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© 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 9–1 Locate the center of mass of the homogeneous rod bent into the shape of a circular arc y 30Њ 300 mm x SOLUTION dL = 300 d u 30Њ ' x = 300 cos u ' y = 300 sin u x = L 2p ' x dL = L L-2p3 300 cos u (300du) 2p dL L-2p3 300d u (300)2 C sin u D -3 2p3 2p = 300 a p b Ans = 124 mm y = Ans (By symmetry) Ans: x = 124 mm y = 879 © 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 9–2 y Determine the location (x, y) of the centroid of the wire ft ft SOLUTION Length and Moment Arm: The length of the differential element is dL = 2dx2 + dy2 = ¢ B + a y = x2 dy dy b ≤ dx and its centroid is ∼ = 2x y = y = x2 Here, dx dx x Centroid: Due to symmetry ∼ Ans x = Applying Eq 9–7 and performing the integration, we have ft ∼ ∼ y = LL ydL = dL LL L-2 ft x 21 + 4x2 dx ft L-2 ft = 21 + 4x2 dx 16.9423 = 1.82 ft 9.2936 Ans Ans: x = y = 1.82 ft 880 © 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 9–3 y Locate the center of gravity x of the homogeneous rod If the rod has a weight per unit length of 100 N>m, determine the vertical reaction at A and the x and y components of reaction at the pin B 1m B y ϭ x2 A 1m x Solution Length And Arm The length of the differential element is dy dy dL = 2dx2 + dy2 = c + a b d dx and its centroid is ~ x = x Here = 2x A dx dx Perform the integration L = Moment LL dL = L0 = 1m L0 = cx 21 + 4x2 dx 1m A A x2 + x2 + = 1.4789 m LL x~ dL = L0 = dx 1 1m + ln ax + x2 + b d 4 A 1m x21 + 4x2 dx L0 1m x A x2 + dx 3>2 m = c ax2 + b d = 0.8484 m2 Centroid x = ~ 0.8484 m2 1L x dL = 0.5736 m = 0.574 m = 1.4789 m 1L dL Ans Equations of Equilibrium Refering to the FBD of the rod shown in Fig a + ΣFx = 0; S Ans Bx = 0 a+ΣMB = 0; 100(1.4789) (0.4264) - Ay(1) = Ans Ay = 63.06 N = 63.1 N a+ΣMA = 0; By(1) - 100(1.4789) (0.5736) = Ans By = 84.84 N = 84.8 N Ans: x = 0.574 m Bx = Ay = 63.1 N By = 84.8 N 881 © 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 *9–4 y Locate the center of gravity y of the homogeneous rod 1m B y ϭ x2 A 1m x Solution Length And Arm The length of the differential element is dy dy dL = 2dx2 + dy2 = c + a b d dx and its centroid is ~ y = y Here = 2x A dx dx Perform the integration, L = Moment LL L0 dL = = 1m 21 + 4x2 dx L0 = cx 1m A A x2 + = 1.4789 m LL y~ dL = L0 = dx 1 1m + ln ax + x2 + b d 4 A 1m x2 21 + 4x2 dx L0 = 2c x2 + 1m x A x2 + dx x 1 1 1m x x2 + ln ax + x2 + b d ax + b 4A 32 A 128 A 4 = 0.6063 m2 Centroid 0.6063 m2 1L y dL y = = 0.40998 m = 0.410 m = 1.4789 m 1L dL ~ Ans Ans: y = 0.410 m 882 © 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 9–5 y Determine the distance y to the center of gravity of the homogeneous rod 1m 2m Solution Length And y ϭ 2x3 x Moment Arm The length of the differential element is dy y = y Here dL = 2dx2 + dy2 = a + a b b dx and its centroid is at ~ A dx dy = 6x2 Evaluate the integral numerically, dx L = LL ~ LL dL = y dL = L0 L0 1m 1m 21 + 36x4 dx = 2.4214 m 2x 21 + 36x4 dx = 2.0747 m2 Centroid Applying Eq 9–7, y = ~ 2.0747 m2 1L y dL = 0.8568 = 0.857 m = 2.4214 m 1L dL Ans Ans: y = 0.857 m 883 © 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 9–6 Locate the centroid y of the area y y 1 – – x2 1m x 2m SOLUTION Area and Moment Arm: The area of the differential element is y 1 ' = a1 - x b dA = ydx = a1 - x2 b dx and its centroid is y = 2 Centroid: Due to symmetry Ans x = Applying Eq 9–4 and performing the integration, we have 2m ' ydA y = LA = 1 ¢ - x ≤ ¢ - x2 ≤ dx 4 L- 2m 2m dA LA L- 2m = ¢ ¢1 - x ≤ dx x x3 x 2m + ≤` 12 160 - 2m x ¢x ≤` 12 - 2m 2m = m Ans Ans: y = 884 m © 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 9–7 Determine the area and the centroid x of the parabolic area y h SOLUTION y Differential Element:The area element parallel to the x axis shown shaded in Fig a will be considered The area of the element is dA = x dy = a h1>2 h x2 –– a2 x a y 1>2 dy x a ' ' Centroid: The centroid of the element is located at x = = y 1>2 and y = y 2h1>2 Area: Integrating, h A = LA LA = dA LA h ' xdA x = dA = L0 ¢ a 1>2 L0 h y 1>2 dy = 2a 3h a y1>2 ≤ ¢ 1>2 y 1>2 dy ≤ 1>2 2h h a ah 1>2 A y3>2 B = h ah Ans a2 y2 h a ¢ ≤` y dy 2h L0 2h = = = a Ans 2 ah ah 3 h Ans: x = 885 a © 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 *9–8 Locate the centroid of the shaded area y y ϭ a cospx L a L x L Solution Area And Moment Arm The area of the differential element shown shaded in y p a p Fig. a is dA = ydx = a cos x dx and its centroid is at y~ = = cos x 2 L Centroid Perform the integration y = ~ 1A y dA 1A dA L>2 = p p a a cos xbaa cos x dxb L L L-L>2 L>2 L-L>2 L>2 = = L-L>2 p x dx L a2 2p x + 1b dx acos L L>2 L-L>2 a cos p x dx L L>2 a2 L 2p a sin x + xb ` 2p L -L>2 a = a cos L>2 aL p sin xb ` p L -L>2 a2 L>4 2aL>p = p a Ans x = 0 Ans Due to Symmetry, Ans: p a x = y = 886 © 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 9–9 y Locate the centroid x of the shaded area 4m yϭ x x 4m Solution Area And Moment Arm The area of the differential element shown shaded in Fig. a is dA = x dy and its centroid is at ~ x = x Here, x = 2y1>2 Centroid Perform the integration x = ~ 1A x dA 1A dA = = L0 4m a2y1>2 ba2y1>2 dyb L0 4m 2y1>2 dy m Ans Ans: x = 887 m © 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 9–10 y Locate the centroid y of the shaded area 4m yϭ x x 4m Solution Area And Moment Arm The area of the differential element shown shaded in Fig. a y = y Here, x = 2y1>2 is dA = x dy and its centroid is at ∼ Centroid Perform the integration y = ~ 1A y dA 1A dA = = = L0 4m L0 y a2y1>2 dyb 4m 2y1>2 dy 4m a y 5>2 b ` 4m a y3>2 b ` 12 m Ans Ans: y = 888 12 m © 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 *9–116. Continued Thus, x = y = LA xp(x, y)dA LA LA = 21.0 kN # m = m = 0.778 m 27.0 kN Ans = 22.5 kN # m = 0.833 m 27.0 kN Ans p(x, y)dA yp(x, y)dA LA p(x, y)dA Ans: FR = 27.0 kN x = 0.778 m y = 0.833 m 995 © 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 9–117 The load over the plate varies linearly along the sides of the plate such that p = 23[x14 - y2] kPa Determine the resultant force and its position 1x, y2 on the plate p kPa y 3m x 4m SOLUTION Resultant Force and its Location: The volume of the differential element is ' ' dV = d FR = pdxdy = (xdx)[(4 - y)dy] and its centroid is at x = x and y = y 3m FR = LFk = dFR = LFR = ' ydFR = LFR = x = L0 2 (x dx) L0 (4 - y)dy Ans 4m (4 - y)dy y2 m x3 m b R = 48.0 kN # m a 4y Ba b 3 3m L0 (xdx) L0 4m y(4 - y)dy y3 m x2 m B a b a 2y2 - b R = 32.0 kN # m 3 ' xdFR LFR = 48.0 = 2.00 m 24.0 Ans = 32.0 = 1.33 m 24.0 Ans dFR LFR ' ydFR y = 4m y2 m x2 m a 4y b R = 24.0 kN Ba b 2 3m xdFR = L0 (xdx) L0 LFR dFR LFR Ans: FR = 24.0 kN x = 2.00 m y = 1.33 m 996 © 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 9–118 p The rectangular plate is subjected to a distributed load over its entire surface The load is defined by the expression p = p0 sin (px>a) sin (py>b), where p0 represents the pressure acting at the center of the plate Determine the magnitude and location of the resultant force acting on the plate p0 y x SOLUTION a b Resultant Force and its Location: The volume of the differential element is py px dy b dx b asin dV = dFR = pdxdy = p0 a sin a b a FR = dFR = p0 LFR L0 a sin = p0 B a = b py px dx b a sin dy b a b L0 a b px a px b b R cos b a - cos p a p b 0 4ab p0 p2 Ans Since the loading is symmetric, the location of the resultant force is at the center of the plate Hence, x = a y = b Ans Ans: FR = 4ab p0 p2 a b y = x = 997 © 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 9–119 A wind loading creates a positive pressure on one side of the chimney and a negative (suction) pressure on the other side, as shown If this pressure loading acts uniformly along the chimney’s length, determine the magnitude of the resultant force created by the wind p p ϭ p0 cos u Solution FRx = p 2l 2p L- u l (p0 cos u) cos u r du = p 2rlp0 p L- p = 2rlp0 a b cos2 u du Ans FRx = plrp0 FRy = 2l p p L- (p0 cos u) sin u r du = Thus, Ans FR = plrp0 Ans: p FRx = 2rlp0 a b FR = plrp0 998 © 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 *9–120 When the tide water A subsides, the tide gate automatically swings open to drain the marsh B For the condition of high tide shown, determine the horizontal reactions developed at the hinge C and stop block D The length of the gate is m and its height is m rw = 1.0 Mg/m3 C 4m A 3m B 2m D SOLUTION Fluid Pressure: The fluid pressure at points D and E can be determined using Eq 9–13, p = rgz pD = 1.01103219.812122 = 19 620 N>m2 = 19.62 kN>m2 pE = 1.01103219.812132 = 29 430 N>m2 = 29.43 kN>m2 Thus, wD = 19.62162 = 117.72 kN>m wE = 29.43162 = 176.58 kN>m Resultant Forces: FR1 = 1176.582132 = 264.87 kN FR2 = 1117.722122 = 117.72 kN Equations of Equilibrium: a + ©MC = 0; 264.87132 - 117.7213.3332 - Dx 142 = Ans Dx = 100.55 kN = 101 kN + ©F = 0; : x 264.87 - 117.72 - 100.55 - Cx = Ans Cx = 46.6 kN Ans: Dx = 101 kN Cx = 46.6 kN 999 © 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 9–121 The tank is filled with water to a depth of d = m 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? ro = 900 kg>m3 and rw = 1000 kg>m3 2m 3m A B d SOLUTION For water At side A: wA = b rw g d = 2(1000)(9.81) (4) = 78 480 N/m FRA = (78 480)(4) = 156 960 N = 157 kN Ans At side B: wB = b rw g d = 3(1000)(9.81)(4) = 117 720 N>m FRB = (117 720)(4) = 235 440 N = 235 kN Ans For oil At side A: wA = b ro g d = 2(900)(9.81)d = 17 658 d FRA = (17 658 d)(d) = 156 960 N Ans d = 4.22 m Ans: For water: FRA = 157 kN FRB = 235 kN For oil: d = 4.22 m 1000 © 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 9–122 The concrete “gravity” dam is held in place by its own weight If the density of concrete is rc = 2.5 Mg>m3, and water has a density of rw = 1.0 Mg>m3, determine the smallest dimension d that will prevent the dam from overturning about its end A 1m 6m A d–1 d Solution Loadings The computation will be based on b = m width of the dam The pressure at the base of the dam is P = rgh = 1000(9.81)(6) = 58.86 ( 103 ) pa = 58.86 kPa Thus w = pb = 58.86(1) = 58.86 kN>m The forces that act on the dam and their respective points of application, shown in Fig a, are W1 = 2500 1(6)(1) (9.81) = 147.15 ( 103 ) N = 147.15 kN W2 = 2500c (d - 1)(6)(1) d (9.81) = 73.575 ( d - )( 103 ) = 73.575(d - 1) kN 2 ( FR )v = 1000c (d - 1)(6)(1) d (9.81) = 29.43 ( d - )( 103 ) = 29.43(d - 1) kN ( FR ) h = (58.86)(6) = 176.58 kN x1 = 0.5 x2 = + y = 1 (d - 1) = (d + 2) x3 = + (d - 1) = (2d + 1) 3 3 (6) = m Equation of Equilibrium Write the moment equation of equilibrium about A by referring to the FBD of the dam, Fig a, a+ΣMA = 0; 147.15(0.5) + [73.575(d - 1)]c (d + 2) d + [29.43(d - 1)] c (2d + 1) d - 176.58(2) = 44.145d + 14.715d - 338.445 = Solving and chose the positive root Ans d = 2.607 m = 2.61 m Ans: d = 2.61 m 1001 © 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 9–123 y The factor of safety for tipping of the concrete dam is defined as the ratio of the stabilizing moment due to the dam’s weight divided by the overturning moment about O due to the water pressure Determine this factor if the concrete has a density of rconc = 2.5 Mg>m3 and for water rw = Mg>m3 1m 6m O x 4m Solution Loadings The computation will be based on b = 1m width of the dam The pressure at the base of the dam is P = pwgh = 1000(9.81)(6) = 58.86(103)pa = 58.86 kPa Thus, w = Pb = 58.86(1) = 58.86 kN>m The forces that act on the dam and their respective points of application, shown in Fig a, are W1 = (2500)[(1(6)(1)](9.81) = 147.15(103)] N = 147.15 kN W2 = (2500) c (3)(6)(1) d (9.81) = 220.725(103) N = 220.725 kN FR = (58.86)(6) = 176.58 kN x1 = + (1) = 3.5 ft x2 = (3) = 2m y = (6) = m 3 Thus, the overturning moment about O is MOT = 176.58(2) = 353.16 kN # m And the stabilizing moment about O is Ms = 147.15(3.5) + 220.725(2) = 956.475 kN # m Thus, the factor of safety is F.S = Ms 956.475 = = 2.7083 = 2.71 MOT 353.16 Ans Ans: F.S = 2.71 1002 © 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 *9–124 The concrete dam in the shape of a quarter circle Determine the magnitude of the resultant hydrostatic force that acts on the dam per meter of length The density of water is rw = Mg>m3 3m SOLUTION Loading: The hydrostatic force acting on the circular surface of the dam consists of the vertical component Fv and the horizontal component Fh as shown in Fig a Resultant Force Component: The vertical component Fv consists of the weight of water contained in the shaded area shown in Fig a For a 1-m length of dam, we have Fv = rgAABCb = (1000)(9.81) B (3)(3) - p (3 ) R (1) = 18947.20 N = 18.95 kN The horizontal component Fh consists of the horizontal hydrostatic pressure Since the width of the dam is constant (1 m), this loading can be represented by a triangular distributed loading with an intensity of wC = rghCb = 1000(9.81)(3)(1) = 29.43 kN>m at point C, Fig a Fh = (29.43)(3) = 44.145 kN Thus, the magnitude of the resultant hydrostatic force acting on the dam is FR = 3F h + F v = 344.1452 + 18.952 = 48.0 kN Ans Ans: FR = 48.0 kN 1003 © 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 9–125 The tank is used to store a liquid having a density of 80 lb>ft3 If it is filled to the top, determine the magnitude of force the liquid exerts on each of its two sides ABDC and BDFE C A D ft ft B F 12 ft ft E Solution w1 = 80(4)(12) = 3840 lb>ft w2 = 80(10)(12) = 9600 lb>ft ABDC : F1 = (3840)(5) = 9.60 kip Ans BDEF : F2 = (9600 - 3840)(6) + 3840(6) = 40.3 kip Ans Ans: F1 = 9.60 kip F2 = 40.3 kip 1004 © 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 9–126 y The parabolic plate is subjected to a fluid pressure that varies linearly from at its top to 100 lb>ft at its bottom B Determine the magnitude of the resultant force and its location on the plate ft ft y ϭ x2 ft x Solution FR = LA p dA = L0 = (100 - 25y)(2x dy) L0 B (100 - 25y)ay dyb = 2c 100 a by - 25a by d = 426.7 lb = 427 lb FR y = LA yp dA; 426.7 y = L0 Ans y(100 - 25y)y dy 426.7 y = 2c 100 a by - 25a by2 d 426.7 y = 731.4 Ans y = 1.71 ft Due to symmetry, Ans x = 0 Ans: FR = 427 lb y = 1.71 ft x = 1005 © 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 9–127 The 2-m-wide rectangular gate is pinned at its center A and is prevented from rotating by the block at B Determine the reactions at these supports due to hydrostatic pressure rw = 1.0 Mg>m3 6m 1.5 m A B SOLUTION 1.5 m w1 = 1000(9.81)(3)(2) = 58 860 N/m w2 = 1000(9.81)(3)(2) = 58 860 N/m F1 = (3)(58 860) = 88 290 F2 = (58 860)(3) = 176 580 a + ©MA = 0; 88 290(0.5) - FB (1.5) = Ans FB = 29 430 N = 29.4 kN + ©F = 0; : x 88 290 + 176 580 - 29 430 - FA = Ans FA = 235 440 N = 235 kN Ans: FB = 29.4 kN FA = 235 kN 1006 © 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 *9–128 The tank is filled with a liquid which has a density of 900 kg>m3 Determine the resultant force that it exerts on the elliptical end plate, and the location of the center of pressure, measured from the x axis y 1m 1m y2 x2 0.5 m x 0.5 m SOLUTION Fluid Pressure: The fluid pressure at an arbitrary point along y axis can be determined using Eq 9–13, p = g(0.5 - y) = 900(9.81)(0.5 - y) = 8829(0.5 - y) Resultant Force and its Location: Here, x = 21 - 4y2 The volume of the differential element is dV = dFR = p(2xdy) = 8829(0.5 - y)[221 - 4y 2] dy Evaluating integrals using Simpson’s rule, we have the 0.5 m FR = LFR d FR = 17658 L-0.5 m (0.5 - y)(21 - 4y2)dy Ans = 6934.2 N = 6.93 kN 0.5 m LFR yd FR = 17658 y(0.5 - y)(21 - 4y2)dy = - 866.7 N # m ' ydFR y = L-0.5 m LFR = dFR - 866.7 = - 0.125 m 6934.2 Ans LFR Ans: FR = 6.93 kN y = - 0.125 m 1007 © 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 9–129 Determine the magnitude of the resultant force acting on the gate ABC due to hydrostatic pressure The gate has a width of 1.5 m rw = 1.0 Mg>m3 1.5 m B 1.25 m C SOLUTION 2m w1 = 1000(9.81)(1.5)(1.5) = 22.072 kN>m 60° A w2 = 1000(9.81)(2)(1.5) = 29.43 kN>m Fx = (29.43)(2) + (22.0725)(2) = 73.58 kN F1 = c (22.072) a 1.25 + F2 = b d = 53.078 kN tan 60° (1.5)(2)a b (1000)(9.81) = 16.99 kN tan 60° Fy = F1 + F2 = 70.069 kN F = 2F2x + F2y = 2(73.58)2 + (70.069)2 = 102 kN Ans Ans: F = 102 kN 1008 © 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 9–130 The semicircular drainage pipe is filled with water Determine the resultant horizontal and vertical force components that the water exerts on the side AB of the pipe per foot of pipe length; gw = 62.4 lb>ft3 A ft B SOLUTION Fluid Pressure: The fluid pressure at the bottom of the drain can be determined using Eq 9–13, p = gz p = 62.4(2) = 124.8 lb>ft2 Thus, w w = 124.8(1) = 124.8 lb>ft Resultant Forces: The area of the quarter circle is A = pr = p(2 2) = p ft2 4 Then, the vertical component of the resultant force is Ans FR v = gV = 62.4[p(1)] = 196 lb and the horizontal component of the resultant force is FR h = (124.8)(2) = 125 lb Ans Ans: FRv = 196 lb FRh = 125 lb 1009 ... 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. .. 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,
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