlb⋅ m 3
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NOTE FOR PROBLEMS 8.75–8.89
(
)
Note to instructors: In this manual, the simplification sin tan−1 µ ≈ µ is NOT used in the solution of journal bearing and axle friction problems. While this approximation may be valid for very small values of µ, there is little if any reason to use it, and the error may be significant. For example, in Problems 8.76–8.79, µ s = 0.50, and the error made by using the approximation is about 11.8%.
PROPRIETARY MATERIAL. © 2007 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 written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1265
