© 2016 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 8–1 P P Determine the maximum force P the connection can support so that no slipping occurs between the plates There are four bolts used for the connection and each is tightened so that it is subjected to a tension of kN The coefficient of static friction between the plates is ms = 0.4 P SOLUTION Free-Body Diagram: The normal reaction acting on the contacting surface is equal to the sum total tension of the bolts Thus, N = 4(4) kN = 16 kN When the plate is on the verge of slipping, the magnitude of the friction force acting on each contact surface can be computed using the friction formula F = msN = 0.4(16) kN As indicated on the free-body diagram of the upper plate, F acts to the right since the plate has a tendency to move to the left Equations of Equilibrium: + ©Fx = 0; : 0.4(16) - P = p = 12.8 kN Ans Ans: P = 12.8 kN 748 © 2016 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 8–2 The tractor exerts a towing force T = 400 lb Determine the normal reactions at each of the two front and two rear tires and the tractive frictional force F on each rear tire needed to pull the load forward at constant velocity The tractor has a weight of 7500 lb and a center of gravity located at GT An additional weight of 600 lb is added to its front having a center of gravity at GA Take ms = 0.4 The front wheels are free to roll GT C Equations of Equilibrium: B ft ft Ans 2NC + (2427.78) - 7500 - 600 = Ans NC = 1622.22 lb = 1.62 kip + ©F = 0; : x F ft 2NB (9) + 400(2.5) - 7500(5) - 600(12) = NB = 2427.78 lb = 2.43 kip + c ©Fy = 0; GA 2.5 ft SOLUTION a + ©MC = A T 2F - 400 = Ans F = 200 lb Friction: The maximum friction force that can be developed between each of the rear tires and the ground is Fmax = ms NC = 0.4 (1622.22) = 648.89 lb Since Fmax F = 200 lb, the rear tires will not slip Hence the tractor is capable of towing the 400 lb load Ans: NB = 2.43 kip NC = 1.62 kip F = 200 lb 749 © 2016 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 8–3 The mine car and its contents have a total mass of Mg and a center of gravity at G If the coefficient of static friction between the wheels and the tracks is ms = 0.4 when the wheels are locked, find the normal force acting on the front wheels at B and the rear wheels at A when the brakes at both A and B are locked Does the car move? 10 kN 0.9 m B SOLUTION 0.15 m 1.5 m NA 11.52 + 1011.052 - 58.8610.62 = Ans NA = 16.544 kN = 16.5 kN + c ©Fy = 0; A 0.6 m Equations of Equilibrium: The normal reactions acting on the wheels at (A and B) are independent as to whether the wheels are locked or not Hence, the normal reactions acting on the wheels are the same for both cases a + ©MB = 0; G NB + 16.544 - 58.86 = Ans NB = 42.316 kN = 42.3 kN When both wheels at A and B are locked, then 1FA2max = msNA = 0.4116.5442 = 6.6176 kN and 1FB2max = msNB = 0.4142.3162 = 16.9264 kN Since 1FA2max + FB max = 23.544 kN 10 kN, the wheels not slip Thus, the mine car does not move Ans Ans: NA = 16.5 kN NB = 42.3 kN It does not move 750 © 2016 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 *8–4 The winch on the truck is used to hoist the garbage bin onto the bed of the truck If the loaded bin has a weight of 8500 lb 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 mA = 0.3 and mB = 0.2, respectively Neglect the height of the support at A 30 G A 10 ft 12 ft B SOLUTION a + ©MB = 0; 8500(12) - NA(22) = NA = 4636.364 lb + ©F = 0; : x T cos 30° - 0.2NB cos 30° - NB sin 30° - 0.3(4636.364) = T(0.86603) - 0.67321 NB = 1390.91 + c ©Fy = 0; 4636.364 - 8500 + T sin 30° + NB cos 30° - 0.2NB sin 30° = T(0.5) + 0.766025 NB = 3863.636 Solving: Ans T = 3666.5 lb = 3.67 kip NB = 2650.6 lb Ans: T = 3.67 kip 751 © 2016 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 8–5 The automobile has a mass of Mg and center of mass at G Determine the towing force F required to move the car if the back brakes are locked, and the front wheels are free to roll Take ms = 0.3 F G 30Њ 0.6 m C 0.3 m 0.75 m A 1m B 1.50 m Solution Equations of Equilibrium Referring to the FBD of the car shown in Fig a, + ΣFx = 0; FB - F cos 30° = 0 S (1) + c ΣFy = 0; NA + NB + F sin 30° - 2000(9.81) = 0 (2) a + ΣMA = 0; F cos 30°(0.3) - F sin 30°(0.75) + (3) NB (2.5) - 2000(9.81)(1) = 0 Friction It is required that the rear wheels are on the verge to slip Thus (4) FB = ms NB = 0.3 NB Solving Eqs (1) to (4), Ans F = 2,762.72 N = 2.76 kN NB = 7975.30 N NA = 10, 263.34 N FB = 2392.59 N Ans: F = 2.76 kN 752 © 2016 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 8–6 The automobile has a mass of Mg and center of mass at G Determine the towing force F required to move the car Both the front and rear brakes are locked Take ms = 0.3 F G 30Њ 0.6 m C 0.3 m 0.75 m A 1m B 1.50 m Solution Equations of Equilibrium Referring to the FBD of the car shown in Fig a, + ΣFx = 0; FA + FB - F cos 30° = 0 S (1) + c ΣFy = 0; F sin 30° + NA + NB - 2000(9.81) = 0 (2) a + ΣMA = 0; F cos 30°(0.3) - F sin 30°(0.75) + NB (2.5) - 200(9.81)(1) = 0 (3) Friction It is required that both the front and rear wheels are on the verge to slip Thus FA = ms NA = 0.3 NA (4) FB = ms NB = 0.3 NB (5) Solving Eqs (1) to (5), Ans F = 5793.16 N = 5.79 kN NB = 8114.93 N NA = 8608.49 N FA = 2582.55 N FB = 2434.48 N Ans: F = 5.79 kN 753 © 2016 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 8–7 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 ms = 0.3, and a torque of N # 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 = 30 N, (b) P = 70 N 5N m 150 mm 50 mm SOLUTION O P A B 200 mm 400 mm To hold lever: a + ©MO = 0; FB (0.15) - = 0; FB = 33.333 N Require NB = 33.333 N = 111.1 N 0.3 Lever, a + ©MA = 0; PReqd (0.6) - 111.1(0.2) - 33.333(0.05) = PReqd = 39.8 N a) P = 30 N 39.8 N No Ans b) P = 70 N 39.8 N Yes Ans Ans: No Yes 754 © 2016 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 *8–8 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 ms = 0.3, and a torque of N # 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 = 30 N, (b) P = 70 N 5N m 150 mm 50 mm SOLUTION O P A B 200 mm 400 mm To hold lever: a + ©MO = 0; - FB(0.15) + = 0; FB = 33.333 N Require NB = 33.333 N = 111.1 N 0.3 Lever, a + ©MA = 0; PReqd (0.6) - 111.1(0.2) + 33.333(0.05) = PReqd = 34.26 N a) P = 30 N 34.26 N No Ans b) P = 70 N 34.26 N Yes Ans Ans: No Yes 755 © 2016 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 8–9 The pipe of weight W is to be pulled up the inclined plane of slope a using a force P If P acts at an angle f, show that for slipping P = W sin(a + u)>cos(f - u), where u is the angle of static friction; u = tan - ms P φ α SOLUTION +a©Fy¿ = 0; N + P sin f - W cos a = +Q©Fx¿ = 0; P cos f - W sin a - tan u(W cos a - P sin f) = P = = N = W cos a - P sin f W(sin a + tan u cos a) cos f + tan u sin f W sin(a + u) W(cos u sin a + sin u cos a) = cos f cos u + sin f sin u cos(f - u) 756 Q.E.D © 2016 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 8–10 Determine the angle f at which the applied force P should act on the pipe so that the magnitude of P is as small as possible for pulling the pipe up the incline What is the corresponding value of P? The pipe weighs W and the slope a is known Express the answer in terms of the angle of kinetic friction, u = tan - mk P φ α SOLUTION +a©Fy¿ = 0; N + P sin f - W cos a = +Q©Fx¿ = 0; P cos f - W sin a - tan u (W cos a - P sin f) = P = N = W cos a - P sin f W(sin a + tan u cos a) cos f + tan u sin f = W(cos u sin a + sin u cos a) cos f cos u + sin f sin u = W sin (a + u) cos (f - u) W sin (a + u) sin (f - u) dP = = df cos2(f - u) W sin (a + u) sin (f - u) = W sin (a + u) = sin (f - u) = f = u P = f - u = Ans W sin (a + u) = W sin (a + u) cos (u - u) Ans Ans: f = u P = W sin (a + u) 757 © 2016 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 8–118 The pivot bearing is subjected to a parabolic pressure distribution at its surface of contact If the coefficient of static friction is ms, determine the torque M required to overcome friction and turn the shaft if it supports an axial force P P M SOLUTION R The differential are dA = (rdu)(dr) r P = L p dA = L p0 ¢ - P = pR p0 dN = pdA = M = L rdF = p0 = 2p R r2 r2 (rdu)(dr) = p0 du r ¢ - ≤ dr 2≤ R R L0 L0 2P pR2 p0 r ) p ϭ p0 (1Ϫ –– R2 r2 2P ¢ ≤ (rdu)(dr) pR2 R2 L ms rdN = = 2ms P 2p pR L0 du R L0 r2 ¢ - r2 ≤ dr R2 m PR 15 s Ans Ans: M = 864 m PR 15 s © 2016 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 8–119 A disk having an outer diameter of 120 mm fits loosely over a fixed shaft having a diameter of 30 mm If the coefficient of static friction between the disk and the shaft is ms = 0.15 and the disk has a mass of 50 kg, determine the smallest vertical force F acting on the rim which must be applied to the disk to cause it to slip over the shaft SOLUTION F Frictional Force on Journal Bearing: Here, fs = tan-1ms = tan-10.15 = 8.531° Then the radius of friction circle is rf = r sin fs = 0.015 sin 8.531° = 2.225110 -32 m Equation of Equilibrium: a + ©MP = 0; 490.512.2252110-32 - F30.06 - 12.2252110-324 = Ans F = 18.9 N Ans: F = 18.9 N 865 © 2016 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 *8–120 The 4-lb pulley has a diameter of ft and the axle has a diameter of in If the coefficient of kinetic friction between the axle and the pulley is mk = 0.20, determine the vertical force P on the rope required to lift the 20-lb block at constant velocity in P Solution Frictional Force on Journal Bearing Here f k = tan - mk = tan - 0.2 = 11.3099° Then the radius of the friction circle is rf = r sin f k = 0.5 sin 11.3099° = 0.09806 in Equations of Equilibrium Referring to the FBD of the pulley shown in Fig a, a + ΣMP = 0; P(6 - 0.09806) - 4(0.09806) - 20(6 + 0.09806) = Ans P = 20.73 = 20.7 lb Ans: P = 20.7 lb 866 © 2016 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 8–121 Solve Prob 8–120 if the force P is applied horizontally to the left in P Solution Frictional Force on Journal Bearing Here f k = tan - 1mk = tan - 0.2 = 11.3099° Then the radius of the friction circle is rf = r sin f k = 0.5 sin 11.3099° = 0.09806 in Equations of Equilibrium Referring to the FBD of the pulley shown in Fig a a + ΣMO = 0; P(6) - 20(6) - R(0.09806) = (1) R = 61.1882 P - 1223.76 + ΣFx = 0; Rx - P = 0 Rx = P S + c ΣFy = 0; Ry - - 20 = 0 Ry = 24 lb Thus, the magnitude of R is R = 2Rx2 + Ry2 = 2P2 + 242 (2) Equating Eqs (1) and (2) 61.1882 P - 1223.76 = 2P2 + 242 3743.00 P2 - 149,760.00 P + 1,497,024.00 = P2 - 40.01 P + 399.95 = chose the root P 20 lb, Ans P = 20.52 lb = 20.5 lb Ans: P = 20.5 lb 867 © 2016 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 8–122 Determine the tension T in the belt needed to overcome the tension of 200 lb created on the other side Also, what are the normal and frictional components of force developed on the collar bushing? The coefficient of static friction is ms = 0.21 in 1.125 in SOLUTION Frictional Force on Journal Bearing: Here, fs = tan-1ms = tan-10.21 = 11.86° Then the radius of friction circle is 200 lb rf = r sin fk = sin 11.86° = 0.2055 in T Equations of Equilibrium: a + ©MP = 0; 20011.125 + 0.20552 - T11.125 - 0.20552 = Ans T = 289.41 lb = 289 lb + c Fy = 0; R - 200 - 289.41 = R = 489.41 lb Thus, the normal and friction force are N = R cos fs = 489.41 cos 11.86° = 479 lb Ans F = R sin fs = 489.41 sin 11.86° = 101 lb Ans Ans: T = 289 lb N = 479 lb F = 101 lb 868 © 2016 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 8–123 If a tension force T = 215 lb is required to pull the 200-lb force around the collar bushing, determine the coefficient of static friction at the contacting surface The belt does not slip on the collar in 1.125 in SOLUTION Equation of Equilibrium: a + ©MP = 0; 20011.125 + rf2 - 21511.125 - rf2 = 200 lb T rf = 0.04066 in Frictional Force on Journal Bearing: The radius of friction circle is rf = r sin fk 0.04066 = sin fk fk = 2.330° and the coefficient of static friction is Ans ms = tan fs = tan 2.330° = 0.0407 Ans: ms = 0.0407 869 © 2016 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 *8–124 The uniform disk fits loosely over a fixed shaft having a diameter of 40 mm If the coefficient of static friction between the disk and the shaft is ms = 0.15, determine the smallest vertical force P, acting on the rim, which must be applied to the disk to cause it to slip on the shaft The disk has a mass of 20 kg 150 mm 40 mm P Solution Frictional Force on Journal Bearing Here, f k = tan - ms = tan - 0.15 = 8.5308° Then the radius of the friction circle is rf = r sin f s = 0.02 sin 8.5308° = 2.9668 ( 10 - ) m Equations of Equilibrium Referring to the FBD of the disk shown in Fig a, a + ΣMP = 0; 20(9.81) 2.9668 ( 10-3 ) - P 0.075 - 2.9668 ( 10 - ) = P = 8.08 N Ans Ans: 8.08 N 870 © 2016 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 8–125 The 5-kg skateboard rolls down the 5° slope at constant speed If the coefficient of kinetic friction between the 12.5-mm diameter axles and the wheels is mk = 0.3, determine the radius of the wheels Neglect rolling resistance of the wheels on the surface The center of mass for the skateboard is at G 75 mm G 250 mm 300 mm SOLUTION Referring to the free-body diagram of the skateboard shown in Fig a, we have ©Fx¿ = 0; Fs - 5(9.81) sin 5° = Fs = 4.275 N ©Fy¿ = 0; N - 5(9.81) cos 5° = N = 48.86 N The effect of the forces acting on the wheels can be represented as if these forces are acting on a single wheel as indicated on the free-body diagram shown in Fig b.We have ©Fx¿ = 0; Rx¿ - 4.275 = Rx¿ = 4.275 N ©Fy¿ = 0; 48.86 - Ry¿ = Ry¿ = 48.86 N Thus, the magnitude of R is R = 2Rx¿ + Ry¿ = 24.2752 + 48.862 = 49.05 N fs = tan-1 ms = tan-1(0.3) = 16.699° Thus, the moment arm of R from point O is (6.25 sin 16.699°) mm Using these results and writing the moment equation about point O, Fig b, we have a + ©MO = 0; 4.275(r) - 49.05(6.25 sin 16.699° = 0) Ans r = 20.6 mm Ans: r = 20.6 mm 871 © 2016 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 8–126 The bell crank fits loosely into a 0.5-in-diameter pin Determine the required force P which is just sufficient to rotate the bell crank clockwise The coefficient of static friction between the pin and the bell crank is ms = 0.3 12 in 50 lb 45Њ P 10 in SOLUTION + ©F = 0; : x P cos 45° - Rx = Rx = 0.7071P + c ©Fy = 0; Ry - P sin 45° - 50 = Ry = 0.7071P + 50 Thus, the magnitude of R is R = 2Rx + Ry = 2(0.7071P)2 + (0.7071P + 50)2 = 2P2 + 70.71P + 2500 We find that fs = tan - ms = tan - 1(0.3) = 16.699° Thus, the moment arm of R from point O is (0.25 sin 16.699°) mm Using these results and writing the moment equation about point O, Fig a, a + ©MO = 0; 50(10) + 2P2 + 70.71P + 2500(0.25 sin 16.699°) - P(12) = Choosing the larger root, Ans P = 42.2 lb Ans: P = 42.2 lb 872 © 2016 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 8–127 The bell crank fits loosely into a 0.5-in-diameter pin If P = 41 lb, the bell crank is then on the verge of rotating counterclockwise Determine the coefficient of static friction between the pin and the bell crank 12 in 50 lb 45Њ P 10 in SOLUTION + ©F = 0; : x 41 cos 45° - Rx = Rx = 28.991 lb + c ©Fy = 0; Ry - 41 sin 45° - 50 = Ry = 78.991 lb Thus, the magnitude of R is R = 2Rx + Ry = 228.9912 + 78.9912 = 84.144 lb We find that the moment arm of R from point O is 0.25 sin fs Using these results and writing the moment equation about point O, Fig a, a + ©MO = 0; 50(10) - 41(12) - 84.144(0.25 sin fs) = fs = 22.35° Thus, Ans ms = tan fs = tan 22.35° = 0.411 Ans: ms = 0.411 873 © 2016 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 *8–128 The vehicle has a weight of 2600 lb and center of gravity at G Determine the horizontal force P that must be applied to overcome the rolling resistance of the wheels The coefficient of rolling resistance is 0.5 in The tires have a diameter of 2.75 ft 2.5 ft ft SOLUTION Rolling Resistance: W = NA + NB = Here, = 2600 lb, a = 0.5 in and r = a P G ft 13000 + 2.5P 5200 - 2.5P + 7 2.75 b 1122 = 16.5 in Applying Eq 8–11, we have P L L Wa r 260010.52 16.5 Ans L 78.8 lb Ans: ≈ 78.8 lb 874 © 2016 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 8–129 The tractor has a weight of 16 000 lb and the coefficient of rolling resistance is a = in Determine the force P needed to overcome rolling resistance at all four wheels and push it forward G P ft SOLUTION Applying Eq 8–11 with W = 16 000 lb, a = a P L Wa = r 16000 a ft b ft and r = ft, we have 12 b 12 ft ft Ans = 1333 lb Ans: P = 1333 lb 875 © 2016 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 8–130 The handcart has wheels with a diameter of in If a crate having a weight of 1500 lb is placed on the cart, determine the force P that must be applied to the handle to overcome the rolling resistance The coefficient of rolling resistance is 0.04 in Neglect the weight of the cart P SOLUTION + c ©Fy = 0; P = Wa , r N - 1500 - P a b = c 1500 + P a b d (0.04) P = 2.4 P = 60 + 0.024 P Ans P = 25.3 lb Ans: P = 25.3 lb 876 © 2016 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 8–131 The cylinder is subjected to a load that has a weight W If the coefficients of rolling resistance for the cylinder’s top and bottom surfaces are aA and aB, respectively, show that a horizontal force having a magnitude of P = [W(aA + aB)]>2r is required to move the load and thereby roll the cylinder forward Neglect the weight of the cylinder W P A r SOLUTION B + ©F = 0; : x (RA)x - P = (RA)x = P + c ©Fy = 0; (RA)y - W = (RA)y = W a + ©MB = 0; (1) P(r cos fA + r cos fB) - W(aA + aB) = Since fA and fB are very small, cos fA - cos fB = Hence, from Eq (1) P = W(aA + aB) 2r (QED) 877 © 2016 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 *8–132 The 1.4-Mg machine is to be moved over a level surface using a series of rollers for which the coefficient of rolling resistance is 0.5 mm at the ground and 0.2 mm at the bottom surface of the machine Determine the appropriate diameter of the rollers so that the machine can be pushed forward with a horizontal force of P = 250 N Hint: Use the result of Prob 8–131 P SOLUTION P = W(aA + aB) 2r 250 = 1400 (9.81) (0.2 + 0.5) 2r r = 19.2 mm Ans d = 38.5 mm Ans: d = 38.5 mm 878 ... 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. .. 2016 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,... 2016 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,