# Statics_Part38.pdf

## Civil Engineering 2110 with Shana at Marquette University *

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Engineering Mechanics - Statics Chapter 8 Given: θ 30 deg= Solution: ΣF x = 0; μ N A cos θ() P+ N A sin θ()− 0= P μ cos θ() sin θ()−()N A = ΣF y = 0; μ N A sin θ() W− N A cos θ()+ 0= W μ sin θ() cos θ()+()N A = ΣΜ Α = 0; W− r sin θ() Pr rcos θ()+()+ 0= W sin θ() P 1 cos θ()+()= μ sin θ() cos θ()+()sin θ() sin θ() μ cos θ()− 1 cos θ()+()= μ sin θ() 1 cos θ()+ = μ 0.27= Problem 8-18 The uniform hoop of weight W is suspended from the peg at A and a horizontal force P is slowly applied at B. If the coefficient of static friction between the hoop and peg is μ s , determine if it is possible for the hoop to reach an angle θ before the hoop begins to slip. 778 © 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: μ s 0.2= θ 30 deg= Solution: ΣF x = 0; μ N A cos θ() P+ N A sin θ()− 0= P μ cos θ() sin θ()−()N A = ΣF y = 0; μ N A sin θ() W− N A cos θ()+ 0= W μ sin θ() cos θ()+()N A = ΣΜ Α = 0; W− r sin θ() Pr rcos θ()+()+ 0= W sin θ() P 1 cos θ()+()= μ sin θ() cos θ()+()sin θ() sin θ() μ cos θ()− 1 cos θ()+()= μ sin θ() 1 cos θ()+ = μ 0.27= If μ s 0.20= < μ 0.27= then it is not possible to reach θ 30.00 deg= . Problem 8-19 The coefficient of static friction between the shoes at A and B of the tongs and the pallet is μ s1 and between the pallet and the floor μ s2 . If a horizontal towing force P is applied to the tongs, determine the largest mass that can be towed. 779 © 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 s1 0.5= a 75 mm= μ s2 0.4= b 20 mm= P 300 N= c 30 mm= g 9.81 m s 2 = θ 60 deg= Solution: Assume that we are on the verge of slipping at every surface. Guesses T 1N= N A 1N= F 1N= N ground 1N= F A 1N= mass 1kg= Given 2 T sin θ() P− 0= T− sin θ() bc+()T cos θ()a− F A b− N A a+ 0= F A μ s1 N A = 2 F A F− 0= N ground mass g− 0= F μ s2 N ground = T N A F A F N ground mass ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find TN A , F A , F, N ground , mass,()= 780 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 T N A F A F N ground ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ 173.21 215.31 107.66 215.31 538.28 ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ N= mass 54.9 kg= Problem *8-20 The pipe is hoisted using the tongs. If the coefficient of static friction at A and B is μ s , determine the smallest dimension b so that any pipe of inner diameter d can be lifted. Solution: W 2 F B − 0= W 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ bN B h− F B d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= Thus F B W 2 = N B W 2 bd−() 4h = Require F B μ s N B ≤ W 2 μ s W 2bd−() 4 h ≤ 2 h μ s 2bd−()≤ b h μ s d 2 +> Problem 8-21 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 to start to pull the bookmark out.The coefficient of static friction between the bookmark and the paper is μ s . Assume the pressure on each page and the bookmark is uniform. 781 © 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: a 1in= W 10 lb= b 8in= c 10 in= μ s 0.7= Solution: Pressure on book mark : P 1 2 W bc = P 0.06 lb in 2 = Normal force on bookmark: NPca= F μ s N= F 0.44 lb= ΣF x = 0; P 2F− 0= P 2F= P 0.88 lb= Problem 8-22 The uniform dresser has weight W and rests on a tile floor for which the coefficient of friction is μ s . If the man pushes on it in the direction θ, determine the smallest magnitude of force F needed to move the dresser. Also, if the man has a weight W man, , determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip. Given: W 90 lb= μ s 0.25= 782 © 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 W man 150 lb= θ 0 deg= Solution: Dresser: Guesses N D 1lb= F 1lb= Given + ↑ Σ F y = 0; N D W− F sin θ()− 0= + → Σ F x = 0; F cos θ() μ s N D − 0= N D F ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find N D F,()= F 22.50 lb= Man: Guesses N m 1lb= μ m 0.2= Given + ↑ N m W man − F sin θ()+ 0= Σ F y = 0; + → Σ F x = 0; F− cos θ() μ m N m + 0= N m μ m ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find N m μ m ,()= μ m 0.15= Problem 8-23 The uniform dresser has weight W and rests on a tile floor for which the coefficient of friction is μ s . If the man pushes on it in the direction θ, determine the smallest magnitude of force F needed to move the dresser. Also, if the man has a weight W man , determine the smallest coefficient of static friction between his shoes and the floor so that he does not slip. Given: W 90 lb= 783 © 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 s 0.25= W man 150 lb= θ 30 deg= Solution: Dresser: Guesses N D 1lb= F 1lb= Given + ↑ Σ F y = 0; N D W− F sin θ()− 0= + → Σ F x = 0; F cos θ() μ s N D − 0= N D F ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find N D F,()= F 30.36 lb= Guesses N m 1lb= μ m 0.2= Man: Given + ↑ N m W man − F sin θ()+ 0= Σ F y = 0; + → Σ F x = 0; F− cos θ() μ m N m + 0= N m μ m ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find N m μ m ,()= μ m 0.195= Problem 8-24 The cam is subjected to a couple moment of M. Determine the minimum force P that should be applied to the follower in order to hold the cam in the position shown.The coefficient of static friction between the cam and the follower is μ s . The guide at A is smooth. 784 © 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: a 10 mm= b 60 mm= M 5Nm⋅= μ s 0.4= Solution: ΣM 0 = 0; M μ s N B b− aN B − 0= N B M μ s ba+ = N B 147.06 N= Follower: ΣF y = 0; N B P− 0= PN B = P 147 N= Problem 8-25 The board can be adjusted vertically by tilting it up and sliding the smooth pin A along the vertical guide G. When placed horizontally, the bottom C then bears along the edge of the guide, where the coefficient of friction is μ s . Determine the largest dimension d which will support any applied force F without causing the board to slip downward. Given: μ s 0.4= a 0.75 in= b 6in= 785 © 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 F y = 0; μ s N C F− 0= ΣM A = 0; F− bdN C + μ s N C a− 0= Solving we find μ s − bd+ μ s a− 0= d μ s ab+()= d 2.70 in= Problem 8-26 The homogeneous semicylinder has a mass m and mass center at G. Determine the largest angle θ of the inclined plane upon which it rests so that it does not slip down the plane. The coefficient of static friction between the plane and the cylinder is μ s . Also, what is the angle φ for this case? Given: μ s 0.3= Solution: The semicylinder is a two-force member: Since F μ N= tan θ() μ s N N = μ S = θ atan μ s ()= 786 © 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 θ 16.7 deg= Law of sines r sin 180 deg φ−() 4r 3π sin θ() = φ asin 3π 4 sin θ() ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = φ 42.6 deg= Problem 8-27 A chain having a length L and weight W rests on a street for which the coefficient of static friction is μ s . If a crane is used to hoist the chain, determine the force P it applies to the chain if the length of chain remaining on the ground begins to slip when the horizontal component is P x . What length of chain remains on the ground? Given: L 20 ft= W 8 lb ft = μ s 0.2= P x 10 lb= Solution: ΣF x = 0; P x − μ s N c + 0= N c P x μ s = N c 50.00 lb= ΣF y = 0; P y WL− N c + 0= 787 © 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 P y WL N c −= P y 110.00 lb= PP x 2 P y 2 += P 110 lb= The length on the ground is supported by N c 50.00 lb= thus L N c W = L 6.25 ft= Problem 8-28 The fork lift has a weight W 1 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 W 2 that the fork lift can push forward. The coefficient of static friction between the wheels and the ground is μ s and between each crate and the ground is μ' s . Given: W 1 2400 lb= W 2 300 lb= μ s 0.4= μ' s 0.35= a 2.5 ft= b 1.25 ft= c 3.50 ft= Solution: Fork lift: ΣM B = 0; W 1 cN A bc+()− 0= N A W 1 c bc+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = N A 1768.4 lb= ΣF x = 0; μ s N A P− 0= 788 © 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 P μ s N A = P 707.37 lb= Crate: N c W 2 − 0= ΣF y = 0; N c W 2 = N c 300.00 lb= ΣF x = 0; P' μ' s N c − 0= P' μ' s N c = P' 105.00 lb= Thus n P P' = n 6.74= n floor n()= n 6.00= Problem 8-29 The brake is to be designed to be self locking, that is, it will not rotate when no load P is applied to it when the disk is subjected to a clockwise couple moment M O . Determine the distance d of the lever that will allow this to happen. The coefficient of static friction at B is μ s . Given: a 1.5 ft= b 1ft= μ s 0.5= Solution: ΣM 0 = 0; M 0 μ s N B b− 0= N B M 0 μ s b = ΣM A = 0; P 2 aN B a− μ s N B d+ 0= 789 © 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 P 0= d a μ s = d 3.00 ft= Problem 8-30 The concrete pipe of weight W is being lowered from the truck bed when it is in the position shown. If the coefficient of static friction at the points of support A and B is μ s determine where it begins to slip first: at A or B, or both at A and B. Given: W 800 lb= a 30 in= μ s 0.4= b 18 in= θ 30 deg= c 5in= r 15 in= Solution: initial guesses are N A 10 lb= N B 10 lb= F A 10 lb= F B 10 lb= Given Assume slipping at A: ΣF x = 0; N A F B + W sin θ()− 0= ΣF y = 0; F A N B + W cos θ()− 0= ΣM 0 = 0; F B rF A r− 0= F A μ s N A = N A N B F A F B ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ Find N A N B , F A , F B ,()= N A N B F A F B ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 285.71 578.53 114.29 114.29 ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ lb= 790 © 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 At B, F Bmax μ s N B = Since F B 114.29 lb= < F Bmax 231.41 lb= then we conclude that slipping begins at A. Problem 8-31 A wedge of mass M is placed in the grooved slot of an inclined plane. Determine the maximum angle θ for the incline without causing the wedge to slip. The coefficient of static friction between the wedge and the surfaces of contact is μ s . Given: M 5kg= μ s 0.2= φ 60 deg= g 9.81 m s 2 = Solution: Initial guesses: N W 10 N= θ 10 deg= Given ΣF x = 0; Mgsin θ() 2 μ s N W − 0= ΣF z = 0; 2 N W sin φ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Mgcos θ()− 0= Solving, N W θ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find N W θ,()= N W 45.5 N= θ 21.8 deg= 791 © 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 8-32 A roll of paper has a uniform weight W and is suspended from the wire hanger so that it rests against the wall. If the hanger has a negligible weight and the bearing at O can be considered frictionless, determine the force P needed to start turning the roll. The coefficient of static friction between the wall and the paper is μ s . Given: W 0.75 lb= θ 30 deg= φ 30 deg= μ s 0.25= a 3in= Solution: Initial guesses: R 100 lb= N A 100 lb= P 100 lb= Given ΣF x = 0; N A R sin φ()− P sin θ()+ 0= ΣF y = 0; R cos φ() W− P cos θ()− μ s N A − 0= ΣM 0 = 0; μ s N A aPa− 0= Solving for P, R N A P ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find RN A , P,()= R 1.14 lb= N A 0.51 lb= P 0.13 lb= Problem 8-33 A roll of paper has a uniform weight W and is suspended from the wire hanger so that it rests against the wall. If the hanger has a negligible weight and the bearing at O can be considered frictionless, determine the minimum force P and the associated angle θ needed to start turning the roll. The coefficient of static friction between the wall and the paper is μ s . 792 © 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: W 0.75 lb= φ 30 deg= μ s 0.25= r 3in= Solution: ΣF x = 0; N A R sin φ()− P sin θ()+ 0= ΣF y = 0; R cos φ() W− P cos θ()− μ s N A − 0= ΣM 0 = 0; μ s N A rPr− 0= Solving for P, P μ s W sin φ() cos φ() μ sin θφ−()+ μ sin φ()− = For minimum P we must have dP dθ μ s 2 − W sin φ()cos θφ−() cos φ() μ s sin θφ−()+ μ s sin φ()− 2 = 0= Implies cos θφ−()0= One answer is θφ90 deg+= θ 120.00 deg= P μ s W sin φ() cos φ() μ s sin θφ−()+ μ s sin φ()− = P 0.0946 lb= Problem 8-34 The door brace AB is to be designed to prevent opening the door. If the brace forms a pin connection under the doorknob and the coefficient of static friction with the floor is μ s determine the largest length L the brace can have to prevent the door from being opened. Neglect the weight of the brace. Given: μ s 0.5= 793 © 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 3ft= Solution: The brace is a two-force member. μ s N N L 2 a 2 − a = μ s aL 2 a 2 −= La1 μ s 2 += L 3.35 ft= Problem 8-35 The man has a weight W, and the coefficient of static friction between his shoes and the floor is μ s . Determine where he should position his center of gravity G at d in order to exert the maximum horizontal force on the door. What is this force? Given: W 200 lb= μ s 0.5= h 3ft= Solution: NW− 0= NW= N 200.00 lb= F max μ s N= F max 100 lb= + → Σ F x = 0; PF max − 0= PF max = P 100 lb= ΣM O = 0; Wd Ph− 0= dP h W = d 1.50 ft= 794 © 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 8-36 In an effort to move the two crates, each of weight W, which are stacked on top of one another, the man pushes horizontally on them at the bottom of crate A as shown. Determine the smallest force P that must be applied in order to cause impending motion. Explain what happens. The coefficient of static friction between the crates is μ s and between the bottom crate and the floor is μ s '. Given: W 100 lb= μ s 0.8= μ' s 0.3= a 2ft= b 3ft= Solution: Assume crate A slips: ΣF y = 0; N A W− 0= N A W= N A 100.00 lb= ΣF x = 0; P μ s N A − 0= P 1 μ s N A = P 1 80.00 lb= Assume crate B slips: ΣF y = 0; N B 2 W− 0= N B 2 W= N B 200.00 lb= ΣF x = 0; P μ' s N B − 0= P 2 μ' s N B = P 2 60.00 lb= Assume both crates A and B tip: ΣM = 0; 2 W a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Pb− 0= P 3 W a b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = P 3 66.7 lb= P min P 1 P 2 , P 3 ,()= P 60.00 lb= Problem 8-37 The man having a weight of W 1 pushes horizontally on the bottom of crate A, which is stacked on top of crate B. Each crate has a weight W 2 . If the coefficient of static friction between each crate is μ s and between the bottom crate, his shoes, and the floor is μ ' s , determine if he can cause impending motion. 795 © 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: W 1 150 lb= W 2 100 lb= a 2ft= b 3ft= μ s 0.8= μ' s 0.3= Assume crate A slips: ΣF y = 0; N A W 2 − 0= N A W 2 = N A 100.00 lb= ΣF x = 0; P μ s N A − 0= P 1 μ s N A = P 1 80.00 lb= Assume crate B slips: ΣF y = 0; N B 2 W 2 − 0= N B 2 W 2 = N B 200.00 lb= ΣF x = 0; P μ' s N B − 0= P 2 μ' s N B = P 2 60.00 lb= Assume both crates A and B tip: ΣM = 0; 2 W 2 a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Pb− 0= P 3 W 2 a b = P 3 66.7 lb= P min min P 1 P 2 , P 3 ,()= P min 60.00 lb= Now check to see if he can create this force ΣF y = 0; N m W 1 − 0= N m W 1 = ΣF x = 0; F m P min − 0= F m P min = F mmax μ' s N m = Since F m 60.00 lb= > F mmax 45.00 lb= then the man cannot create the motion. 796 © 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 8-38 The crate has a weight W and a center of gravity at G. Determine the horizontal force P required to tow it.Also, determine the location of the resultant normal force measured from A. Given: a 3.5 ft= b 3ft= c 2ft= W 200 lb= h 4ft= μ s 0.4= Solution: ΣF x = 0; PF O = ΣF y = 0; N O W= N O 200.00 lb= ΣM o = 0; P− hWx+ 0= F O μ s N O = F O 80.00 lb= PF O = P 80.00 lb= xP h W = x 1.60 ft= The distance of N O from A is cx− 0.40 ft= 797 © 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 8-39 The crate has a weight W and a center of gravity at G. Determine the height h of the tow rope so that the crate slips and tips at the same time. What horizontal force P is required to do this? Given: a 3.5 ft= b 3ft= c 2ft= W 200 lb= h 4ft= μ s 0.4= Solution: ΣF y = 0; N A W= N A 200.00 lb= ΣF x = 0; PF A = F s = μ s N; F A μ s W= F A 80.00 lb= P 80 lb= ΣM A = 0; P− hWc+ 0= hW c P = h 5.00 ft= Problem 8-40 Determine the smallest force the man must exert on the rope in order to move the crate of mass M. Also, what is the angle θ at this moment? The coefficient of static friction between the crate and the floor is μ s . Given: M 80 kg= 798 © 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. dhanesh_h Mathcad - CombinedMathcads

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