Engineering Mechanics - Statics Chapter 8 Frictional Forces on Screw: Here θ atan h 2 π d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ = θ 6.06 deg= φ s atan μ s () = φ s 16.70 deg= Applying Eq.8-3, we have MP d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ tan θφ s + () = P 2 M d tan θφ s + () ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = P 127.15 N= Note since φ s 16.70 deg= > θ 6.06 deg= , the screw is self-locking. It will not unscrew even if the moment M is removed. Equations of Equilibrium and Friction: Σ M c = 0; c b 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Pb F E cos β () b− F E sin β () a− 0= 841 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 8 F E Pcb b 2 c 2 + cos β () b sin β () a+ () = F E 72.7 N= The equilibrium of clamped block requires that F D F E = F D 72.7 N= Problem 8-82 The clamp provides pressure from several directions on the edges of the board. If the square-threaded screw has a lead h, radius r, and the coefficient of static friction is μ s , determine the horizontal force developed on the board at A and the vertical forces developed at B and C if a torque M is applied to the handle to tighten it further. The blocks at B and C are pin-connected to the board. Given: h 3mm= r 10 mm= μ s 0.4= M 1.5 N m⋅= β 45 deg= Solution: φ s atan μ s () = φ s 21.801 deg= θ atan h 2 π r ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 2.734 deg= MA x rtan φ s θ + () = A x M r tan φ s θ + () = A x 329 N= + → Σ F x = 0; A x 2Tcos β () − 0= T 1 2 A x cos β () ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T 232.36 N= C y Tsin β () = B y C y = B y C y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 164.3 164.3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= 842 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 8 Problem 8-83 The two blocks under the double wedge are brought together using a left and right square-threaded screw. If the mean diameter is d, the lead is r l , and the coefficient of static friction is μ s , determine the torque needed to draw the blocks together. The coefficient of static friction between each block and its surfaces of contact is μ ' s . Units Used: kN 10 3 N= Given: F 5kN= θ 20 deg= d 20 mm= r l 5mm= μ s 0.4= μ ' s 0.4= Solution: Top block: F− 2N 1 cos θ () + 2 μ ' s N 1 sin θ () − 0= N 1 F 2 cos θ () μ ' s sin θ () − () = N 1 3.1138 kN= Bottom block: N' N 1 cos θ () − μ ' s N 1 sin θ () + 0= N' N 1 cos θ () μ ' s N 1 sin θ () −= N' 2.50 kN= N 1 − sin θ () μ ' s N 1 cos θ () − T+ μ ' s N'− 0= TN 1 sin θ () μ ' s N 1 cos θ () + μ ' s N'+= T 3.2354kN= 843 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 8 φ atan μ s () = φ 21.80 deg= θ atan r l πd ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 4.55 deg= Since there are two blocks, M 2T d 2 tan θφ + () = M 32N m⋅= Problem 8-84 The two blocks under the double wedge are brought together using a left and right square-threaded screw. If the mean diameter is d, the lead is r l , and the coefficient of static friction is μ s , determine the torque needed to spread the blocks apart. The coefficient of static friction between each block and its surfaces of contact is μ ' s . Units Used: kN 10 3 N= Given: F 5kN= θ 20 deg= d 20 mm= r l 5mm= μ s 0.4= μ ' s 0.4= Solution: Top block: F− 2N 1 cos θ () + 2 μ ' s N 1 sin θ () + 0= N 1 F 2 cos θ () μ ' s sin θ () + () = N 1 2.3223 kN= 844 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 8 Bottom block : N' N 1 cos θ () − μ ' s N 1 sin θ () − 0= N' N 1 cos θ () μ ' s N 1 sin θ () += N' 2.50 kN= N 1 − sin θ () μ ' s N 1 cos θ () + T− μ ' s N'+ 0= TN 1 − sin θ () μ ' s N 1 cos θ () + μ ' s N'+= T 1.0786kN= φ atan μ s () = φ 21.80 deg= θ atan r l π d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 4.55 deg= Since there are two blocks, M 2T d 2 tan φθ − () = M 6.7 N m⋅= Problem 8-85 The cord supporting the cylinder of mass M passes around three pegs, A, B, C, where the coefficient of friction is μ s . Determine the range of values for the magnitude of the horizontal force P for which the cylinder will not move up or down. Given: M 6kg= θ 45 deg= μ s 0.2= g 9.81 m s 2 = Solution: Total angle β 5 2 π 4 θ −= β 270.00 deg= 845 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 8 Forces P min Mge μ s − β = P max Mge μ s β = Answer P min 15.9 N= < P < P max 217.4 N= Problem 8-86 The truck, which has mass m t , is to be lowered down the slope by a rope that is wrapped around a tree. If the wheels are free to roll and the man at A can resist a pull P, determine the minimum number of turns the rope should be wrapped around the tree to lower the truck at a constant speed. The coefficient of kinetic friction between the tree and rope is μ k . Units Used: Mg 1000 kg= Given: m t 3.4 Mg= P 300 N= θ 20 deg= μ k 0.3= g 9.81 m s 2 = Solution: Σ F x = 0; T 2 m t gsin θ () − 0= T 2 m t gsin θ () = T 2 11407.74 N= T 2 Pe μ k β = β ln T 2 P ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ μ k = β 694.86 deg= Use ceil β 360 deg ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2.00= turns 846 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 8 Problem 8-87 The wheel is subjected to a torque M . If the coefficient of kinetic friction between the band brake and the rim of the wheel is μ k , determine the smallest horizontal force P that must be applied to the lever to stop the wheel. Given: a 400 mm= d 25 mm= b 100 mm= r 150 mm= c 50 mm= M 50 N m⋅= μ k 0.3= Solution: Initial guesses: T 1 5N= T 2 10 N= Given Wheel: Σ M 0 = 0; T 2 − rT 1 r+ M+ 0= T 2 T 1 e μ k 3 π 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T 1 T 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find T 1 T 2 , () = T 1 54.66 N= Link: Σ M B = 0; T 1 cFd− 0= FT 1 c d ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F 109.32 N= Lever: Σ M A = 0; P− aFb+ 0= PF b a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = P 27.3 N= Problem 8-88 A cylinder A has a mass M. Determine the smallest force P applied to the handle of the lever required for equilibrium. The coefficient of static friction between the belt and the wheel is μ s . The drum is pin connected at its center, B. 847 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 8 Given: M 75 kg= a 700 mm= b 25 mm= c 300 mm= d 200 mm= e 1 60 mm= μ s 0.3= e 2.718= Solution: Initial guesses: T 1 1N= T 2 1N= P 1N= Given Drum: T 2 T 1 e μ s 3 π 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T 2 − cT 1 c+ Mgd+ 0= Lever: T 1 − e 1 T 2 b+ Pa− 0= T 1 T 2 P ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find T 1 T 2 , P, () = T 1 T 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 157.7 648.2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= P 9.63 N= Problem 8-89 Determine the largest mass of cylinder A that can be supported from the drum if a force P is applied to the handle of the lever. The coefficient of static friction between the belt and the wheel is μ s . The drum is pin supported at its center, B. Given: P 20 N= 848 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 8 a 700 mm= b 25 mm= c 300 mm= d 200 mm= e 1 60 mm= μ s 0.3= e 2.718= Solution: Initial guesses: T 1 1N= T 2 1N= M 1kg= Given Drum: T 2 T 1 e μ s 3 π 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T 2 − cT 1 c+ Mgd+ 0= Lever: T 1 − e 1 T 2 b+ Pa− 0= T 1 T 2 M ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find T 1 T 2 , M, () = T 1 T 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 327.4 1.3 10 3 × ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= M 155.7 kg= Problem 8-90 The uniform bar AB is supported by a rope that passes over a frictionless pulley at C and a fixed peg at D. If the coefficient of static friction between the rope and the peg is μ D , determine the smallest distance x from the end of the bar at which a force F may be placed and not cause the bar to move. Given: F 20 N= a 1m= 849 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 8 μ D 0.3= Solution: Initial guesses: T A 5N= T B 10 N= x 10 m= Given Σ M A = 0; F− xT B a+ 0= Σ F y = 0; T A T B + F− 0= T A T B e μ D π 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T A T B x ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find T A T B , x, () = x 0.38 m= Problem 8-91 Determine the smallest lever force P needed to prevent the wheel from rotating if it is subjected to a torque M . The coefficient of static friction between the belt and the wheel is μ s . The wheel is pin-connected at its center, B. Given: M 250 N m= μ s 0.3= r 400 mm= a 200 mm= b 750 mm= Solution: Σ M A = 0; F− aPab+()+ 0= 850 © 2007 R. C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. [...]... - Statics ⌠ ⎮ P= ⎮ ⌡ 2 0 R2 ⌠ ⎮ ⎮ ⌡R Chapter 8 ⎛ R2 ⎞ ⎟ r dr dθ ⎝ r ⎠ p0 ⎜ 1 P = 2 p0 R2 ( R 2 − R 1 ) p0 = P 2 R 2 ( R 2 − R 1 ) 2 ⌠ ⌠ M = ⎮ r dF = ⎮ ⌡ ⌡ 0 A ⌠ ⎮ M= ⎮ ⌡ 2 0 ⌠ ⎮ ⎮ ⌡ 2 R2 ⌠ ⎮ ⌡R rμ s p0 r dr dθ 1 ( ⎛ R2 ⎞ 2 2 2 ⎟ r dr dθ = π μ s p0 R 2 R 2 − R 1 r ⎠ ⎝ μ s p0 ⎜ ) 0 ( P 2 2 ⎤ ⎡ M = π μ s⎢ ⎥ R2 R2 − R1 2 R2( R2 − R1 )⎦ ) M= μ s P(R2 + R1) 2 Problem 8-1 18 A disk having an outer... Solution: ⎤ 2 ⎡c − b a −b μ s⎢ kP + ( 1 − k)P⎥ 2 2 2 ⎥ 3 ⎢ 2 a −b ⎣c − b ⎦ 3 M = 3 3 3 M = 16.1 N⋅ m Problem 8-1 10 The annular ring bearing is subjected to a thrust P If the coefficient of static friction is μs, determine the torque M that must be applied to overcome friction Given: P = 800 lb μ s = 0 .35 a = 0.75 in b = 1 in c = 2 in Solution: ⎛b − c ⎟ ⎞ 2 M = μ s P⎜ ⎜ b2 − c2 ⎟ 3 ⎝ ⎠ 3 3 M = 36 .3 lb⋅ ft... from the publisher Engineering Mechanics - Statics Chapter 8 ⎛ − f ⎞T + μ N = 0 sAB B ⎜ 2 2⎟ ⎝ f +d ⎠ ⎛ NB ⎞ ⎜ ⎟ ⎜ T ⎟ = Find ( N , T , m , x) B A ⎜ mA ⎟ ⎜ ⎟ ⎝ x ⎠ Since x = 51.6 mm < b 2 f d ⎛ ⎛ ⎞ ⎞ b ⎜ 2 2 ⎟ T a − ⎜ 2 2 ⎟ T 2 − NB x = 0 ⎝ f +d ⎠ ⎝ f +d ⎠ ⎛ NB ⎞ ⎛ 64. 63 ⎞ ⎜ ⎟=⎜ ⎟N ⎝ T ⎠ ⎝ 20 .20 ⎠ mA = 7. 82 kg x = 0.0 52 m = 125 mm our assumption is correct mA = 7. 82 kg Problem 8-1 07 The collar bearing... may be reproduced, in any form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 8 ⎛d − c ⎟ ⎞ 2 M2 = μ s P' ⎜ 3 ⎜ d2 − c2 ⎟ ⎝ ⎠ P' = P' b − P a = 0 P = P' 3 3M2 ⎛ d2 − c2 ⎞ ⎜ ⎟ 3 P' = 88.5 N 2 s ⎜ d3 − c3 ⎟ ⎝ ⎠ ⎛ b⎞ ⎜ ⎟ ⎝ a⎠ P = 118 N Problem 8-1 13 The shaft of diameter b is held in the hole such that the normal pressure acting around the shaft... - Statics Chapter 8 Given: a = 2 in b = 3 in P = 500 lb M = 3 lb ft Solution: ⎡⎛ b ⎞ 3 ⎛ a ⎞ 3 ⎢⎜ ⎟ − ⎜ ⎟ ⎥ 3 3⎞ a ⎢ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎥ = ⎛ aμ k P ⎞ ⎛ b − a ⎟ ⎜ M = μk P ⎜ ⎟ 2 ⎝ 2b ⎠ ⎜ 2 2⎟ b ⎢ b 2 a⎞ ⎝b − a ⎠ ⎞ ⎢⎛ ⎟ − ⎛ ⎟ ⎥ ⎜ ⎜ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦ (2 2 2M b b − a μk = (3 aP b − a 3 ) ) μ k = 0.0568 Problem 8-1 09 The double-collar bearing is subjected to an axial force P Assuming that collar A supports kP... by any means, without permission in writing from the publisher P πR 2 Engineering Mechanics - Statics dN = pdA = 2P ⎛ 2 rdr ⎞ = rdr ⎜ ⎟ 2 ⎝ cos ( θ ) ⎠ 2 (θ ) πR R cos P ⌠ ⌠ ⎮ M = ⎮ r dF = rμ s dN = ⎮ ⌡ ⌡ M= Chapter 8 R ⎛ 2 s P ⎞ ⌠ 2 ⎜ 2 ⎟ ⎮ r dr ⌡ ⎝ R cos ( θ ) ⎠ 0 ⎛ 2 s P ⎞ R3 2 s P R = ⎜ 2 ⎟ (θ ) ⎠ 3 3 cos (θ ) ⎝ R cos Problem 8-1 16 The tractor is used to push the pipe of weight W To do this it... Given: r = 3 in d = 0.5 in W = 18 lb F 1 = 5 lb F 2 = 5.5 lb Solution: + ↑Σ Fy = 0; ΣMO = 0; R − W − F1 − F2 = 0 R = W + F1 + F2 −F 2 r + F1 r + R rf = 0 rf = r⎜ ⎛ F2 − F1 ⎞ ⎟ ⎝ R ⎠ ⎛ rf ⎞ ⎝ d⎠ d sin ( φ k) 2 R = 28 .5 lb rf = 0.0 52 63 in φ k = asin ⎜ 2 ⎟ φ k = 12. 15 deg μ = tan ( φ k) rf = μ = 0 .21 5 Also, ΣMO = 0; ⎛ d⎞ = 0 ⎟ ⎝ 2 −F 2 r + F1 r + F ⎜ ⎛ F2 − F1 ⎞ ⎟ ⎝ d ⎠ F = 2 r⎜ F = 6 lb 875 © 20 07 R C... 0 .25 μ sC = 0 .3 a = 0 .3 m b = 0 .25 m c = 0.4 m d = 3 f = 4 g = 9.81 m 2 s Solution: Assume block A slips but does not tip The initial guesses: Given NB = 100 N T = 50 N f⎞ β = π − atan ⎛ ⎟ ⎜ ⎝ d⎠ mD = 1 kg x = 10 mm μ sCβ mD g = T e d ⎛ ⎞ ⎜ 2 2 ⎟ T − mA g + NB = 0 ⎝ f +d ⎠ ⎛ − f ⎞T + μ N = 0 sAB B ⎜ 2 2⎟ ⎝ f +d ⎠ f d ⎛ ⎛ ⎞ ⎞ ⎛ b⎞ ⎜ 2 2 ⎟ T a − ⎜ 2 2 ⎟ T⎜ 2 ⎟ − NB x = 0 ⎝ ⎠ ⎝ f +d ⎠ ⎝ f +d ⎠ 864 © 20 07... static friction is μs, determine the torque M required to overcome friction Given: a = 2 in b = 3 in P = 800 lb μ s = 0 .3 Solution: ⎡ ⎛ b ⎞ 3 ⎛ a ⎞ 3 ⎢⎜ ⎟ − ⎜ ⎟ ⎥ a ⎝ 2 ⎝ 2 ⎥ M = μ s P⎢ 2 b ⎢ b 2 ⎛ ⎞ − ⎛ a⎞ ⎥ ⎢⎜ ⎟ ⎜ ⎟ ⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎦ M = ⎡1 a μ s P⎢ b 2 ⎛ a2 + a b + b2 ⎞⎤ ⎜ ⎟⎥ ⎝ a + b ⎠⎦ M = 30 4.00 lb⋅ in Problem 8-1 08 The collar bearing uniformly supports an axial force P If a torque M is applied... Mechanics - Statics Chapter 8 ⎛ dθ ⎞ − T sin ⎛ dθ ⎞ + 2dN sin ⎛ α ⎞ = 0 ⎟ ⎜ ⎟ ⎜ ⎟ 2 2 2 −( T + dT) sin ⎜ ΣF y = 0; Since dθ, dN, and dT are small, these become ⎛α⎞ ⎟ 2 Tdθ = 2dN sin ⎜ dT = 2 dN dT =μ T Combine dθ ⎛α⎞ ⎟ 2 sin⎜ θ = 0 , T = T1 Integrate from to θ = β , T = T2 μβ We get, T2 = T1 e ⎛α⎞ ⎟ 2 sin⎜ Q.E.D Problem 8-9 6 A V-fan-belt (V-angle θ) of an automobile engine passes around the . publisher. Engineering Mechanics - Statics Chapter 8 a 700 mm= b 25 mm= c 30 0 mm= d 20 0 mm= e 1 60 mm= μ s 0 .3= e 2. 718= Solution: Initial guesses: T 1 1N= T 2 1N= M 1kg= Given Drum: T 2 T 1 e μ s 3 π 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T 2 −. tan φ s θ + () = A x 32 9 N= + → Σ F x = 0; A x 2Tcos β () − 0= T 1 2 A x cos β () ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T 23 2 .36 N= C y Tsin β () = B y C y = B y C y ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 164 .3 164 .3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= 8 42 © 20 07 R. C. Hibbeler publisher. Engineering Mechanics - Statics Chapter 8 Given: M 75 kg= a 700 mm= b 25 mm= c 30 0 mm= d 20 0 mm= e 1 60 mm= μ s 0 .3= e 2. 718= Solution: Initial guesses: T 1 1N= T 2 1N= P 1N= Given Drum: T 2 T 1 e μ s 3 π 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T 2 −