Engineering Mechanics - Statics Chapter 8 Check: If F A 604 N= < F Amax 664 N= then our no-slip assumption is good. Problem 8-10 The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple moment M 0 If the coefficient of static friction between the wheel and the block is μ s , determine the smallest force P that should be applied. Solution: Σ M C = 0; Pa Nb− μ s Nc+ 0= N Pa b μ s c− = μ s Nr M O − 0= Σ M O = 0; μ s Par b μ s c− M O = P M O b μ s c− () μ s ra = Problem 8-11 The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple 771 © 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 moment M 0 If the coefficient of static friction between the wheel and the block is μ s , show that the brake is self locking, i. e., P 0≤ , provided b c μ s ≤ Solution: Σ M C = 0; Pa Nb− μ s Nc+ 0= N Pa b μ s c− = Σ M O = 0; μ s Nr M O − 0= μ s Par b μ s c− M O = P M O b μ s c− () μ s ra = P < 0 if b μ s c− () 0< i.e. if b c μ s < Problem 8-12 The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple moment M 0 If the coefficient of static friction between the wheel and the block is μ s , determine the smallest force P that should be applied if the couple moment M O is applied counterclockwise . 772 © 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 C = 0; Pa Nb− μ s Nc− 0= N Pa b μ s c+ = Σ M O = 0; μ s − Nr M O + 0= μ s Par b μ s c+ M O = P M O b μ s c+ () μ s ra = Problem 8-13 The block brake consists of a pin-connected lever and friction block at B. The coefficient of static friction between the wheel and the lever is μ s and a torque M is applied to the wheel. Determine if the brake can hold the wheel stationary when the force applied to the lever is (a) P 1 (b) P 2 . 773 © 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 0.3= M 5Nm⋅= a 50 mm= b 200 mm= c 400 mm= r 150 mm= P 1 30 N= P 2 70 N= Solution: To hold lever: Σ M O = 0; F B rM− 0= F B M r = F B 33.333 N= Require N B F B μ s = N B 111.1 N= Lever, Σ M A = 0; P Reqd bc+()N B b− F B a− 0= P Reqd N B bF B a+ bc+ = P Reqd 39.8 N= (a) If P 1 30.00 N= > P Reqd 39.81 N= then the break will hold the wheel (b) If P 2 70.00 N= > P Reqd 39.81 N= then the break will hold the wheel Problem 8-14 The block brake consists of a pin-connected lever and friction block at B. The coefficient of static friction between the wheel and the lever is μ s and a torque M is applied to the wheel. Determine if the brake can hold the wheel stationary when the force applied to the lever is (a) P 1 (b) P 2. Assume that the torque M is applied counter-clockwise. 774 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 μ s 0.3= M 5Nm⋅= a 50 mm= b 200 mm= c 400 mm= r 150 mm= P 1 30 N= P 2 70 N= Solution: To hold lever: Σ M O = 0; F B rM− 0= F B M r = F B 33.333 N= Require N B F B μ s = N B 111.1 N= Lever, Σ M A = 0; P Reqd bc+()N B b− F B a+ 0= P Reqd N B bF B a− bc+ = P Reqd 34.3 N= (a) If P 1 30.00 N= > P Reqd 34.26 N= then the break will hold the wheel (b) If P 2 70.00 N= > P Reqd 34.26 N= then the break will hold the wheel 775 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 Problem 8-15 The doorstop of negligible weight is pin connected at A and the coefficient of static friction at B is μ s . Determine the required distance s from A to the floor so that the stop will resist opening of the door for any force P applied to the handle. Given: μ s 0.3= a 1.5 in= Solution: Σ F y = 0; N B s s 2 a 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F A − 0= Σ F x = 0; μ s N B a s 2 a 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F A − 0= μ s s s 2 a 2 + ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ F A a s 2 a 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ F A − 0= μ s sa= s a μ s = s 5.00 in= Problem 8-16 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 μ A , determine the smallest force P that must be applied to the handle to open the door. 776 © 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 A 0.3= a 1.20 ft= b 0.75 ft= c 3ft= θ 30 deg= W 10 lb= Solution: Guesses B y 1lb= N A 1lb= P 1lb= Given Σ F x = 0; P− μ A N A + 0= Σ F y = 0; N A W− B y − 0= Σ Μ Β = 0; μ A N A c cos θ () N A c sin θ () − Wca−( )sin θ () bcos θ () + ⎡ ⎣ ⎤ ⎦ + 0= B y N A P ⎛ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎠ Find B y N A , P, () = B y 11.5 lb= N A 21.5 lb= P 6.45 lb= Problem 8-17 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 hoop begins to slip at A when the angle is θ , determine the coefficient of static friction between the hoop and the peg. 777 © 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: θ 30 deg= Solution: Σ F x = 0; μ N A cos θ () P+ N A sin θ () − 0= P μ cos θ () sin θ () − () N A = Σ F y = 0; μ N A sin θ () W− N A cos θ () + 0= W μ sin θ () cos θ () + () N A = Σ Μ 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: Σ F x = 0; μ N A cos θ () P+ N A sin θ () − 0= P μ cos θ () sin θ () − () N A = Σ F y = 0; μ N A sin θ () W− N A cos θ () + 0= W μ sin θ () cos θ () + () N A = Σ Μ 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 massg− 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. [...]... 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 = 3 ft Solution: The brace is a two-force member μs N N 2 L −a a = 2 μs a = L −a L = a 2 2 1 + μs 2 L = 3.35 ft Problem 8-3 5 The man has a weight W, and the coefficient of static friction between his shoes and the floor is μs Determine where... form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 8 Given: a = 1 in W = 10 lb b = 8 in c = 10 in μ s = 0.7 Solution: Pressure on book mark : P = 1 W 2 bc P = 0.06 in Normal force on bookmark: F = μs N ΣF x = 0; lb 2 N = Pca F = 0.44 lb P − 2F = 0 P = 2F P = 0.88 lb Problem 8-2 2 The uniform dresser has weight W and rests on a tile floor for which... 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 = 5 N⋅ m μ s = 0.4 Solution: ΣM0 = 0; M − μ s NB b − a NB = 0 NB = M μs b + a NB = 147.06 N Follower: ΣF y = 0; NB − P = 0 P = NB P = 147 N Problem 8-2 5 The board can be adjusted vertically by tilting it up and sliding the smooth pin A along... 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 Fy = 0; μ s NC − F = 0 ΣMA = 0; −F b + d NC − μ s NC a = 0 Solving we find −μ s b + d − μ s a = 0 d = μ s ( a + b) d = 2.70 in Problem 8-2 6 The homogeneous semicylinder has a mass m and mass center at G Determine the largest angle θ of the... two-force member: Since F=μ N tan ( θ ) = μs N N = μS 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. .. 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 = 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... 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... 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 μs d = 3.00 ft Problem 8-3 0 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... 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 NB Since F B = 114.29 lb < F Bmax = 231.41 lb then we conclude that slipping begins at A Problem 8-3 1 A wedge of mass M is placed in the grooved slot of an inclined plane Determine the maximum angle θ for the incline without... 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 8-3 2 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 . Engineering Mechanics - Statics Chapter 8 Check: If F A 604 N= < F Amax 664 N= then our no-slip assumption is good. Problem 8-1 0 The block brake is used to. publisher. Engineering Mechanics - Statics Chapter 8 Solution: Σ M C = 0; Pa Nb− μ s Nc− 0= N Pa b μ s c+ = Σ M O = 0; μ s − Nr M O + 0= μ s Par b μ s c+ M O = P M O b μ s c+ () μ s ra = Problem 8-1 3 The. form or by any means, without permission in writing from the publisher. Engineering Mechanics - Statics Chapter 8 Problem 8-1 5 The doorstop of negligible weight is pin connected at A and the coefficient