Engineering Mechanics - Statics Chapter 7 pp gg y sag is h, determine the maximum tension in the chain. Given: w 0.5 lb ft = L 60 ft= h 3 ft= Solution: Form Example 7-15 y F H w cosh wx F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = x y d d sinh wx F H ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Guess F H 1 lb= Given h F H w cosh w F H L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = F H Find F H () = F H 75.2 lb= θ max atan sinh w F H L 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = T max F H cos θ max () = T max 76.7 lb= Problem 7-123 Draw the shear and moment diagrams for the beam. Units Used: kN 10 3 N= Given: w 2 kN m = a 5 m= b 5 m= M 50 kN m⋅= Solution: Guesses A 1 N= C 1 N= 761 © 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 7 Given wa b a 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Aa b+()− M− 0= AC+ wa− 0= A C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find AC,()= x 1 0 0.01 a, a = V 1 x 1 () Awx 1 − () 1 kN = M 1 x 1 () Ax 1 wx 1 x 1 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 kN m⋅ = x 2 a 1.01a, ab+ = V 2 x 2 () C− 1 kN = M 2 x 2 () M− Ca b+ x 2 − () + ⎡ ⎣ ⎤ ⎦ 1 kN m⋅ = 0246810 10 5 0 5 Distance (m) Force (kN) V 1 x 1 () V 2 x 2 () x 1 x 2 , 0246810 60 40 20 0 20 Distance (m) Moment (kN-m) M 1 x 1 () M 2 x 2 () x 1 x 2 , 762 © 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 8-1 The horizontal force is P . Determine the normal and frictional forces acting on the crate of weight W. The friction coefficients are μ k and μ s . Given: W 300 lb= P 80 lb= μ s 0.3= μ k 0.2= θ 20 deg= Solution: Assume no slipping: Σ F x = 0; P cos θ () W sin θ () − F c + 0= F c P− cos θ () W sin θ () += F c 27.4 lb= Σ F y = 0; N c W cos θ () − P sin θ () − 0= N c W cos θ () P sin θ () ⋅+= N c 309lb= Check F cmax μ s N c = F cmax 92.8 lb= F cmax F c > Problem 8-2 Determine the magnitude of force P needed to start towing the crate of mass M. Also determine the location of the resultant normal force acting on the crate, measured from point A. Given: M 40 kg= c 200 mm= μ s 0.3= d 3= a 400 mm= e 4= b 800 mm= 763 © 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 Initial guesses: N C 200 N= P 50 N = Given d d 2 e 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ P μ s N C − 0= Σ F x = 0; Σ F y = 0; N C Mg− eP d 2 e 2 + + 0= N C P ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find N C P, () = N C 280.2 N= P 140 N= Σ M O = 0; μ s − N C a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N 1 x− eP d 2 e 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 0= x 1− 2 μ s N C ad 2 e 2 + ePb− N C d 2 e 2 + = x 123.51 mm= Thus, the distance from A is Ax b 2 += A 523.51 mm= Problem 8-3 Determine the friction force on the crate of mass M, and the resultant normal force and its position x, measured from point A, if the force is P . Given: M 40 kg= μ s 0.5= a 400 mm= μ k 0.2= b 800 mm= d 3= c 200 mm= e 4= P 300 N= Solution: Initial guesses: F C 25 N= N C 100 N= 764 Solution: © 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 Σ F x = 0; P d d 2 e 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F C − 0= N C Mg− P e d 2 e 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 0= Σ F y = 0; F C N C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find F C N C , () = F Cmax μ s N C = Since F C 180.00 N= > F Cmax 76.13 N= then the crate slips F C μ k N C = F C N C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 30.5 152.3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= Σ M O = 0; N C − xP e d 2 e 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a+ P d d 2 e 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ c− 0= xP− e− adc+ N C d 2 e 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = Since x 0.39 m= < b 2 0.40 m= Then the block does not tip. x 1 ax+= x 1 0.79 m= Problem 8-4 The loose-fitting collar is supported by the pipe for which the coefficient of static friction at the points of contact A and B is μ s . Determine the smallest dimension d so the rod will not slip when the load P is applied. Given: μ s 0.2= 765 © 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 Solution: Σ M A = 0; N B dPL d 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − μ s N B d− 0= N B PL d 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 μ s − () d = Σ M B = 0; N A d μ s N A d+ PL d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= N A PL d 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 μ s + () d = Σ F y = 0; μ s N A N B + () P− 0= μ s P d L d 2 + 1 μ s + L d 2 − 1 μ s − + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ P= Thus, d 2 μ s L= k 2 μ s = dkL= Problem 8-5 The spool of wire having a mass M rests on the ground at A and against the wall at B. Determine the force P required to begin pulling the wire horizontally off the spool. The coefficient of static friction between the spool and its points of contact is μ s . Units Used: kN 10 3 N= Given: M 150 kg= μ s 0.25= a 0.45 m= b 0.25 m= Solution: Initial guesses: P 100 N= F A 10 N= N A 20 N= N B 30 N= F B 10 N= 766 © 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 Σ F y = 0; N A F B + Mg− 0= Σ F x = 0; F A N B − P+ 0= Σ M B = 0; P− bMga+ N A a− F A a+ 0= F A μ s N A = F B μ s N B = P F A F B N A N B ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find PF A , F B , N A , N B , () = F A N A F B N B ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 0.28 1.12 0.36 1.42 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ kN= P 1.14kN= Problem 8-6 The spool of wire having a mass M rests on the ground at A and against the wall at B. Determine the forces acting on the spool at A and B for the given force P. The coefficient of static friction between the spool and the ground at point A is μ s . The wall at B is smooth. Units Used: kN 10 3 N= Given: P 800 N= a 0.45 m= M 150 kg= b 0.25 m= μ s 0.35= Solution: Assume no slipping Initial guesses : F A 10N= N A 10N= N B 10N= F Amax 10N= 767 Given © 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 Σ F x = 0; F A N B − P+ 0= Σ F y = 0; N A Mg− 0= Σ M 0 = 0; P− bF A a+ 0= F Amax μ s N A = F A F Amax N A N B ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find F A F Amax , N A , N B , () = F A F Amax ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 444 515 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N= If F A 444 N= < F Amax 515 N= then our no-slip assumption is good. N A F A ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1.47 0.44 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kN= N B 1.24 kN= Problem 8-7 The crate has a mass M and is subjected to a towing force P acting at an angle θ 1 with the horizontal. If the coefficient of static friction is μ s , determine the magnitude of P to just start the crate moving down the plane. Given: M 350 kg= θ 1 20 deg= θ 2 10 deg= μ s 0.5= g 9.81 m s 2 = Solution: Initial guesses: N C 10N= P 20N= 768 © 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 Σ F x = 0; P cos θ 1 θ 2 + () μ s N C − Mgsin θ 2 () + 0= Σ F y = 0; N C Mgcos θ 2 () − P sin θ 1 θ 2 + () + 0= N C P ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find N C P, () = N C 2891 N= P 981 N= Problem 8-8 The winch on the truck is used to hoist the garbage bin onto the bed of the truck. If the loaded bin has weight W and center of gravity at G, determine the force in the cable needed to begin the lift. The coefficients of static friction at A and B are μ Α and μ B respectively. Neglect the height of the support at A. Units Used: kip 10 3 lb= Given: W 8.5 kip= μ A 0.3= μ B 0.2= a 10 ft= b 12 ft= θ 30 deg= Solution: The initial guesses are T 1lb= N B 1lb= N A 1lb= Given Σ M B = 0; Wb N A ab+()− 0= 769 © 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 + → Σ F x = 0; T cos θ () μ B N B cos θ () − N B sin θ () − μ A N A − 0= + ↑ Σ F y = 0; N A W− T sin θ () + N B cos θ () + μ B N B sin θ () − 0= T N A N B ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find TN A , N B , () = N A N B ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4.64 2.65 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ kip= T 3.67kip= Problem 8-9 The motorcyclist travels with constant velocity along a straight, horizontal, banked road. If he aligns his bike so that the tires are perpendicular to the road at A, determine the frictional force at A. The man has a mass M C and a mass center at G C , and the motorcycle has a mass M m and a mass center at G m . If the coefficient of static friction at A is μ A , will the bike slip? Given: M C 60 kg= M m 120 kg= μ A 0.4= θ 20 deg= g 9.81 m s 2 = Solution: Assume no slipping Σ F y = 0; N A M m M C + () g cos θ () − 0= N A M m M C + () g cos θ () = N A 1659 N= Σ F x = 0; F A M m M C + () gsin θ () − F A M m M C + () gsin θ () = F A 604 N= F Amax μ A N A = 770 © 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: W1 = 150 lb W2 = 100 lb a = 2 ft b = 3 ft μ s = 0.8 μ's = 0.3 Assume crate A slips: ΣF y = 0; NA − W2 = 0 NA = W2 NA = 100 .00 lb ΣF x = 0; P − μ s NA = 0 P 1 = μ s NA P 1 = 80.00 lb Assume crate B slips: ΣF y = 0; NB − 2 W2 = 0 NB = 2 W2 NB = 20 0.00 lb ΣF x = 0; P − μ's NB = 0 P 2 = μ's NB P 2 = 60.00 lb Assume both crates A and B tip: ΣM = 0; 2 W2... publisher Engineering Mechanics - Statics Chapter 8 P y = W L − Nc P y = 110. 00 lb 2 P = Px + Py 2 P = 110 lb The length on the ground is supported by L = Nc = 50.00 lbthus Nc W L = 6 .25 ft Problem 8 -2 8 The fork lift has a weight W1 and center of gravity at G If the rear wheels are powered, whereas the front wheels are free to roll, determine the maximum number of crates, each of weight W2 that the... publisher Engineering Mechanics - Statics Chapter 8 ⎛ T ⎞ ⎛ 173 .21 ⎞ ⎜ ⎟ NA ⎟ ⎜ 21 5.31 ⎟ ⎜ ⎜ ⎟ ⎜ FA ⎟ = ⎜ 107 .66 ⎟ N ⎜ ⎟ ⎜ ⎟ ⎜ F ⎟ ⎜ 21 5.31 ⎟ ⎜ Nground ⎟ ⎝ 538 .28 ⎠ ⎝ ⎠ mass = 54.9 kg Problem * 8 -2 0 The pipe is hoisted using the tongs If the coefficient of static friction at A and B is μs, determine the smallest dimension b so that any pipe of inner diameter d can be lifted Solution: W − 2 FB = 0 ⎛... Given: W = 100 lb μ s = 0.8 μ's = 0.3 a = 2 ft b = 3 ft Solution: Assume crate A slips: ΣF y = 0; NA − W = 0 NA = W NA = 100 .00 lb ΣF x = 0; P − μ s NA = 0 P 1 = μ s NA P 1 = 80.00 lb Assume crate B slips: ΣF y = 0; NB − 2 W = 0 NB = 2 W NB = 20 0.00 lb ΣF x = 0; P − μ's NB = 0 P 2 = μ's NB P 2 = 60.00 lb Assume both crates A and B tip: ΣM = 0; 2W ⎛ a⎞ − P b = 0 ⎜ ⎟ ⎝ 2 P = min ( P 1 , P 2 , P 3 ) P3... the smallest dimension b so that any pipe of inner diameter d can be lifted Solution: W − 2 FB = 0 ⎛ W⎞ b − N h − F ⎛ d⎞ = 0 ⎜ ⎟ B B⎜ ⎟ 2 ⎝ 2 Thus FB = NB = W 2 W ( 2 b − d) 4h Require F B ≤ μ s NB W 2 ≤ μ s W ( 2b − d) 2 h ≤ μ s ( 2b − d) 4h b> h μs + d 2 Problem 8 -2 1 A very thin bookmark having a width a is in the middle of a dictionary of weight W If the pages are b by c, determine the force P needed... between the wedge and the surfaces of contact is μs Given: M = 5 kg μ s = 0 .2 φ = 60 deg g = 9.81 m 2 s Solution: Initial guesses: NW = 10 N θ = 10 deg Given ΣF x = 0; M g sin ( θ ) − 2 μ s NW = 0 ΣF z = 0; 2 NW sin ⎜ ⎛ φ ⎞ − M g cos ( θ ) = 0 ⎟ 2 Solving, ⎛ NW ⎞ ⎜ ⎟ = Find ( NW , θ ) ⎝ θ ⎠ NW = 45.5 N θ = 21 .8 deg 791 © 20 07 R C Hibbeler Published by Pearson Education, Inc., Upper Saddle River,... from the publisher Engineering Mechanics - Statics Chapter 8 Given: μ s1 = 0.5 a = 75 mm μ s2 = 0.4 b = 20 mm P = 300 N c = 30 mm m θ = 60 deg g = 9.81 2 s Solution: Assume that we are on the verge of slipping at every surface Guesses T = 1N NA = 1N F = 1N Nground = 1N F A = 1N mass = 1kg Given 2 T sin ( θ ) − P = 0 −T sin ( θ ) ( b + c) − T cos ( θ ) a − FA b + NA a = 0 F A = μ s1 NA 2 FA − F = 0 Nground... reproduced, in any form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics P = μ s NA Chapter 8 P = 707.37 lb Crate: Nc − W2 = 0 ΣF y = 0; Nc = W2 ΣF x = 0; Nc = 300.00 lb P' − μ's Nc = 0 P' = μ's Nc Thus n = P P' P' = 105 .00 lb n = 6.74 n = floor ( n) n = 6.00 Problem 8 -2 9 The brake is to be designed to be self locking, that is, it will not rotate when no... 0; μ s NB − ⎛ a ⎞F =0 ⎜ 2 2⎟ A ⎝ s +a ⎠ ⎛ μs s ⎟ ⎞ ⎛ a ⎞F = 0 ⎜ FA − ⎜ A 2 2⎟ ⎜ s2 + a2 ⎟ ⎝ ⎠ ⎝ s +a ⎠ μs s = a s = a μs s = 5.00 in Problem 8-1 6 The chair has a weight W and center of gravity at G It is propped against the door as shown If the coefficient of static friction at A is μA, determine the smallest force P that must be applied to the handle to open the door 776 © 20 07 R C Hibbeler Published... not slip Given: W = 90 lb μ s = 0 .25 7 82 © 20 07 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 Wman = 150 lb θ . M 1 x 1 () Ax 1 wx 1 x 1 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 kN m⋅ = x 2 a 1.01a, ab+ = V 2 x 2 () C− 1 kN = M 2 x 2 () M− Ca b+ x 2 − () + ⎡ ⎣ ⎤ ⎦ 1 kN m⋅ = 024 6 810 10 5 0 5 Distance (m) Force (kN) V 1 x 1 () V 2 x 2 () x 1 x 2 , 024 6 810 60 40 20 0 20 Distance. N C P, () = N C 28 0 .2 N= P 140 N= Σ M O = 0; μ s − N C a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ N 1 x− eP d 2 e 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 0= x 1− 2 μ s N C ad 2 e 2 + ePb− N C d 2 e 2 + = x 123 .51 mm= Thus,. the publisher. Engineering Mechanics - Statics Chapter 8 Initial guesses: N C 20 0 N= P 50 N = Given d d 2 e 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ P μ s N C − 0= Σ F x = 0; Σ F y = 0; N C Mg− eP d 2 e 2 + + 0= N C P ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find