Engineering Mechanics - Statics Chapter 9 r 3 r 2 r 1 −= Solution: V 2 π 2 π 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r 2 2 4r 2 3 π ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2r 2 2r 3 () r 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 2 π 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r 3 2 r 2 4r 3 3 π − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = V 1.40 10 3 × in 3 = Problem 9-94 A circular sea wall is made of concrete. Determine the total weight of the wall if the concrete has a specific weight γ c . Given: γ c 150 lb ft 3 = a 60 ft= b 15 ft= c 8ft= d 30 ft= θ 50 deg= Solution: W γ c θ a 1 2 db c−() ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 3 bc−() 1 2 db c−() ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ + ab+ c 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ dc+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = W 3.12 10 6 × lb= Problem 9-95 Determine the surface area of the tank, which consists of a cylinder and hemispherical cap. 961 © 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 9 Given: a 4m= b 8m= Solution: A 2 π ab 2a π π a 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = A 302 m 2 = Problem 9-96 Determine the volume of the tank, which consists of a cylinder and hemispherical cap. Given: a 4m= b 8m= Solution: V 2 π 4a 3 π π a 2 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ a 2 ba()+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = V 536 m 3 = 962 © 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 9 Problem 9-97 Determine the surface area of the silo which consists of a cylinder and hemispherical cap. Neglect the thickness of the plates. Given: a 10 ft= b 10 ft= c 80 ft= Solution: A 2 π 2a π π a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ac+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = A 5.65 10 3 × ft 2 = Problem 9-98 Determine the volume of the silo which consists of a cylinder and hemispherical cap. Neglect the thickness of the plates. Given: a 10 ft= b 10 ft= c 80 ft= Solution: V 2 π 4a 3 π π a 2 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ca a 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = V 27.2 10 3 × ft 3 = 963 © 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 9 Problem 9-99 The process tank is used to store liquids during manufacturing. Estimate both the volume of the tank and its surface area. The tank has a flat top and the plates from which the tank is made have negligible thickness. Given: a 4m= b 6m= c 3m= Solution: V 2 π c 3 ca 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ c 2 cb()+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = V 207m 3 = A 2 π c 2 ccb+ c 2 a 2 c 2 ++ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = A 188 m 2 = Problem 9-100 Determine the height h to which liquid should be poured into the cup so that it contacts half the surface area on the inside of the cup. Neglect the cup's thickness for the calculation. Given: a 30 mm= b 50 mm= c 10 mm= Solution: Total area A total 2 π c c 2 ac+ 2 b 2 ac−() 2 ++ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = Guess h 1mm= e 1mm= 964 © 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 9 Given ac− b ec− h = A total 2 2 π c c 2 ec+ 2 h 2 ec−() 2 ++ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = e h ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Find eh,()= e 21.942 mm= h 29.9 mm= Problem 9-101 Using integration, compute both the area and the centroidal distance x c of the shaded region. Then, using the second theorem of Pappus–Guldinus, compute the volume of the solid generated by revolving the shaded area about the aa axis. Given: a 8in= b 8in= Solution: A 0 a xb x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⌠ ⎮ ⎮ ⌡ d= x c 2a 1 A 0 a xxb x a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⌠ ⎮ ⎮ ⌡ d−= A 21.333 in 2 = x c 10in= V 2 π Ax c = V 1.34 10 3 × in 3 = 965 © 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 9 Problem 9-102 Using integration, determine the area and the centroidal distance y c of the shaded area. Then, using the second theorem of Pappus–Guldinus, determine the volume of a solid formed by revolving the area about the x axis. Given: a 0.5 ft= b 2ft= c 1ft= Solution: A a b x c 2 x ⌠ ⎮ ⎮ ⌡ d= A 1.386 ft 2 = y c 1 A a b x 1 2 c 2 x ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⌠ ⎮ ⎮ ⎮ ⌡ d= y c 0.541 ft= V 2 π Ay c = V 4.71 ft 3 = Problem 9-103 Determine the surface area of the roof of the structure if it is formed by rotating the parabola about the y axis. Given: a 16 m= b 16 m= 966 © 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 9 Solution: Centroid : The length of the differential element is dL dx 2 dy 2 += 1 dy dx ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ dx= and its centroid is x c x= Here, dy dx 2− bx a 2 = Evaluating the integrals, we have L 0 a x1 4b 2 x 2 a 4 + ⌠ ⎮ ⎮ ⎮ ⌡ d= L 23.663 m= x c 1 L 0 a xx 1 4b 2 x 2 a 4 + ⌠ ⎮ ⎮ ⎮ ⌡ d= x c 9.178 m= A 2 π x c L= A 1.365 10 3 × m 2 = Problem 9-104 The suspension bunker is made from plates which are curved to the natural shape which a completely flexible membrane would take if subjected to a full load of coal.This curve may be approximated by a parabola, y/b = (x/a) 2 . Determine the weight of coal which the bunker would contain when completely filled. Coal has a specific weight of γ , and assume there is a fraction loss p in volume due to air voids. Solve the problem by integration to determine the cross-sectional area of ABC; then use the second theorem of Pappus–Guldinus to find the volume. Units Used: kip 10 3 lb= Given: a 10 ft= b 20 ft= 967 © 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 9 γ 50 lb ft 3 = p 0.2= Solution: A 0 b ya y b ⌠ ⎮ ⎮ ⌡ d= A 133.3 ft 2 = x c 1 A 0 b y 1 2 a y b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⌠ ⎮ ⎮ ⌡ d= x c 3.75 ft= V 2 π Ax c = V 3.142 10 3 × ft 3 = W 1 p−() γ V= W 125.7 kip= Problem 9-105 Determine the interior surface area of the brake piston. It consists of a full circular part. Its cross section is shown in the figure. Given: a 40 mm= b 30 mm= c 20 mm= d 20 mm= e 80 mm= 968 © 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 9 f 60 mm= g 40 mm= Solution: A 2 π a 2 aa b 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ b 2 e 2 ++ ca b+ c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + ab+ c+()f+ ab+ 3c 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ c+ ab+ 2c+()g+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ = A 119 10 3 × mm 2 = Problem 9-106 Determine the magnitude of the resultant hydrostatic force acting on the dam and its location H, measured from the top surface of the water. The width of the dam is w; the mass density is ρ w . Units Used: Mg 10 3 kg= MN 10 6 N= Given: w 8m= ρ w 1 Mg m 3 = h 6m= g 9.81 m s 2 = Solution: ph ρ w g= p 58860 N m 2 = F 1 2 hwp= F 1.41 MN= H 2 3 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ h= H 4m= 969 © 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 9 Problem 9-107 The tank is filled with water to a depth d. Determine the resultant force the water exerts on side A and side B of the tank. If oil instead of water is placed in the tank, to what depth d should it reach so that it creates the same resultant forces? The densities are ρ 0 and ρ w . Given: kN 10 3 N= d 4m= a 3m= b 2m= ρ o 900 kg m 3 = ρ w 1000 kg m 3 = g 9.81 m s 2 = Solution: For water At side A: W A b ρ w gd= W A 78480 N m = F RA 1 2 W A d= F RA 157kN= At side B: W B a ρ w gd= W B 117720 N m = F RB 1 2 W B d= F RB 235kN= For oil At side A: F RA 1 2 b ρ o gd 1 d 1 = d 1 2F RA b ρ o g = d 1 4.216 m= 970 © 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. [...]... the publisher Engineering Mechanics - Statics Chapter 9 Given: r = 600 mm a = 25 mm b = 50 mm c = 75 mm Solution: ⎡⎛ ⎣⎝ V = 2π ⎢⎜ r + c ⎞ ⎛ 1⎞ ⎛ c⎞ ⎤ ⎟ 2⎜ ⎟ a c + ⎜ r + ⎟ b c⎥ 3 ⎝ 2⎠ ⎝ 2⎠ ⎦ V = 22.4 × 10 3 m 3 Problem 9-1 25 A circular V-belt has an inner radius r and a cross-sectional area as shown Determine the surface area of the belt Given: r = 600 mm a = 25 mm b = 50 mm c = 75 mm Solution: ⎡... writing from the publisher Engineering Mechanics - Statics Chapter 9 Units Used: 3 Mg = 10 kg 3 kN = 10 N Given: L = 8m Mg ρ w = 1.0 m 3 a = 3m b = 2m g = 9.81 m 2 s Solution: F3 = ρ w g a b L F 3 = 470.88 kN F2 = ρ w g a b L F 2 = 470.88 kN b F1 = ρ w g b L 2 F 1 = 156 .96 kN ⎛ W = ⎜b − 2 2 πb ⎞ ⎝ 4 ⎟ Lρ w g ⎠ W = 67 .36 8 kN Fx = F1 + F2 F x = 628 kN Fy = F2 + W F y = 53 8 kN Problem 9-1 17 The rectangular... publisher Engineering Mechanics - Statics Chapter 9 Given: a = 1 .5 ft lb γ = 55 ft 3 Solution: a ⌠ 2 2 F R = ⎮ γ 2 a − y ( a − y) d y ⌡− a F R = 5 83 lb a 1 ⌠ ⎮ yγ 2 a2 − y2 ( a − y) d y d = a− FR ⌡− a d = 1.8 75 ft Problem 9-1 21 The gasoline tank is constructed with elliptical ends on each side of the tank Determine the resultant force and its location on these ends if the tank is half full Given: a = 3 ft... Units Used: 3 kip = 10 lb Given: a = 12 ft f = 1 ft b = 9 ft g = 2 ft c = 1 .5 ft L = 10 ft d = 5. 5 ft γ = 150 e = 1 .5 ft lb ft 3 Solution: A = bc + a f + 1 a( e − f ) 2 W = γAL W = 42.8 kip xc = 1⎡ b f⎞ 1 e − f ⎞⎤ ⎛ ⎛ ⎢b c + a f⎜g + ⎟ + a( e − f) ⎜g + f + ⎟⎥ A⎣ 2 3 ⎠⎦ ⎝ 2⎠ 2 ⎝ xc = 3. 52 ft yc = 1⎡ c ⎛ a⎞ 1 ⎛ a ⎞⎤ ⎢b c + a f⎜ c + ⎟ + a( e − f) ⎜c + ⎟⎥ A⎣ 2 ⎝ 2⎠ 2 ⎝ 3 ⎠⎦ yc = 4.09 ft Problem 9-1 30 The hopper... writing from the publisher Engineering Mechanics - Statics Chapter 9 Problem 9-1 33 The load over the plate varies linearly along the sides of the plate such that p = 2 x( 4 − y) kPa 3 Determine the resultant force and its position (xc, yc) on the plate Solution: ⌠ F = ⎮ ⎮ ⌡ 3 0 ⌠ ⎮ ⎮ ⌡ 4 2 x( 4 − y) d y dx 3 F = 24 kN 0 3 4 1⌠ ⌠ ⎮ ⎮ x 2 x( 4 − y) d y dx xc = F⎮ ⎮ 3 ⌡ ⌡ 0 3 xc = 2 m 0 4 1⌠ ⌠ ⎮ ⎮ y 2... permission in writing from the publisher Engineering Mechanics - Statics Chapter 10 Alternatively h ⌠ 2⎞ ⎮ 2⎛ ⎜ b − b y ⎟ d y = 2 b h3 Ix = ⎮ y 2⎟ ⎜ 15 ⎮ h ⎠ ⎝ ⌡0 Ix = 2 3 bh 15 Ix = ab 3( 1 + 3n) Problem 1 0-7 Determine the moment of inertia for the shaded area about the x axis Solution: b ⌠ 1⎤ ⎮ ⎡ ⎢ ⎮ n⎥ ⎞ ⎮ A y2 ⎢a − a ⎛ y ⎟ ⎥ d y Ix = ⎜ ⎮ ⎣ ⎝ b⎠ ⎦ ⌡ 3 0 Problem 1 0-8 Determine the moment of inertia for... = 15 ft c = 9 ft γ w = 62.4 lb ft 3 lb γ conc = 150 ft 3 Solution: For a 1-ft thick section: W = γ w b( 1ft) W = 936 1 Wb 2 F = lb ft F = 7020 lb W1 = γ conc( 1ft)a b W1 = 6 750 lb 1 W2 = γ conc ( c − a)b( 1ft ) 2 W2 = 6 750 lb Moment to overturn: 1 MO = F b 3 MO = 35 100 lb ft Moment to stabilize: ⎡ ⎣ MS = W1 ⎢( c − a) + Fs = a⎤ ⎤ ⎡2 ⎥ + W2⎢ ( c − a)⎥ 2⎦ 3 ⎦ MS = 776 25 lb⋅ ft MS F s = 2.21 MO 971 © 2007... permission in writing from the publisher Engineering Mechanics - Statics Chapter 9 Given Fv = d−a b wρ w g 4 Fh = 1 ρw g b wb 2 ⎡ ⎣ W = ρ c g w⎢a b + W ⎛ d − a ⎞ b⎤ ⎜ ⎟⎥ ⎝ 2 ⎠⎦ d ⎛ d − a⎞ − F b = 0 + F v⎜ d − ⎟ h 3 2 6 ⎠ ⎝ ⎛ Fv ⎞ ⎜ ⎟ ⎜ Fh ⎟ = Find ( F , F , W , d) v h ⎜W⎟ ⎜ ⎟ ⎝d ⎠ ⎛ Fv ⎞ ⎛ 0 .37 9 ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ Fh ⎟ = ⎜ 3. 178 ⎟ MN ⎜ W ⎟ ⎝ 4 .54 5 ⎠ ⎝ ⎠ d = 3. 65 m Problem 9-1 12 The tank is used to store a liquid... Engineering Mechanics - Statics Chapter 9 c = 3m g = 9.81 m 2 s Solution: Fluid Pressure: The fluid pressure at points A and B can be determined using Eq 9-1 5, pA = ρ w g( a + b) pA = 88.29 kN m pB = ρ w g a pB = 49. 05 wA = pA w wB = pB w 2 kN m 2 wA = 706 .32 wB = 39 2.4 kN m kN m Equilibrium 2 wB 2 2 2 b +c 1 b +c + ( wA − wB) − Ay c = 0 2 2 3 2 wB 2 b +c 2 Ay = Ay − wB c − + (wA − wB) 2 1 (2 2 2b +c ) 3 Ay... without permission in writing from the publisher Engineering Mechanics - Statics w = 2m g = 9.81 Chapter 9 m 2 s Solution: w1 = ρ w g( b − 2a)w w1 = 59 kN m w2 = ρ w g2a w w2 = 59 kN m F1 = 1 2a w1 2 F 2 = w2 2a F 1 = 88 kN F 2 = 177 kN a − FB a = 0 3 ΣMA = 0; F1 ΣF x = 0; F1 + F2 − FB − FA = 0 FB = 1 F1 3 FA = F1 + F2 − FB F B = 29.4 kN F A = 2 35 kN Problem 9-1 14 The gate AB has width w Determine the horizontal . publisher. Engineering Mechanics - Statics Chapter 9 γ 50 lb ft 3 = p 0.2= Solution: A 0 b ya y b ⌠ ⎮ ⎮ ⌡ d= A 133 .3 ft 2 = x c 1 A 0 b y 1 2 a y b ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⌠ ⎮ ⎮ ⌡ d= x c 3. 75 ft= V 2 π Ax c = V 3. 142. Engineering Mechanics - Statics Chapter 9 r 3 r 2 r 1 −= Solution: V 2 π 2 π 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r 2 2 4r 2 3 π ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2r 2 2r 3 () r 2 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ + 2 π 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ r 3 2 r 2 4r 3 3 π − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = V. the publisher. Engineering Mechanics - Statics Chapter 9 Units Used: Mg 10 3 kg= kN 10 3 N= Given: L 8m= ρ w 1.0 Mg m 3 = a 3m= b 2m= g 9.81 m s 2 = Solution: F 3 ρ w gabL= F 3 470.88 kN= F 2 ρ w gabL=