Engineering Mechanics - Statics Chapter 8 of the belt so that the collar rotates clockwise with a constant angular velocity. Assume that the belt does not slip on the collar; rather, the collar slips on the shaft. Neglect the weight and thickness of the belt and collar. The radius, measured from the center of the collar to the mean thickness of the belt is R. Given: r 2in= μ k 0.3= R 2.25 in= F 20 lb= Solution: φ k atan μ k () = φ k 16.699 deg= r f rsin φ k () = r f 0.5747 in= Equilibrium: + ↑ Σ F y = 0; R y F− 0= R y F= R y 20.00 lb= + → Σ F x = 0; PR x − 0= R x P= RR x 2 R y 2 += P 2 F 2 += Guess P 1lb= Given P 2 F 2 + () r f FR+ PR− 0= P Find P()= P 29.00 lb= Problem 8-127 The connecting rod is attached to the piston by a pin at B of diameter d 1 and to the crank shaft by a bearing A of diameter d 2 . If the piston is moving downwards, and the coefficient of static friction at these points is μ s , determine the radius of the friction circle at each connection. Given: d 1 0.75 in= 881 © 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 d 2 2in= μ s 0.2= Solution: r fA 1 2 d 2 μ s = r fA 0.2 in= r fB 1 2 d 1 μ s = r fB 0.075 in= Problem 8-128 The connecting rod is attached to the piston by a pin at B of diameter d 1 and to the crank shaft by a bearing A of diameter d 2 . If the piston is moving upwards, and the coefficient of static friction at these points is μ s , determine the radius of the friction circle at each connection. Given: d 1 20 mm= d 2 50 mm= μ s 0.3= Solution: r fA 1 2 d 2 μ s = r fA 7.50 mm= r fB 1 2 d 1 μ s = r fB 3mm= 882 © 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 The lawn roller has mass M. If the arm BA is held at angle θ from the horizontal and the coefficient of rolling resistance for the roller is r, determine the force P needed to push the roller at constant speed. Neglect friction developed at the axle, A, and assume that the resultant force P acting on the handle is applied along arm BA. Given: M 80 kg= θ 30 deg= a 250 mm= r 25 mm= Solution: θ 1 asin r a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ 1 5.74 deg= Σ M 0 = 0; r− Mg Psin θ () r− P cos θ () acos θ 1 () + 0= P rMg sin θ () − r cos θ () a cos θ 1 () + = P 96.7 N= Problem 8-130 The handcart has wheels with a diameter D. If a crate having a weight W 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 μ . Neglect the weight of the cart. 883 Problem 8-129 © 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 1500 lb= D 6in= a 0.04 in= c 3= b 4= Solution: Guesses N 1lb= P 1lb= Given NW− P c c 2 b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= b b 2 c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ PN 2a D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = N P ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find NP,()= N 1515lb= P 25.3 lb= Problem 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 a A and a B respectively, show that a force having a magnitude of P = [W(a A + a B )]/2r is required to move the load and thereby roll the cylinder forward. Neglect the weight of the cylinder. 884 © 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 x = 0; R Ax P− 0= R Ax P= + ↑ Σ F y = 0; R Ay W− 0= R Ay P= Σ M B = 0; Prcos φ A () rcos φ B () + () Wa A a B + () − 0= Since φ Α and φ B are very small, cos φ A () cos φ B () = 1= Hence from Eq.(1) P Wa A a B + () 2r = (QED) Problem 8-132 A steel beam of mass M is moved over a level surface using a series of rollers of diameter D for which the coefficient of rolling resistance is a g at the ground and a s at the bottom surface of the beam. Determine the horizontal force P needed to push the beam forward at a constant speed. Hint: Use the result of Prob. 8–131. Units Used: Mg 1000 kg= 885 © 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 1.2 Mg= D 30 mm= a g 0.4 mm= a s 0.2 mm= Solution: P Mg a g a s + () 2 D 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = P 235N= Problem 8-133 A machine of mass M is to be moved over a level surface using a series of rollers for which the coefficient of rolling resistance is a g at the ground and a m 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 P . Hint: Use the result of Prob. 8-131. Units Used: Mg 1000 kg= Given: M 1.4 Mg= a g 0.5 mm= a m 0.2 mm= P 250 N= Solution: P Mga g a m + () 2 r = rMg a g a m + 2 P ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = r 19.2 mm= d 2 r= d 38.5mm= 886 © 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 8-134 A single force P is applied to the handle of the drawer. If friction is neglected at the bottom and the coefficient of static friction along the sides is μ s determine the largest spacing s between the symmetrically placed handles so that the drawer does not bind at the corners A and B when the force P is applied to one of the handles. Given: μ s 0.4= a 0.3 m= b 1.25 m= Solution: Equation of Equilibrium and Friction : If the drawer does not bind at corners A and B, slipping would have to occur at points A and B. Hence, F A = μ N A and F B = μ N B + → Σ F x = 0; N B N A − 0= N A N B = N= + ↑ Σ F y = 0; μ s N A μ s N B + P− 0= P 2 μ s N= Σ M B = 0; Na μ s Nb+ P sb+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − 0= a μ s b+ 2 μ s sb+ 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ N 0= a μ s b+ μ s sb+()− 0= s a μ s = s 0.750 m= Problem 8-135 The truck has mass M and a center of mass at G. Determine the greatest load it can pull if (a) the truck has rear-wheel drive while the front wheels are free to roll, and ( b ) the truck has 887 © 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 () four-wheel drive. The coefficient of static friction between the wheels and the ground is μ st and between the crate and the ground, it is μ sc . Units Used: kN 10 3 N= Mg 1000 kg= Given: M 1.25 Mg= μ st 0.5= a 600 mm= b 1.5 m= μ sc 0.4= c 1m= g 9.81 m s 2 = d 800 mm= Solution: Guesses N A 1N= N B 1N= T 1N= N C 1N= W 1N= (a) Rear wheel drive Given T− μ st N A + 0= N A N B + Mg− 0= M− gb N B bc+()+ Ta+ 0= T μ sc N C − 0= N C W− 0= N A N B T N C W ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find N A N B , T, N C , W, () = W 6.97 kN= (b) Four wheel drive Given T− μ st N A + μ st N B + 0= 888 © 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 N A N B + Mg− 0= M− gb N B bc+()+ Ta+ 0= T μ sc N C − 0= N C W− 0= N A N B T N C W ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find N A N B , T, N C , W, () = W 15.33 kN= Problem 8-136 The truck has M and a center of mass at G. The truck is traveling up an incline of angle θ . Determine the greatest load it can pull if (a) the truck has rear-wheel drive while the front wheels are free to roll, and (b) the truck has four-wheel drive. The coefficient of static friction between the wheels and the ground is μ st and between the crate and the ground, it is μ sc . Units Used: kN 10 3 N= Mg 1000 kg= Given: θ 10 deg= 889 © 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 M 1.25 Mg= a 600 mm= μ st 0.5= b 1.5 m= μ sc 0.4= c 1m= d 800 mm= g 9.81 m s 2 = Solution: Guesses N A 1N= N B 1N= T 1N= N C 1N= W 1N= (a) Rear wheel drive Given T− μ st N A + Mgsin θ () − 0= N A N B + Mgcos θ () − 0= M− gbcos θ () Mgdsin θ () + N B bc+()+ Ta+ 0= T μ sc N C − W sin θ () − 0= N C W cos θ () − 0= N A N B T N C W ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find N A N B , T, N C , W, () = W 1.25 kN= (b) Four wheel drive Given T− μ st N A + μ st N B + Mgsin θ () − 0= N A N B + Mgcos θ () − 0= M− gbcos θ () Mgdsin θ () + N B bc+()+ Ta+ 0= T μ sc N C − W sin θ () − 0= N C W cos θ () − 0= 890 © 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. [...]... reproduced, in any form or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 9 Solution: 3 a ⌠ A=⎮ ⌡0 ( 2) 2 2 2 a a y dy = 3 a 3 ⌠ ⎮ xc = 2⎮ 2a ⌡ 2a A= 3 a xc = 3 a 8 yc = 1 3 a y dy = a 2 8 3 a 5 0 5 a yc = ( ) 2 3 ⌠ 3 2 ⎮ y a y dy = a 2⌡ 4 2a 0 5a Problem 9-1 7 Locate the centroid of the quarter elliptical area 907 © 2007 R C Hibbeler Published... or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 9 dA = x dy xc = x 2 yc = y 1 h ⌠ A=⎮ b ⎮ ⌡ y ⎛ y⎞ dy = h b ⎜ ⎟ h ⎝ h⎠ 2 0 ⌠ 3 ⎮ xc = 2h b ⎮ ⌡ h 1⎛ ⎜b 2⎝ 2 y⎞ ⎟ dy = h⎠ 0 3 bh 2 2 8h 3 b 8 xc = 5 h ⌠ 3 ⎮ yc = yb 2h b ⎮ ⌡ y 3 h dy = h 5 ⎛ h⎞ ⎜ ⎟ ⎝ h⎠ 2 0 yc = 3 h 5 Problem 9-8 Locate the centroid yc of the shaded area Given: a = 100 mm b = 100... means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 9 Problem 9-1 0 Determine the location (xc, yc) of the centroid of the triangular area Solution: a ⌠ 1 2 A = ⎮ m x dx = a m ⌡0 2 a xc = 2 ⌠ 2 ⎮ x m x dx = a 2⌡ 3 0 ma xc = 2 a 3 yc = m a 3 a ⌠ 1 1 2 ⎮ yc = ( m x) dx = a m 2⎮ 2 3 ma ⌡ 2 0 Problem 9-1 1 Determine the location (xc, yc) of the center of gravity... from the publisher Engineering Mechanics - Statics Chapter 9 y b x=a b ⌠ ⎮ y2a ⎮ ⌡ y dy b 0 yc = yc = 60 mm b ⌠ ⎮ 2a ⎮ ⌡0 y b dy Problem 9-9 Locate the centroid xc of the shaded area Solution: dA = ydx xc = x yc = y 2 b xc = yc = ⌠ h 2 ⎮ x x dx 2 ⎮ b ⌡ 0 b ⌠ 2 ⎮ x h dx ⎮ 2 ⎮ b ⌡0 ⌠ ⎮ ⎮ ⎮ ⌡ 4 3 b = 4 3 b xc = 3 b 4 yc = 3 h 10 b 2 1 ⎛ h 2⎞ x dx 2 ⎜ b2 ⎟ ⎝ ⎠ 0 ⌠ ⎮ ⎮ ⌡ b h 2 x dx 2 b 2 3 5 h = b 5 10 b... publisher Engineering Mechanics - Statics Chapter 9 Problem 9-2 1 Locate the centroid yc of the shaded area Solve the problem by evaluating the integrals using Simpson's rule Given: a = 2 ft 1 b = a 5 2 + 2a 3 Solution: a ⌠ ⎮ ⎮ A = ⎮ ⌡0 1 5⎞ ⎛ ⎜ 2 3 ⎝b − x + 2x ⎠ dx ⌠ ⎮ 1 ⎮ yc = A⎮ ⌡ a A = 2.177 ft 1 5 ⎞⎛ 1 5⎞ ⎛ ⎜ ⎟⎜ 1 2 3 2 3 ⎝b + x + 2x ⎠ ⎝ b − x + 2x ⎠ dx 2 2 yc = 2.04 ft 0 Problem 9-2 2 The steel... or by any means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 9 Solution: a ⌠ 2 ⎮ ⎛ x ⎞ dx A = ⎮ b 1−⎜ ⎟ ⎝ a⎠ ⌡ A= πa b 4 0 a ⌠ 2 4 ⎮ 4 ⎛x⎞ xc = x b 1 − ⎜ ⎟ dx = a ⎮ πa b ⌡ 3 ⎝ a⎠ xc = 4 3 a 0 a ⌠ 2 ⎮ 2 4 ⎮ 1⎡ x⎞ ⎤ ⎢b 1 − ⎛ ⎟ ⎥ dx = 4 b yc = ⎜ πa b ⎮ 2 ⎣ 3 ⎝ a⎠ ⎦ ⌡ yc = 4 3 b 0 Problem 9-1 8 Locate the centroid xc of the triangular area Solution: A= bh... W( b + xc) = 0 ⎛ Ax ⎞ ⎛ 0.00 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Ay ⎟ = ⎜ 0.00 ⎟ kN ⎜ A ⎟ ⎝ 52 .32 ⎠ ⎝ z⎠ MA = 52 .32 kN⋅ m Problem 9-1 4 Locate the centroid (xc, yc) of the exparabolic segment of area Solution: 0 ⌠ 1 b 2 A=⎮ x dx = ab 2 3 ⎮ a ⌡− a 0 3 ⌠ 3 b 2 ⎮ x xc = x dx = a 2 ab ⎮ 4 a ⌡− a xc = 3 a 4 yc = 3 b 10 0 ⌠ 2 3 ⎮ 1 ⎛ b 2⎞ 3 yc = b ⎮ − ⎜ x ⎟ dx = ab ⎮ 2 2 10 ⎝a ⎠ ⌡− a 905 © 2007 R C Hibbeler Published by Pearson... permission in writing from the publisher Engineering Mechanics - Statics Chapter 9 Given: a = 4 in b = 2 in c = 3 in Solution: ⌠ A = ⎮ ⎮ ⌡ a+ b bc dy y A = 6.592 in 2 b 1 ⌠ ⎮ yc = A⎮ ⌡ a+ b ⎛ b c⎞ dy ⎟ ⎝ y⎠ y⎜ yc = 3. 64 in b Problem 9 -3 1 Determine the location rc of the centroid C of the cardioid, r = a(1 − cosθ) Solution: 2π ⌠ A=⎮ ⌡0 a ( 1 − cos( θ ) ) ⌠ ⎮ ⌡0 r dr dθ = 3 2 a π 2 916 © 2007 R C Hibbeler Published... writing from the publisher Engineering Mechanics - Statics Chapter 9 Solution: ⌠ A = ⎮ ⎮ ⌡ a+ b x dx a+b c a ⌠ 1 ⎮ yc = A⎮ ⌡ a+ b 1⎛ ⎜c 2⎝ 2 x ⎞ ⎟ dx a + b⎠ yc = 0.804 in a Problem 9-2 9 Locate the centroid xc of the shaded area Given: a = 4 in b = 2 in c = 3 in Solution: ⌠ A = ⎮ ⎮ ⌡ a+ b bc dy y A = 6.592 in 2 b ⌠ 1 ⎮ xc = A⎮ ⌡ a+ b 2 1 ⎛ b c⎞ ⎜ ⎟ dy 2⎝ y⎠ xc = 0.910 in b Problem 9 -3 0 Locate the centroid... means, without permission in writing from the publisher Engineering Mechanics - Statics Chapter 9 Problem 9-2 7 Locate the centroid xc of the shaded area Given: a = 1 in b = 3 in c = 2 in Solution: ⌠ A = ⎮ ⎮ ⌡ a+ b c x dx a+b a ⌠ 1 ⎮ xc = A⎮ ⌡ a+ b xc x dx a+b xc = 2.66 in a Problem 9-2 8 Locate the centroid yc of the shaded area Given: a = 1 in b = 3 in c = 2 in 914 © 2007 R C Hibbeler Published by Pearson . the publisher. Engineering Mechanics - Statics Chapter 8 Given: M 1.2 Mg= D 30 mm= a g 0.4 mm= a s 0.2 mm= Solution: P Mg a g a s + () 2 D 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = P 235 N= Problem 8-1 33 A machine of mass. publisher. Engineering Mechanics - Statics Chapter 9 dAxdy= x c x 2 = y c y= A 0 h yb y h ⌠ ⎮ ⎮ ⌡ d= hb y h ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 2 = x c 3 2hb 0 h y 1 2 b y h ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⌠ ⎮ ⎮ ⌡ d= 3 8 h 2 bh 2 = x c 3 8 b= y c 3 2hb 0 h yyb y h ⌠ ⎮ ⎮ ⌡ d= 3 5 h h h ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 5 2 = y c 3 5 h= Problem. writing from the publisher. Engineering Mechanics - Statics Chapter 8 N A N B T N C W ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ Find N A N B , T, N C , W, () = W 6.89 kN= Problem 8-1 37 A roofer, having a mass