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5.1 PROBLEM Locate the centroid of the plane area shown. SOLUTION A, in 2 x , in. y , in. xA, in 3 yA, in 3 1 2 8 6 = 48 16 12 = 192 -4 9 6 -192 432 1152 1584 8 1536 1344 240 xA 1344 in 3 = A 240 in 2 yA 1584 in 3 = A 240 in 2 Then X = Y = or X = 5.60 in. or Y = 6.60 in. and PROBLEM 5.2 Locate the centroid of the plane area shown. SOLUTION A, mm 2 x , mm 40 112.5 y , mm xA, mm3 yA, mm3 56 250 295 300 351 600 1 2 1 60 75 = 2250 2 105 75 = 7875 10 125 25 37.5 90 000 885 900 975 900 Then and X = Y = xA 975 900 mm3 = A 10 125 mm 2 yA 351 600 mm3 = A 10 125 mm 2 or X = 96.4 mm or Y = 34.7 mm PROBLEM 5.3 Locate the centroid of the plane area shown. SOLUTION For the area as a whole, it can be concluded by observation that Y = A, in 2 2 ( 24 in.) 3 x , in. xA, in 3 or Y = 16.00 in. 1 2 1 24 10 = 120 2 1 24 16 = 192 2 312 2 (10 ) = 6.667 3 10 + 1 (16 ) = 15.333 3 800 2944 3744 Then X = xA 3744 in 3 = A 312 in 2 or X = 12.00 in. PROBLEM 5.4 Locate the centroid of the plane area shown. SOLUTION A, mm 2 x , mm 1.5 -6 8 y , mm xA, mm3 yA, mm3 5082 -54 -72 4956 1 2 3 21 22 = 462 - - 1 ( 6 )( 9 ) = -27 2 1 ( 6 )(12 ) = -36 2 399 11 2 2 693 162 -288 567 Then and X = Y = xA 567 mm3 = A 399 mm 2 yA 4956 mm3 = A 399 mm 2 or X = 1.421 mm or Y = 12.42 mm PROBLEM 5.5 Locate the centroid of the plane area shown. SOLUTION A, mm 2 x , mm 60 94.5 y , mm xA, mm3 yA, mm3 2 880 000 -678 600 2 201 400 1 2 120 200 = 24 000 - 120 120 1 440 000 -534 600 905 400 ( 60 ) 2 2 = -5654.9 18 345 Then X = Y = xA 905 400 mm3 = A 18 345 mm 2 yA 2 201 400 mm3 = A 18 345 mm 2 or X = 49.4 mm or Y = 93.8 mm and PROBLEM 5.6 Locate the centroid of the plane area shown. SOLUTION A, in 2 x , in. y , in. 3.8917 3 x A, in 3 -243 337.5 94.5 y A, in 3 243 202.5 445.5 1 2 (9) 4 2 = 63.617 -4 ( 9 ) = -3.8917 ( 3 ) 5 1 (15)( 9 ) = 67.5 2 131.1 Then X = xA 94.5 in 3 = A 131.1 in 2 or X = 0.721 in. or Y = 3.40 in. and Y = yA 445.5 in 3 = A 131.1 in 2 PROBLEM 5.7 Locate the centroid of the plane area shown. SOLUTION First note that symmetry implies X = Y A, mm 2 x , mm 20 16.98 xA, mm3 1 2 40 40 = 1600 - 32 000 -21 330 10 667 (40) 2 4 = -1257 343 xA 10 667 mm3 = A 343 mm 2 Then X = or X = 31.1 mm and Y = X = 31.1 mm PROBLEM 5.8 Locate the centroid of the plane area shown. SOLUTION First note that symmetry implies X =0 A, in 2 y , in. 1.6977 yA, in 3 -42.67 1 - ( 4) 2 2 = -25.13 2 2 ( 6) 2 = 56.55 2.546 144 101.33 31.42 yA 101.33 in 3 = A 31.42 in 2 Then Y = or Y = 3.23 in. PROBLEM 5.9 For the area of Problem 5.8, determine the ratio r2 /r1 so that y = 3r1/4. SOLUTION A 1 2 - r12 2 y yA 2 - r13 3 2 3 r2 3 2 3 r2 - r13 3 4r1 3 4r2 3 2 r22 - r12 (r 2 2 2 ) ) ( ) ( ) Then or Y A = y A 3 2 2 3 r1 r2 - r12 = r2 - r13 4 2 3 ( 2 r 3 9 r2 - 1 = 2 - 1 16 r1 r1 Let p= r2 r1 9 [ ( p + 1)( p - 1)] = ( p - 1)( p 2 + p + 1) 16 or 16 p 2 + (16 - 9) p + (16 - 9) = 0 PROBLEM 5.9 CONTINUED Then or Taking the positive root p= -(16 - 9) (16 - 9) 2 - 4(16)(16 - 9) 2(16) p = -0.5726 p = 1.3397 r2 = 1.340 r1 PROBLEM 5.10 Show that as r1 approaches r2 , the location of the centroid approaches that of a circular arc of radius ( r1 + r2 ) / 2. SOLUTION First, determine the location of the centroid. From Fig. 5.8A: y2 = 2 sin 2 - r2 3 - 2 ( ( ) ) A2 = ( 2 - ) r22 = Similarly Then yA = 2 cos r2 3 - 2 ( ) A1 = y1 = 2 cos r1 3 - 2 ( ) ( 2 - ) r12 ( ) and Now 2 cos 2 cos - r22 - r1 - r12 r2 2 3 3 - - 2 2 2 2 3 = r2 - r13 cos 3 A = - r22 - - r12 2 2 2 2 = - r2 - r1 2 Y A = yA 2 3 Y - r22 - r12 = r2 - r13 cos 2 3 2 r23 - r13 cos Y = 3 r22 - r12 - 2 ( ( ) ) ( ) ( ) ( ) ( ) ( ) PROBLEM 5.10 CONTINUED Using Figure 5.8B, Y of an arc of radius 1 ( r1 + r2 ) is 2 Y = sin - 1 ( r1 + r2 ) 2 2 - 2 ( ( ) ) (1) = 1 cos (r1 + r2 ) 2 - 2 ( Now ( r2 - r1 ) r22 + r1 r2 + r12 r23 - r13 = r22 - r12 ( r2 - r1 )( r2 + r1 ) 2 r + r1 r2 + r12 = 2 r2 + r1 r2 = r + r1 = r - 1 r = ( r1 + r2 ) 2 2 ( ) ) Let Then ( r + ) + ( r + )( r - ) + ( r - ) r23 - r13 and = 2 2 r2 - r1 (r + ) + (r - ) 3r 2 + 2 = 2r In the limit as 0 (i.e., r1 = r2 ), then r23 - r13 3 = r 2 2 2 r2 - r1 3 1 = (r1 + r2 ) 2 2 2 3 cos Y = ( r1 + r2 ) 3 4 - 2 2 so that Which agrees with Eq. (1). or Y = 1 cos ( r1 + r2 ) 2 - 2 PROBLEM 5.11 Locate the centroid of the plane area shown. SOLUTION First note that symmetry implies X =0 r2 = 2 2 in., = 45 y2 = 2r sin 2 2 2 sin = 3 3 4 ( ) ( ) = 1.6977 in. ( ) 4 A, in 2 y , in. y A, in 3 1 1 ( 4)(3) = 6 2 1 6 1.8997 -2.667 5.2330 2 (2 2 ) 4 - 2 = 6.283 2 - y = 0.3024 0.6667 3 1 ( 4)( 2) = -4 2 8.283 Then Y A = yA Y 8.283 in 2 = 5.2330 in 3 ( ) or Y = 0.632 in. PROBLEM 5.12 Locate the centroid of the plane area shown. SOLUTION A, mm 2 x , mm -15 10 y , mm xA, mm3 yA, mm3 72 000 -10 125 1 2 (40)(90) = 3600 20 -15 -54 000 6750 ( 40)( 60) 4 = 2121 3 Then 1 (30)( 45) = 675 2 6396 -25.47 -19.099 -54 000 -101 250 -40 500 21 375 XA = xA X 6396 mm 2 = -101 250 mm3 and YA = yA ( ) or X = -15.83 mm Y 6396 mm 2 = 21 375 mm3 ( ) or Y = 3.34 mm PROBLEM 5.13 Locate the centroid of the plane area shown. SOLUTION A, mm 2 x , mm 48 53.33 y , mm xA, mm3 yA, mm3 32 000 -21 330 10 667 1 2 2 ( 40)(80) = 2133 3 - 1 ( 40)(80) = -1600 2 533.3 15 13.333 102 400 -85 330 17 067 Then X A = XA X 533.3 mm 2 = 17 067 mm3 and Y A = yA ( ) or X = 32.0 mm Y 533.3 mm 2 = 10 667 mm3 ( ) or Y = 20.0 mm PROBLEM 5.14 Locate the centroid of the plane area shown. SOLUTION A, mm 2 x , mm 56.25 50 y , mm xA, mm3 yA, mm3 2 304 000 -360 000 1 944 000 1 2 2 (150 )( 240 ) = 24 000 3 - 1 (150)(120) = -9000 2 15 000 96 40 1 350 000 -450 000 900 000 Then X A = xA X 15 000 mm 2 = 900 000 mm3 and Y A = yA ( ) or X = 60.0 mm Y 15 000 mm 2 = 1 944 000 ( ) or Y = 129.6 mm PROBLEM 5.15 Locate the centroid of the plane area shown. SOLUTION A, in 2 x , in. y , in. 7.5 16.366 xA, in 3 yA, in 3 375 2892 3267 1 2 1 (10)(15) = 50 3 4.5 6.366 225 1125 1350 4 (15)2 = 176.71 226.71 X A = x A Then X 226.71 in 2 = 1350 in 3 and Y A = y A ( ) or X = 5.95 in. Y 226.71 in 2 = 3267 in 3 ( ) or Y = 14.41 in. PROBLEM 5.16 Locate the centroid of the plane area shown. SOLUTION A, in 2 x , in. y , in. 2.8 -0.8 xA, in 3 yA, in 3 119.47 4.267 123.73 1 2 2 (8)(8) = 42.67 3 - 2 ( 4)( 2) = -5.333 3 37.33 3 1.5 128 -8 120 Then X A = x A X 37.33 in 2 = 120 in 3 and Y A = y A ( ) or X = 3.21 in. Y 37.33 in 2 = 123.73 in 3 ( ) or Y = 3.31 in. PROBLEM 5.17 The horizontal x axis is drawn through the centroid C of the area shown and divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x axis, and explain the results obtained. SOLUTION Note that Then and Qx = yA (Qx )1 = 5 3 ( Qx )2 1 m 6 5 m 2 2 or ( Qx )1 = 25.0 103 mm3 2 1 1 1 = - 2.5 m 9 2.5 m 2 + - 2.5 m 6 2.5 m 2 3 2 3 2 or ( Qx ) 2 = -25.0 103 mm3 Now Qx = ( Qx )1 + ( Qx ) 2 = 0 This result is expected since x is a centroidal axis ( thus y = 0 ) and Qx = y A = Y A (y = 0 Qx = 0 ) PROBLEM 5.18 The horizontal x axis is drawn through the centroid C of the area shown and divides the area into two component areas A1 and A2. Determine the first moment of each component area with respect to the x axis, and explain the results obtained. SOLUTION First, locate the position y of the figure. A, mm 2 y , mm yA, mm3 7 200 000 -2 560 000 1 2 160 300 = 48 000 -150 80 = -16 000 32 000 Y A = y A 150 160 4 640 000 Then Y 32 000 mm 2 = 4 640 000 mm3 or ( ) Y = 145.0 mm PROBLEM 5.18 CONTINUED A I: Q I = yA 155 115 - ( 80 115) (160 155) + = 2 2 6 3 = 1.393 10 mm A II : Q II = yA 145 =- (160 145) - 2 = -1.393 106 mm3 85 - 2 ( 80 85 ) ( Qarea ) x = QI + QII = 0 Which is expected since Qx = yA = yA and y = 0 , since x is a centroidal axis. PROBLEM 5.19 The first moment of the shaded area with respect to the x axis is denoted by Qx . (a) Express Qx in terms of r and . (b) For what value of is Qx maximum, and what is the maximum value? SOLUTION (a) With Qx = yA and using Fig. 5.8 A, 2 r sin - r 2 - - Qx = 3 2 2 - 2 2 = r 3 cos - cos sin 2 3 ( ) ( 2 ) ( 3 r sin ) 1 2r cos r sin 2 ( ) or Qx = (b) By observation, Qx is maximum when and then 2 3 r cos3 3 =0 Qx = 2 3 r 3 PROBLEM 5.20 A composite beam is constructed by bolting four plates to four 2 2 3/8-in. angles as shown. The bolts are equally spaced along the beam, and the beam supports a vertical load. As proved in mechanics of materials, the shearing forces exerted on the bolts at A and B are proportional to the first moments with respect to the centroidal x axis of the red shaded areas shown, respectively, in parts a and b of the figure. Knowing that the force exerted on the bolt at A is 70 lb, determine the force exerted on the bolt at B. SOLUTION From the problem statement: F Qx FA FB so that = (Qx ) A (Qx ) B and Now So FB = (Qx ) B F (Qx ) A A Qx = yA ( Qx ) A ( Qx )B 0.375 = 7.5 in. + in. 10 in. ( 0.375 in.) = 28.82 in 3 2 0.375 = ( Qx ) A + 2 7.5 in. - in. (1.625 in.)( 0.375 in.) 2 + 2 ( 7.5 in. - 1 in.) ( 2 in.)( 0.375 in.) and = 28.82 in 3 + 8.921 in 3 + 9.75 in 3 = 47.49 in 3 Then FB = 47.49 in 3 ( 70 lb) = 115.3 lb 28.82 in 3 PROBLEM 5.21 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. L, in. x , in. y , in. xL, in 2 yL, in 2 1 2 16 12 24 8 16 4 0 6 12 128 102 96 -48 -32 0 72 288 54 48 18 480 3 4 6 8 6 72 -8 -4 9 6 3 5 6 0 0 336 Then X L = x L X ( 72 in.) = 336 in 2 and Y L = y L or X = 4.67 in. Y (72 in.) = 480 in 2 or Y = 6.67 in. PROBLEM 5.22 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. L, mm x , mm 82.5 165 112.5 30 y , mm xL, mm 2 yL, mm 2 0 2812 7875 3602 14 289 1 2 3 4 Then 165 75 105 602 + 752 = 96.05 441.05 0 37.5 75 37.5 13 612 12 375 11 812 2881 40 680 X L = x L X (441.05 mm) = 40 680 mm 2 and Y L = y L or X = 92.2 mm Y (441.05 mm) = 14 289 mm 2 Y = 32.4 mm PROBLEM 5.23 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. L, mm x , mm 6 12 1.5 -9 -4.5 y , mm xL, mm 2 yL, mm 2 40.25 224 462 224 32.45 982.7 1 2 3 4 5 Then 122 + 62 = 13.416 16 21 16 62 + 92 = 10.817 77.233 3 14 22 14 3 80.50 192 31.50 -144 -48.67 111.32 X L = x L X (77.233 mm) = 111.32 mm 2 and Y L = y L or X = 1.441 mm Y (77.233 mm) = 982.7 mm 2 or Y = 12.72 mm PROBLEM 5.24 A thin, homogeneous wire is bent to form the perimeter of the figure indicated. Locate the center of gravity of the wire figure thus formed. SOLUTION First note that because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. By symmetry L, in. X =0 y , in. 0 2 ( 6) yL, in 2 0 72 0 32 104 1 2 3 4 2 ( 6) 2 2 ( 4) = 3.820 0 ( 4) 35.416 = 2.546 Then Y L = y L Y (35.416 in.) = 104 in 2 or Y = 2.94 in. PROBLEM 5.25 A 750 = g uniform steel rod is bent into a circular arc of radius 500 mm as shown. The rod is supported by a pin at A and the cord BC. Determine (a) the tension in the cord, (b) the reaction at A. SOLUTION First note, from Figure 5.8B: X = ( 0.5 m ) sin 30 /6 = Then 1.5 m W = mg = ( 0.75 kg ) 9.81 m/s 2 = 7.358 N ( ) Also note that ABD is an equilateral triangle. Equilibrium then requires (a) M A = 0: 1.5 m cos 30 ( 7.358 N ) - ( 0.5 m ) sin 60 TBC = 0 0.5 m - or TBC = 1.4698 N or TBC = 1.470 N (b) Fx = 0: Ax + (1.4698 N ) cos 60 = 0 or Ax = -0.7349 N Fy = 0: Ay - 7.358 N + (1.4698 N ) sin 60 = 0 or Ay = 6.085 N thus A = 6.13 N 83.1 PROBLEM 5.26 The homogeneous wire ABCD is bent as shown and is supported by a pin at B. Knowing that l = 8 in., determine the angle for which portion BC of the wire is horizontal. SOLUTION First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through B. Further, because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. Thus M B = 0, which implies that x = 0 or xL = 0 Hence - 2(6 in.) ( 6 in.) + 8 in. ( 8 in.) 2 6 in. + 8 in. - cos ( 6 in.) = 0 2 Then cos = 4 9 or = 63.6 PROBLEM 5.27 The homogeneous wire ABCD is bent as shown and is supported by a pin at B. Knowing that = 30 , determine the length l for which portion CD of the wire is horizontal. SOLUTION First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through B. Further, because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. Thus M B = 0, which implies that x = 0 Hence or xi Li = 0 2 ( 6 in.) - cos 30 + ( 6 in.) sin 30 ( 6 in.) ( l in.) + cos 30 ( l in.) 2 6 in. + ( l in.) cos 30 - ( 6 in.) = 0 2 or l 2 + 12.0l - 316.16 = 0 with roots l1 = 12.77 and -24.77. Taking the positive root l = 12.77 in. PROBLEM 5.28 The homogeneous wire ABCD is bent as shown and is attached to a hinge at C. Determine the length L for which the portion BCD of the wire is horizontal. SOLUTION First note that for equilibrium, the center of gravity of the wire must lie on a vertical line through C. Further, because the wire is homogeneous, its center of gravity will coincide with the centroid of the corresponding line. Thus M C = 0, which implies that x = 0 or Hence or xi Li = 0 L ( L ) + ( -4 in.)(8 in.) + ( -4 in.)(10 in.) = 0 2 L2 = 144 in 2 or L = 12.00 in. PROBLEM 5.29 Determine the distance h so that the centroid of the shaded area is as close to line BB as possible when (a) k = 0.2, (b) k = 0.6. SOLUTION Then y = yA A or (a + h) a ( ab ) - kb ( a - h ) 2 2 y = ba - kb ( a - h ) = Let Then 2 2 1 a (1 - k ) + kh 2 a(1 - k ) + kh c =1- k y = and a c + k 2 2 c + k = h a (1) Now find a value of (or h) for which y is minimum: 2 dy a 2k ( c + k ) - k c + k = =0 d 2 ( c + k ) 2 ( ) or 2 ( c + k ) - c + k 2 = 0 ( ) (2) PROBLEM 5.29 CONTINUED Expanding (2) Then 2c + 2 2 - c - k -2c 2 =0 or k 2 + 2c - c = 0 = ( 2c) 2 - 4 ( k ) ( c ) 2k Taking the positive root, since h > 0 (hence > 0 ) h=a -2 (1 - k ) + 4 (1 - k ) + 4k (1 - k ) 2 2k 4 (1 - 0.2 ) + 4 ( 0.2 )(1 - 0.2 ) 2 (a) k = 0.2: h=a -2 (1 - 0.2 ) + 2 ( 0.2 ) or h = 0.472a (b) k = 0.6: h=a -2 (1 - 0.6 ) + 4 (1 - 0.6 ) + 4 ( 0.6 )(1 - 0.6 ) 2 2 ( 0.6 ) or h = 0.387a PROBLEM 5.30 Show when the distance h is selected to minimize the distance y from line BB to the centroid of the shaded area that y = h. SOLUTION From Problem 5.29, note that Eq. (2) yields the value of that minimizes h. Then from Eq. (2) We see 2 = c + k 2 c + k (3) Then, replacing the right-hand side of (1) by 2 , from Eq. (3) We obtain y = a ( 2) 2 h a Q.E.D. But So = y =h PROBLEM 5.31 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION For the element of area (EL) shown and dA = ( h - y ) dx x = h 1 - dx a y = h x a Then xEL = x yEL = 1 (h + 2 h = 1 + 2 y) x a a Then area A = dA = a x h 1 0 a h 1 0 x x2 1 - dx = h x - = 2 ah 2a a 0 a and xEL dA = x 2 x3 x 1 2 - dx = h 2 - 3a = 6 a h a 0 yEL dA = 0 a h x x h2 a x2 1 + h 1 - dx = 1 - 2 dx a 2 2 0 a a a h2 x3 = x- 2 2 3a = 0 1 2 ah 3 Hence xA = xEL dA 1 1 x ah = a 2h 2 6 x = yA = yEL dA 1 1 y ah = ah 2 2 3 y = 1 a 3 2 h 3 PROBLEM 5.32 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION For the element (EL) shown At Then Now x = a, y = h : h = ka3 x= or a 1/3 y h1/3 k = h a3 dA = xdy = a 1/3 y dy h1/3 xEL = 1 1 a 1/3 x= y , yEL = y 2 2 h1/ 3 h Then A = dA =0 h a 1/3 3 a y dy = y 4/3 1/3 4 h1/3 h ( ) h = 0 3 ah 4 and xEL dA = 0 1 a 1/3 a 1/3 1 a 3 5/3 3 2 y 1/3 y dy = a h y = 2 h1/3 2 h 2/3 5 10 h 0 h h a 3 7/3 3 2 h a 1/3 yEL dA = 0 y h1/3 y dy = h1/3 7 y = 7 ah 0 Hence 3 2 3 xA = xEL dA : x ah = a h 4 10 3 3 yA = yEL dA: y ah = ah 2 4 7 x = y = 2 a 5 4 h 7 PROBLEM 5.33 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION For the element (EL) shown At x = a, y = h: h = k1a3 a = k 2 h3 Hence, on line 1 or k2 = a h3 k1 = h a3 or y = and on line 2 y = Then h h dA = 1/3 x1/3 - 3 x3 dx a a h 3 x a3 h 1/3 x a1/3 and yEL = 1 h 1/3 h 3 1/3 x + 3 x a 2 a a h 1 1 a h 3 A = dA = 0 1/3 x1/3 - 3 x3 dx = h 1/3 x 4/3 - 3 x 4 = ah 2 a a 4a 4a 0 xELdA = a x 0 h 1/3 h 1 8 2 3 x - 3 x3 dx = h 1/3 x7/3 - 3 x5 = a h 1/3 5a a a 7a 0 35 1 h h h h a a 1/3 3 1/3 3 yEL dA = 0 2 a1/3 x + a3 x a1/3 x - a3 x dx h 2 a x 2/3 x 6 h 2 3 x5/3 1 x 6 8 2 = 2/3 - 6 dx = 0 a 5 a5/3 - 7 a 6 = 35 ah 2 2 a 0 a From and 8 2 ah xA = xEL dA: x = a h 2 35 8 2 ah yA = yEL dA: y = ah 2 35 or x = or y = 16 a 35 16 h 35 PROBLEM 5.34 Determine by direct integration the centroid of the area shown. SOLUTION First note that symmetry implies For the element (EL) shown 2r x =0 yEL = dA = rd r r2 rd r r1 (Figure 5.8B) Then A = dA = r2 2 r2 - r12 = = 2 2 r1 r2 ( ) and 2 yEL dA = r1 r 2r ( rd r ) = 2 r 3 = r23 - r13 3 r 3 1 1 r2 2 ( ) So 2 3 yA = yEL dA: y r22 - r12 = r2 - r13 2 3 4 r23 - r13 or y = 3 r22 - r12 ( ) ( ) PROBLEM 5.35 Determine by direct integration the centroid of the area shown. SOLUTION First note that symmetry implies For the element (EL) shown x =0 y = R cos , x = R sin dx = R cos d dA = ydx = R 2 cos 2 d Hence 1 2 sin 2 A = dA = 20 R 2 cos 2 d = 2 R 2 + = R ( 2 sin 2 ) 2 4 0 2 R 2 2 2 31 2 yEL dA = 20 2 cos R cos d = R 3 cos sin + 3 sin 0 ( ) = R3 cos 2 sin + 2sin 3 ( ) But yA = yEL dA so R3 cos 2 sin + 2sin 3 y = R2 ( 2 + sin 2 ) 2 ( ) or cos 2 + 2 2 y = R sin 3 ( 2 + sin 2 ) ( ) Alternatively, y = 2 3 - sin 2 R sin 3 2 + sin 2 PROBLEM 5.36 Determine by direct integration the centroid of the area shown. SOLUTION For the element (EL) shown y = and b 2 a - x2 a xEL Then ) 1 b = x; y = ( y + b ) = a+ a -x ) 2 2a ( b A = dA = ( a - a - x ) dx a = b a - a 2 - x 2 dx a dA = ( b - y ) dx ( EL 2 2 a 0 2 2 To integrate, let x = a sin : a 2 - x 2 = a cos , dx = a cos d Then A = 0 /2 b a ( a - a cos )( a cos d ) /2 b 2 = a 2 sin - a 2 + sin a 4 0 2 = ab 1 - 4 1 a b b a and xEL dA = 0 x a - a 2 - x 2 dx = x 2 + a 2 - x 2 a a 2 3 ( ) ( ) 3/2 /2 0 = 1 3 ab 6 a b 2 2 b 2 2 yEL dA = 0 2a a + a - x a a - a - x dx ( ) ( ) b2 a 2 b 2 x3 x dx = = 2 0 2a 2a 2 3 ( ) a = 0 1 2 ab 6 or x = or y = 2a 3 ( 4 - ) 2b 3 ( 4 - ) xA = xEL dA: yA = yEL dA: 1 x ab 1 - = a 2b 4 6 1 y ab 1 - = ab 2 4 6 PROBLEM 5.37 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION For the element (EL) shown on line 1 at x = a, b = k2a 2 y = On line 2 at or b 2 x a2 or k2 = -2b a3 k2 = b a2 x = a, -2b = k1a3 y = -2b 3 x a3 2b b dA = 2 x 2 + 3 x3 dx a a Then b 2 x3 b x3 2 x 4 A = dA = 2 x 2 + dx = 2 + x 4a a a 3 a 0 a 0 1 1 5 = ab + = ab 3 2 6 and xEL dA = = a x 0 b 2 2b 3 b x 4 2 x5 2 2 1 x + 3 x dx = 2 2 4 + 5a = a b 4 + 5 a a a 0 a 13 2 ab 20 2b 3 b 2 2b 3 a1 b 2 yEL dA = 0 2 a 2 x - a3 x a 2 x + a3 x dx 1 b 2b = 2 x 2 - 3 x 3 2 a a a 0 2 2 b 2 x5 2 7 dx = 4 5 - 7a 2 x 2a 0 a 2 13 1 = b 2a5 - = - ab 2 70 10 7 Then xA = xEL dA: yA = yEL dA: 5 13 2 x ab = ab 6 20 5 13 2 y ab - ab 6 70 or x = or y = - 39 a 50 39 b 175 PROBLEM 5.38 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION At x = 0, y = b b = k (0 - a) 2 or k = b a2 Then Now and Then y = b ( x - a )2 a2 y b 2 = x - a) 2( 2 2a b ( x - a )2 dx a2 xEL = x, yEL = dA = ydx = A = dA = 0 a a b b 1 2 3 x - a ) dx = 2 ( x - a ) = ab 2( 0 3 3a a and 2 a a 3 2 2 xEL dA = 0 x a 2 ( x - a ) dx = a 2 0 ( x - 2ax + a x )dx b b = b x4 2 3 a2 2 1 2 2 4 - 3 ax + 2 x = 12 a b a b b b2 1 a 2 2 5 a yEL dA = 0 2a 2 ( x - a ) a 2 ( x - a ) dx = 2a 4 5 ( x - a ) 0 = Hence 1 2 ab 10 1 2 1 xA = xEL dA: x ab = ab 3 12 x = y = 1 a 4 3 b 10 1 2 1 yA = yEL dA: y ab = ab 3 10 PROBLEM 5.39 Determine by direct integration the centroid of the area shown. SOLUTION Have yEL = xEL = x 1 a x x2 y = 1 - + 2 2 2 L L x x2 dA = ydx = a 1 - + 2 dx L L Then A = dA = 2L a 1 0 x x2 x2 x3 8 - + 2 dx = a x - + 2 = aL L L 2 L 3L 0 3 2L 2L and x2 x x2 x3 x4 2L xEL dA = 0 x a 1 - L + L2 dx = a 2 - 3L + 4L2 0 10 2 = aL 3 yEL dA = 0 = 2L a x x2 x x2 1 - + 2 a 1 - + 2 dx L L L L 2 a 2 EL x x2 x3 x 4 0 1 - 2 L + 3 L2 - 2 L3 + L4 dx 2 a2 2 x 2 x3 x4 x5 11 + 2 - 3 + 4 = a2L x - L 5 L 2L 5L 0 2L = Hence 8 10 2 xA = xEL dA: x aL = aL 3 3 1 11 yA = yEL dA: y a = a 2 5 8 x = y = 5 L 4 33 a 40 PROBLEM 5.40 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION For y1 at x = a, y = 2b y1 = 2b = ka 2 2b 2 x a2 or k = 2b a2 Then By observation Now and for 0 x a : y2 = - b x ( x + 2b) = b 2 - a a xEL = x yEL = For a x 2a : yEL = Then 1 b y1 = 2 x 2 2 a and dA = y1dx = 2b 2 x dx a2 1 b x y2 = 2 - 2 2 a A = dA = 0 a and x dA = y2dx = b 2 - dx a 2b 2 x 2a x dx + a b 2 - dx 2 a a 2a a 2 a 2b x3 x 7 = 2 + b - 2 - = ab a 6 a 3 0 2 0 and x a 2b 2 2a xEL dA = 0 x a 2 x dx + a x b 2 - a dx = = = 2 x3 2b x 4 + b x - 3a 0 a2 4 0 a 2a 1 2 1 2 2 2 3 a b + b ( 2a ) - ( a ) + 3a 2a - ( a ) 2 7 2 ab 6 { ( ) PROBLEM 5.40 CONTINUED x x a b 2 2b 2 2a b yEL dA = 0 a 2 x a 2 x dx + 0 2 2 - a b 2 - a dx 2b 2 = 4 a = Hence a 3 x5 b2 a x + - 2 - 2 3 a 5 0 a 2a 17 2 ab 30 7 7 xA = xEL dA: x ab = a 2b 6 6 x =a y = 17 b 35 7 17 2 yA = yEL dA: y ab = ab 6 30 PROBLEM 5.41 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and b. SOLUTION For y2 Then Now and for 0 x at x = a, y = b : a = kb 2 y2 = b 1/2 x a or k = a b2 xEL = x a y b x1/2 x1/2 : yEL = 2 = , dA = y2dx = b dx 2 2 2 a a For a 1 b x 1 x1/2 x a : yEL = ( y1 + y2 ) = - + 2 2 2a 2 a x1/2 x 1 dA = ( y2 - y1 ) dx = b - + dx a a 2 Then A = dA = 0 b b = a = a/2 x1/2 x 1 x1/2 a dx + a/2 b a - a + 2 dx a a a/2 2 x3/2 x2 1 2 3/2 x + b - + x 3 2a 2 a/2 0 3 a 3/2 3/2 2 b a 3/2 a + ( a ) - 3 a 2 2 2 1 a 1 a 2 + b - a - + ( a ) - 2 2 2 2a ( ) = 13 ab 24 PROBLEM 5.41 CONTINUED and 1/2 x1/2 x 1 a/2 x a - + dx xEL dA = 0 x b dx + a/2 x b a a a 2 a/2 2 x5/2 x3 x 4 b 2 5/2 = - + x + b 4 a/2 a 5 0 5 a 3a a = 5/2 5/2 2 b a 5/2 a + ( a ) - 5 a 2 2 1 3 a 3 1 2 a 2 + b - ( a ) - + ( a ) - 2 3a 2 4 = 71 2 ab 240 a/2 b dx b yEL dA = 0 2 a a x1/2 x1/2 1 x1/2 x1/2 x 1 a b x + a/2 - + - + dx b 2 a 2 a a a 2 a/2 3 b2 1 2 b 2 x 2 1 x 1 x = + - - 2a 2 0 2 2a 3a a 2 a/2 a = = b a 2 a + ( a ) - 4a 2 2 2 2 b2 a 1 - - 6a 2 2 3 11 2 ab 48 x = Hence 71 2 13 xA = xEL dA: x ab = ab 24 240 13 11 2 yA = yEL dA: y ab = ab 24 48 17 a = 0.546a 130 11 b = 0.423b 26 y = PROBLEM 5.42 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. Express your answer in terms of a. SOLUTION First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line Have at Thus x = a, y = a : a = ka 2 y = or dy = k = 1 a 1 2 x a 2 and 2 xdx a 2 Then dy 2 dL = 1 + dx = 1 + x dx dx a a 0 L = dL = = x a 2 4 4x2 4x2 1 + 2 x 2 dx = 1 + 2 + ln x + 1 + 2 4 a 2 a a a 0 a a a 5 + ln 2 + 5 = 1.4789a 2 4 a x 0 ( ) xELdL = = 3/2 4x2 2 a2 4 1 + 2 dx = 1 + 2 x 2 3 8 a a 0 a Then a 2 3/2 5 - 1 = 0.8484a 2 12 xL = xEL dL: x (1.4789a ) = 0.8484a 2 ( ) x = 0.574a PROBLEM 5.43 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. SOLUTION First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line Now Then and xEL = r cos 7 /4 and dL = rd 7 /4 L = dL = /4 rd = r [ ] /4 = 7 /4 xELdL = /4 r cos ( rd ) 3 r 2 1 1 7 /4 2 = r 2 [sin ] /4 = r 2 - - = -r 2 2 2 Thus 3 xL = xdL : x r = -r 2 2 2 x =- 2 2 r 3 PROBLEM 5.44 A homogeneous wire is bent into the shape shown. Determine by direct integration the x coordinate of its centroid. SOLUTION First note that because the wire is homogeneous, its center of gravity coincides with the centroid of the corresponding line Now Where xEL = a cos3 and dL = dx 2 + dy 2 x = a cos3 : dx = -3a cos 2 sin d y = a sin 3 : dy = 3a sin 2 cos d Then dL = -3a cos 2 sin d ( ) + (3a sin cos d ) 2 2 2 1/2 = 3a cos sin cos 2 + sin 2 = 3a cos sin d L = dL = 0 = and 3 a 2 /2 ( ) 1/2 d /2 1 3a cos sin d = 3a sin 2 2 0 /2 3 xEL dL = 0 a cos ( 3a cos sin d ) 1 = 3a 2 - cos5 5 0 Hence /2 = 3 2 a 5 x = 2 a 5 3 3 xL = xEL dL : x a = a 2 2 5 PROBLEM 5.44 CONTINUED Alternative solution x x = a cos3 cos 2 = a y y = a sin sin = a 3 2 2/3 2/3 x a Then Now and 2/3 y + a 2/3 =1 or 1/2 y = a 2/3 - x 2/3 ( ) 3/2 dy = a 2/3 - x 2/3 dx xEL = x dy dL = 1 + dx 2 ( ) (-x ) -1/3 dx = 1 + a 2/3 - x 2/3 ( ) (-x ) 1/2 -1/3 1/2 2 dx Then L = dL = 0 a a1/3 3 3 dx = a1/3 x 2/3 = a 1/ 3 2 2 x 0 a 3 1/3 3 5/3 dx = a x = a 2 5 0 5 a and xELdL = 1/3 a a x 1/3 0 x Hence 3 3 xL = xEL dL : x a = a 2 2 5 x = 2 a 5 PROBLEM 5.45 Determine by direct integration the centroid of the area shown. SOLUTION Have xEL = yEL 2 2 r cos = ae cos 3 3 2 2 = r sin = ae sin 3 3 1 1 ( r )( rd ) = a 2e2 d 2 2 and Then dA = 1 1 1 1 A = dA = 0 a 2e 2 d = a 2 e 2 = a 2 e2 - 1 = 133.623a 2 2 2 2 4 0 ( ) and 2 2 3 3 xELdA = 0 3ae cos 2 a e d = 3 a 0 e cos d 2 1 1 To proceed, use integration by parts, with u = e3 and du = 3e3 d and v = sin dv = cos d Then 3 3 3 e cos d = e sin - sin ( 3e d ) Now let u = e3 then then du = 3e3 d v = - cos dv = sin d , Then So that 3 3 -3 3 e sin d = e sin - 3 -e cos - ( - cos ) ( 3e d ) 3 e cos d = 10 ( sin + 3cos ) e3 1 e3 a3 xEL dA = a3 ( sin + 3cos ) = -3e3 - 3 = -1239.26a3 3 10 0 30 ( ) Also 2 2 3 3 yEL dA = 0 3 ae sin 2 a e d = 3 a 0 e sin d 2 1 1 PROBLEM 5.45 CONTINUED Using integration by parts, as above, with u = e3 and du = 3e3 d and v = - cos dv = sin d Then 3 3 3 e sin d = -e cos - ( - cos ) ( 3e d ) So that e3 sin d = e3 ( - cos + 3sin ) 10 1 e3 a3 yEL dA = a3 ( - cos + 3sin ) = e3 + 1 = 413.09a3 3 10 0 30 ( ) Hence xA = xEL dA: x 133.623a 2 = -1239.26a3 yA = yEL dA: 2 3 ( ) y (133.623a ) = 413.09a or x = -9.27a or y = 3.09a PROBLEM 5.46 Determine by direct integration the centroid of the area shown. SOLUTION Have and xEL = x, yEL = 1 x x sin 2 L dA = ydx A = dA = 0 x sin L/2 x L2 x L x - x cos dx = 2 sin L L L 0 L/2 = L2 2 and x L/2 x = xEL dA = 0 x x sin dx L 2 L2 x 2 L3 x L 2 x = 2 x sin + 3 cos - x sin L L L 0 L/2 = L3 2 -2 L3 3 Also y = yEL dA = 0 L/2 1 2 x sin x x dx x sin L L L/2 1 2 L2 x L 2 L3 x = 2 x sin - x - 3 cos 2 L L 0 = L2 L L3 1 1 L3 6 + 2 - ( -1) = 2 2 6 8 4 2 2 96 ( ) PROBLEM 5.46 CONTINUED Hence L2 z 1 xA = xEL dA: x 2 = L3 2 - 3 or x = 0.363L L2 L3 1 2 yA = yEL dA: y 2 = 96 2 2 - 3 or y = 0.1653L PROBLEM 5.47 Determine the volume and the surface area of the solid obtained by rotating the area of Problem. 5.2 about (a) the x axis, (b) the line x = 165 mm. SOLUTION From the solution to Problem 5.2: A = 10 125 mm 2 , X area = 96.4 mm, Yarea = 34.7 mm From the solution to Problem 5.22: L = 441.05 mm X line = 92.2 mm, Yline = 32.4 mm Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the x axis: Area = 2 Yline L = 2 ( 32.4 mm )( 441.05 mm ) = 89.786 103 mm 2 A = 89.8 103 mm 2 Volume = 2 Yarea A = 2 ( 34.7 mm )(10 125 mm ) = 2.2075 106 mm3 V = 2.21 106 mm3 (b) Rotation about x = 165 mm: Area = 2 165 - X line L = 2 (165 - 92.2 ) mm ( 441.05 mm ) = 2.01774 105 mm 2 A = 0.202 106 mm 2 Volume = 2 165 - X area A = 2 (165 - 96.4 ) mm (10 125 mm ) = 4.3641 106 mm3 V = 4.36 106 mm3 ( Area ) ( Line ) ( ) ( ) PROBLEM 5.48 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.4 about (a) the line y = 22 mm, (b) the line x = 12 mm. SOLUTION From the solution to Problem 5.4: A = 399 mm 2 , X area = 1.421 mm, Yarea = 12.42 mm From the solution to Problem 5.23: L = 77.233 mm, X line = 1.441 mm, Yline = 12.72 mm (Line) (Area) Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the line y = 22 mm: Area = 2 22 - Yline L = 2 ( 22 - 12.72 ) mm ( 77.233 mm ) = 4503 mm 2 A = 4.50 103 mm 2 Volume = 2 22 - Yarea A = 2 ( 22 - 12.42 ) mm 399 mm 2 = 24 016.97 mm3 V = 24.0 103 mm3 (b) Rotation about line x = 12 mm: Area = 2 12 - X line L = 2 (12 - 1.441) mm ( 77.233 mm ) = 5124.45 mm 2 A = 5.12 103 mm 2 Volume = 2 (12 - 1.421) A = 2 (12 - 1.421) mm 399 mm 2 = 26 521.46 mm3 V = 26.5 103 mm3 ( ) ( ) ( ) ( ) ( ) PROBLEM 5.49 Determine the volume and the surface area of the solid obtained by rotating the area of Problem 5.1 about (a) the x axis, (b) the line x = 16 in. SOLUTION From the solution to Problem 5.1: A = 240 in 2 , X area = 5.60 in., Yarea = 6.60 in. From the solution to Problem 5.21: L = 72 in., X line = 4.67 in., Yline = 6.67 in. (Area) Applying the theorems of Pappus-Guldinus, we have (a) Rotation about the x axis: Ax = 2 Yline L = 2 ( 6.67 in.)( 72 in.) = 3017.4 in 2 A = 3020 in 2 Vx = 2 Yarea A = 2 ( 6.60 in.) 240 in 2 = 9952.6 in 3 V = 9950 in 3 (b) Rotation about x = 16 in.: Ax =16 = 2 16 - X line L = 2 (16 - 4.67 ) in. ( 72 in.) = 5125.6 in 2 Ax =16 = 5130 in 2 Vx =16 = 2 16 - X area A = 2 (16 - 5.60 ) in. 240 in 2 = 15 682.8 in 3 Vx =16 = 15.68 103 in 3 ( ) ( ) ( ) ( ) PROBLEM 5.50 Determine the volume of the solid generated by rotating the semielliptical area shown about (a) the axis AA, (b) the axis BB, (c) the y axis. SOLUTION Applying the second theorem of Pappus-Guldinus, we have (a) Rotation about axis AA: ab 2 2 Volume = 2 yA = 2 ( a ) = ab 2 (b) Rotation about axis BB: ab 2 2 Volume = 2 yA = 2 ( 2a ) = 2 a b 2 (c) Rotation about y-axis: 4a ab 2 2 Volume = 2 yA = 2 = a b 3 2 3 V = 2 2 a b 3 V = 2 2a 2b V = 2a 2b PROBLEM 5.51 Determine the volume and the surface area of the chain link shown, which is made from a 2-in.-diameter bar, if R = 3 in. and L = 10 in. SOLUTION First note that the area A and the circumference C of the cross section of the bar are A= 4 d2 and C = d Observe that the semicircular ends of the link can be obtained by rotating the cross section through a horizontal semicircular arc of radius R. Then, applying the theorems of Pappus-Guldinus, we have Volume = 2 (Vside ) + 2 (Vend ) = 2 ( AL ) + 2 ( RA) = 2 ( L + R ) A 2 = 2 10 in. + ( 3 in.) ( 2 in.) 4 = 122.049 in 3 V = 122.0 in 3 Area = 2 ( Aside ) + 2 ( Aend ) = 2 ( CL ) + 2 ( RC ) = 2 ( L + R ) C = 2 10 in. + ( 3 in.) ( 4 in.) = 488.198 in 2 A = 488 in 2 PROBLEM 5.52 Verify that the expressions for the volumes of the first four shapes in Figure 5.21 on page 261 are correct. SOLUTION Following the second theorem of Pappus-Guldinus, in each case a specific generating area A will be rotated about the x axis to produce the given shape. Values of y are from Fig. 5.8A. (1) Hemisphere: the generating area is a quarter circle Have 4a V = 2 yA = 2 a 2 3 4 or V = 2 3 a 3 (2) Semiellipsoid of revolution: the generating area is a quarter ellipse Have 4a V = 2 yA = 2 ha 3 4 or V = 2 2 a h 3 (3) Paraboloid of revolution: the generating area is a quarter parabola Have 3 2 V = 2 yA = 2 a ah 8 3 or V = 1 2 a h 2 (4) Cone: the generating area is a triangle Have a 1 V = 2 yA = 2 ha 3 2 or V = 1 2 a h 3 PROBLEM 5.53 A 15-mm-diameter hole is drilled in a piece of 20-mm-thick steel; the hole is then countersunk as shown. Determine the volume of steel removed during the countersinking process. SOLUTION The required volume can be generated by rotating the area shown about the y axis. Applying the second theorem of Pappus-Guldinus, we have 5 1 V = 2 xA = 2 + 7.5 mm 5 mm 5 mm 3 2 or V = 720 mm3 PROBLEM 5.54 Three different drive belt profiles are to be studied. If at any given time each belt makes contact with one-half of the circumference of its pulley, determine the contact area between the belt and the pulley for each design. SOLUTION Applying the first theorem of Pappus-Guldinus, the contact area AC of a belt is given by AC = yL = yL Where the individual lengths are the "Lengths" of the belt cross section that are in contact with the pulley Have AC = 2 ( y1L1 ) + y2 L2 2.5 mm 2.5 = 2 60 - mm 2 cos 20 + ( 60 - 2.5 ) mm (12.5 mm ) or AC = 3.24 103 mm 2 Have AC = 2 ( y1L1 ) 7.5 mm 7.5 = 2 60 - 1.6 - mm 2 cos 20 or AC = 2.74 103 mm 2 Have 2 5 AC = ( y1L1 ) = 60 - mm ( 5 mm ) or AC = 2.80 103 mm 2 PROBLEM 5.55 Determine the capacity, in gallons, of the punch bowl shown if R = 12 in. SOLUTION The volume can be generated by rotating the triangle and circular sector shown about the y axis. Applying the second theorem of Pappus-Guldinus and using Fig. 5.8A, we have V = 2 xA = 2xA = 2 ( x1 A1 + x2 A2 ) 1 1 1 1 2 3 2 R sin 30 = 2 R R cos30 R R + 2 3 3 2 2 2 6 6 R3 R3 3 3 3 = 2 + 16 3 2 3 = 8 R = 3 3 3 (12 in.) = 3526.03 in 3 8 Since 1 gal = 231 in 3 V = 3526.03 in 3 = 15.26 gal 231 in 3/gal V = 15.26 gal PROBLEM 5.56 The aluminum shade for a small high-intensity lamp has a uniform thickness of 3/32 in. Knowing that the specific weight of aluminum is 0.101 lb/in 3 , determine the weight of the shade. SOLUTION The weight of the lamp shade is given by W = V = At where A is the surface area of the shade. This area can be generated by rotating the line shown about the x axis. Applying the first theorem of Pappus-Guldinus, we have A = 2 yL = 2yL = 2 ( y1L1 + y2 L2 + y3 L3 + y4 L4 ) 0.6 mm 0.60 + 0.75 = 2 ( 0.6 mm ) + mm 2 2 ( 0.15 mm )2 + (1.5 mm )2 0.75 + 1.25 + mm 2 1.25 + 1.5 + mm 2 = 22.5607 in Then 2 ( 0.50 mm )2 + ( 0.40 mm )2 ( 0.25 mm )2 + (1.25 mm )2 3 in. = 0.21362 lb 32 W = 0.101 lb/in 3 22.5607 in 2 W = 0.214 lb PROBLEM 5.57 The top of a round wooden table has the edge profile shown. Knowing that the diameter of the top is 1100 mm before shaping and that the density of the wood is 690 kg/m3 , determine the weight of the waste wood resulting from the production of 5000 tops. SOLUTION All dimensions are in mm Have Vwaste = Vblank - Vtop Vblank = ( 550 mm ) ( 30 mm ) = 9.075 106 mm3 2 Vtop = V1 + V2 + V3 + V4 Applying the second theorem of Pappus-Guldinus to parts 3 and 4 2 2 Vtop = ( 529 mm ) (18 mm ) + ( 535 mm ) (12 mm ) 4 12 2 + 2 535 + mm (12 mm ) 3 4 4 18 2 + 2 529 + mm (18 mm ) 3 4 = ( 5.0371 + 3.347 + 0.1222 + 0.2731) 106 mm3 = 8.8671 106 mm3 Vwaste = ( 9.0750 - 8.8671) 106 mm3 = 0.2079 10-3 m3 Finally Wwaste = wood Vwaste g N tops = 690 kg/m3 0.2079 10-3 m3 9.81 m/s 2 5000 ( tops ) ( ) or Wwaste = 2.21 kN PROBLEM 5.58 The top of a round wooden table has the shape shown. Determine how many liters of lacquer are required to finish 5000 tops knowing that each top is given three coats of lacquer and that 1 liter of lacquer covers 12 m2. SOLUTION Referring to the figure in solution of Problem 5.57 and using the first theorem of Pappus-Guldinus, we have Asurface = Atop circle + Abottom circle + Aedge 2 2 = ( 535 mm ) + ( 529 mm ) 2 12 + 2 535 + mm (12 mm ) 2 2 18 + 2 529 + mm (18 mm ) 2 = 617.115 103 mm 2 Then # liters = Asurface Coverage N tops N coats = 617.115 10-3 m 2 1 liter 5000 3 12 m 2 or # liters = 2424 L PROBLEM 5.59 The escutcheon (a decorative plate placed on a pipe where the pipe exits from a wall) shown is cast from yellow brass. Knowing that the specific 3 weight of yellow brass is 0.306 lb/in . determine the weight of the escutcheon. SOLUTION The weight of the escutcheon is given by W = (specific weight)V where V is the volume of the plate. V can be generated by rotating the area A about the x axis. Have and Then a = 3.0755 in. - 2.958 in. = 0.1175 in. sin = 0.5 = 0.16745 R = 9.5941 3 or 2 = 26 - 9.5941 = 16.4059 = 8.20295 = 0.143169 rad The area A can be obtained by combining the following four areas, as indicated. Applying the second theorem of Pappus-Guldinus and then using Figure 5.8A, we have V = 2 yA = 2 yA PROBLEM 5.59 CONTINUED A, in 2 y , in. yA, in 3 1 2 3 4 1 ( 3.0755)(1.5) = 2.3066 2 - ( 3) = -1.28851 2 - 1 ( 2.958)( 0.5) = -0.7395 2 - ( 0.1755 )( 0.5 ) = -0.05875 2 ( 3) sin sin ( + ) = 0.60921 3 1 ( 0.5 ) = 0.16667 3 1 ( 0.5) = 0.25 2 1 (1.5) = 0.5 3 1.1533 0.78497 0.12325 0.14688 yA = 0.44296 in 3 Then so that V = 2 0.44296 in 3 = 1.4476 in 3 W = 1.4476 in 3 0.306 lb/in 3 = 0.44296 lb ( ) ( ) W = 0.443 lb PROBLEM 5.60 The reflector of a small flashlight has the parabolic shape shown. Determine the surface area of the inside of the reflector. SOLUTION First note that the required surface area A can be generated by rotating the parabolic cross section through 2 radians about the x axis. Applying the first theorem of Pappus-Guldinus, we have A = 2 yL Now, since x = ky , 2 at x = a : a = k ( 7.5 ) 2 or At or Then a = 56.25 k x = ( a + 15 ) mm: a + 15 = k (12.5 ) 2 (1) a + 15 = 156.25k (2) Eq. (2) a + 15 156.25k : = Eq. (1) 56.25k a or a = 8.4375 mm 1 mm dx = 0.3 y dy Eq. (1) k = 0.15 x = 0.15 y 2 and 2 Now So dx dL = 1 + dy = 1 + 0.09 y 2 dy dy A = 2 yL 12.5 and yL = ydL A = 2 7.5 y 1 + 0.09 y 2 dy 2 1 2 = 2 1 + 0.09 y 3 0.18 ( ) 12.5 3/2 7.5 = 1013 mm 2 or A = 1013 mm 2 PROBLEM 5.61 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports. SOLUTION Resultant (a) Have R = R1 + R2 R1 = ( 40 lb/ft )(18 ft ) = 720 lb R2 = 1 (120 lb/ft )(18 ft ) = 1080 lb 2 R = 1800 lb or M A: The resultant is located at the centroid C of the distributed load x Have or (1800 lb ) x = ( 40 lb/ft )(18 ft )( 9 ft ) + x = 10.80 ft 1 (120 lb/ft )(18 ft )(12 ft ) 2 R = 1800 lb x = 10.80 ft (b) Fx = 0: Ax = 0 Fy = 0: Ay - 1800 lb = 0, Ay = 1800 lb M A = 0: M A - (1800 lb )(10.8 ft ) = 0 M A = 19.444 lb ft A = 1800 lb or M A = 19.44 kip ft PROBLEM 5.62 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports. SOLUTION (a) Have RI = ( 300 N/m )( 6 m ) = 1800 N RII = 1 ( 6 m )( 900 N/m ) = 1800 N 3 Then or Fy : - R = - RI - RII R = 1800 N + 1800 N = 3600 N M A : - x ( 3600 N ) = - ( 3 m )(1800 N ) - ( 4.5 m )(1800 N ) or x = 3.75 m R = 3600 N x = 3.75 m (b) Reactions Fx = 0: Ax = 0 M A = 0: or ( 6 m ) By - ( 3600 N )( 3.75 m ) = 0 By = 2250 N B = 2250 N Fy = 0: Ay + 2250 N = 3600 N or Ay = 1350 N A = 1350 N PROBLEM 5.63 Determine the reactions at the beam supports for the given loading. SOLUTION Have RI = (100 lb/ft )( 4 ft ) = 400 lb RII = RIII 1 ( 200 lb/ft )( 6 ft ) = 600 lb 2 = ( 200 lb/ft )( 4 ft ) = 800 lb Fx = 0: Ax = 0 M A = 0: Then ( 2 ft )( 400 lb ) - ( 4 ft )( 600 lb ) - (12 ft )(800 lb ) + (10 ft ) By By = 800 lb Fy = 0: Ay + 800 lb - 400 lb - 600 lb - 800 = 0 =0 or B = 800 lb or Ay = 1000 lb A = 1000 lb PROBLEM 5.64 Determine the reactions at the beam supports for the given loading. SOLUTION Have RI = ( 9 ft )( 200 lb/ft ) = 1800 lb RII = 1 ( 3 ft )( 200 lb/ft ) = 300 lb 2 Then Fx = 0: Ax = 0 M A = 0: - ( 4.5 ft )(1800 lb ) - (10 ft )( 300 lb ) + ( 9 ft ) By = 0 or By = 1233.3 lb Fy = 0: Ay - 1800 lb - 300 lb + 1233.3 lb = 0 B = 1233 lb or Ay = 866.7 lb A = 867 lb PROBLEM 5.65 Determine the reactions at the beam supports for the given loading. SOLUTION Have RI = 1 ( 200 N/m )( 0.12 m ) = 12 N 2 RII = ( 200 N/m )( 0.2 m ) = 40 N Then Fx = 0: Ax = 0 Fy = 0: Ay + 18 N - 12 N - 40 N = 0 or Ay = 34 N M A = 0: M A - ( 0.8 m )(12 N ) - ( 0.22 m )( 40 N ) + ( 0.38 m )(18 N ) or M A = 2.92 N m A = 34.0 N M A = 2.92 N m PROBLEM 5.66 Determine the reactions at the beam supports for the given loading. SOLUTION First, replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a linear relation between load and distance, and the values at the end points are the same. Have RI = (1.8 m )( 2000 N/m ) = 3600 N RII = 1 (1.8 m )( 4500 N/m ) = 4050 N 2 Then Fx = 0: Ax = 0 M B = 0: - ( 3 m ) Ay - ( 2.1 m )( 3600 N ) + ( 2.4 m )( 4050 N ) or Ay = 270 N Fy = 0: 270 N - 3600 N + 4050 N - By = 0 A = 270 N or By = 720 N B = 720 N PROBLEM 5.67 Determine the reactions at the beam supports for the given loading. SOLUTION Have RI = RII = 1 ( 4 m )( 2000 kN/m ) = 2667 N 3 1 ( 2 m )(1000 kN/m ) = 666.7 N 3 Fx = 0: Ax = 0 Then Fy = 0: Ay - 2667 N - 666.7 N = 0 or Ay = 3334 N M A = 0: M A - (1 m )( 2667 N ) - ( 5.5 m )( 666.7 N ) A = 3.33 kN or M A = 6334 N m M A = 6.33 kN m PROBLEM 5.68 Determine the reactions at the beam supports for the given loading. SOLUTION First, replace the given loading with the loading shown below. The two loadings are equivalent because both are defined by a parabolic relation between load and distance, and the values at end points are the same. Have RI = ( 8 ft )(100 lb/ft ) = 800 lb RII = 2 (8 ft )( 600 lb/ft ) = 3200 lb 3 Fx = 0: Ax = 0 Then M A = 0: 11B + ( 5 ft )( 800 lb ) - ( 4 ft )( 3200 ) lb = 0 or B = 800 lb Fy = 0: Ay - 3200 lb + 800 lb + 800 lb = 0 or A = 1600 lb PROBLEM 5.69 Determine (a) the distance a so that the vertical reactions at supports A and B are equal, (b) the corresponding reactions at the supports. SOLUTION (a) Have RI = RII = 1 ( a ft )(120 lb/ft ) = ( 60a ) lb 2 1 (12 - a )( 40 lb/ft ) = ( 240 - 20a ) lb 2 Then or Now Also or Equating Eqs. (1) and (2) Fy = 0: Ay - 60a - ( 240 - 2a ) + By = 0 Ay + By = 240 + 40a Ay = By Ay = By = 120 + 20a a 1 M B = 0: - (12 m ) Ay + ( 60a ) lb 12 - ft + (12 - a ) ft ( 240 - 20a ) lb = 0 3 3 Ay = 80 - (1) 140 10 2 a- a 3 9 140 10 a - a2 3 9 (2) 120 + 20a = 80 - or Then Now (b) Have Eq. (1) 40 2 a - 320a + 480 = 0 3 a = 1.6077 ft, a = 22.392 a = 1.608 ft a 12 ft Fx = 0: Ax = 0 Ay = By = 120 + 20 (1.61) = 152.2 lb A = B = 152.2 lb PROBLEM 5.70 Determine (a) the distance a so that the vertical reaction at support B is minimum, (b) the corresponding reactions at the supports. SOLUTION (a) Have RI = RII = 1 ( a ft )(120 lb/ft ) = 60a lb 2 1 (12 - a ) ft ( 40 lb/ft ) = ( 240 - 20a ) lb 2 Then or Then (b) Eq. (1) a a M A = 0: - ft ( 60a lb ) - ( 240 - 20a ) lb 8 + ft + (12 ft ) By = 0 3 3 By = dBy da By = 10 2 20 a - a + 160 9 3 = (1) or a = 3.00 ft 20 20 a- =0 9 3 10 20 ( 3.00 )2 - ( 3.00 ) + 160 9 3 B = 150.0 lb Fx = 0: Ax = 0 = 150 lb and Fy = 0: Ay - 60 ( 3.00 ) lb - 240 - 20 ( 3.00 ) lb + 150 lb = 0 or Ay = 210 lb A = 210 lb PROBLEM 5.71 Determine the reactions at the beam supports for the given loading when w0 = 1.5 kN/m. SOLUTION Have RI = 1 ( 9 m )( 2 kN/m ) = 9 kN 2 RII = ( 9 m )(1.5 kN/m ) = 13.5 kN Then Fx = 0: Cx = 0 M B = 0: - 50 kN m - (1 m )( 9 kN ) - ( 2.5 m )(13.5 kN ) + ( 6 m ) C y = 0 or C y = 15.4583 kN Fy = 0: By - 9 kN - 13.5 kN + 15.4583 = 0 C = 15.46 kN or By = 7.0417 kN B = 7.04 kN PROBLEM 5.72 Determine (a) the distributed load w0 at the end D of the beam ABCD for which the reaction at B is zero, (b) the corresponding reactions at C. SOLUTION Have RI = 1 ( 9 m ) ( 3.5 - w0 ) kN/m = 4.5 ( 3.5 - w0 ) kN 2 RII = ( 9 m ) ( w0 kN/m ) = 9w0 kN (a) Then or so M C = 0: - 50 kN m + ( 5 m ) 4.5 ( 3.5 - w0 ) kN + ( 3.5 m ) ( 9w0 kN ) = 0 9w0 + 28.75 = 0 w0 = -3.1944 kN/m w0 = 3.19 kN/m Note: the negative sign means that the distributed force w0 is upward. (b) Fx = 0: Cx = 0 Fy = 0: - 4.5 ( 3.5 + 3.19 ) kN + 9 ( 3.19 ) kN + C y = 0 or C y = 1.375 kN C = 1.375 kN PROBLEM 5.73 A grade beam AB supports three concentrated loads and rests on soil and the top of a large rock. The soil exerts an upward distributed load, and the rock exerts a concentrated load RR as shown. Knowing that P = 4 kN and wB = 1 wA , determine the values of wA and RR corresponding to 2 equilibrium. SOLUTION Have RI = (1.2 m )( wA kN/m ) = 1.2 wA kN RII = 1 1 (1.8 m ) wA kN/m = 0.45 wA kN 2 2 1 RIII = (1.8 m ) wA kN/m = 0.9 wA kN 2 Then M C = 0: - ( 0.6 m ) (1.2 wA ) kN + ( 0.6 m ) ( 0.45 wA ) kN/m + ( 0.9 m ) ( 0.9 wA ) kN/m - (1.2 m )( 4 kN/m ) - ( 0.8 m )(18 kN/m ) + ( 0.7 m )( 24 kN/m ) = 0 or and wA = 6.667 kN/m wA = 6.67 kN/m Fy = 0: RR + (1.2 m )( 6.67 kN/m ) + ( 0.45 m )( 6.67 kN/m ) + ( 0.9 m )( 6.67 kN/m ) - 24 kN - 18 kN - 4 kN or RR = 29.0 kN RR = 29.0 kN PROBLEM 5.74 A grade beam AB supports three concentrated loads and rests on soil and the top of a large rock. The soil exerts an upward distributed load, and the rock exerts a concentrated load RR as shown. Knowing that wB = 0.4wA, determine (a) the largest value of P for which the beam is in equilibrium, (b) the corresponding value of wA. In the following problems, use = 62.4 lb/ft3 for the specific weight of fresh water and c = 150 lb/ft3 for the specific weight of concrete if U.S. customary units are used. With SI units, use = 103 kg/m3 for the density of fresh water and c = 2.40 103 kg/m3 for the density of concrete. (See the footnote on page 222 for how to determine the specific weight of a material given its density.) j SOLUTION Have RI = (1.2 m )( wA kN/m ) = 1.2 wA kN RII = 1 (1.8 m )( 0.6 wA kN/m ) = 0.54 wA kN 2 RIII = (1.8 m )( 0.4 wA kN/m ) = 0.72 wA kN (a) Then M A = 0: ( 0.6 m ) (1.2 wA ) kN + (1.2 m ) RR + (1.8 m ) ( 0.54 wA ) kN + ( 2.1 m ) ( 0.72 wA ) kN - ( 0.5 m )( 24 kN ) - ( 2.0 m )(18 kN ) + ( 2.4 m ) P = 0 or and or 3.204 wA + 1.2 RR - 2.4 P = 48 Fy = 0: RR + 1.2 WA + 0.54 WA + 0.72 WA - 24 - 18 - P = 0 RR + 2.46 WA - P = 42 (2) (1) Now combine Eqs. (1) and (2) to eliminate wA : ( 3.204 ) Eq. 2 - ( 2.46 ) Eq. 1 or (b) Then, from Eq. (2): 0.252 RR = 16.488 - 2.7 P P = 6.11 kN or WA = 19.56 kN/m Since RR must be 0, the maximum acceptable value of P is that for which R = 0, P = 6.1067 kN 2.46 WA - 6.1067 = 42 PROBLEM 5.75 The cross section of a concrete dam is as shown. For a dam section of unit width, determine (a) the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of the reaction forces of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam. In the following problems, use = 62.4 lb/ft3 for the specific weight of fresh water and c = 150 lb/ft3 for the specific weight of concrete if U.S. customary units are used. With SI units, use = 103 kg/m3 for the density of fresh water and c = 2.40 103 kg/m3 for the density of concrete. (See the footnote on page 222 for how to determine the specific weight of a material given its density.) SOLUTION The free body shown consists of a 1-m thick section of the dam and the triangular section BCD of the water behind the dam. Note: X1 = 6 m X 2 = ( 9 + 3) m = 12 m X 3 = (15 + 2 ) m = 17 m X 4 = (15 + 4 ) m = 19 m (a) Now W = gV so that 1 W1 = 2400 kg/m3 9.81 m/s 2 ( 9 m )(15 m )(1 m ) = 1589 kN 2 ( )( ) W2 = 2400 kg/m3 9.81 m/s 2 ( 6 m )(18 m )(1 m ) = 2543 kN 1 W3 = 2400 kg/m3 9.81 m/s 2 ( 6 m )(18 m )(1 m ) = 1271 kN 2 1 W4 = 2400 kg/m3 9.81 m/s 2 ( 6 m )(18 m )(1 m ) = 529.7 kN 2 Also P= 1 1 Ap = (18 m )(1 m ) 103 kg/m3 9.81 m/s 2 (18 m ) 2 2 = 1589 kN Then or Fx = 0: H - 1589 kN = 0 H = 1589 kN ( )( ) ( )( ) ( )( ) ( )( ) H = 1589 kN Fy = 0: V - 1589 kN - 2543 kN - 1271 kN - 529.7 kN or V = 5933 kN V = 5.93 MN PROBLEM 5.75 CONTINUED (b) Have M A = 0: X ( 5933 kN ) + ( 6 m )(1589 kN ) - ( 6 m )(1589 kN ) - (12 m )( 2543 kN ) - (17 m )(1271 kN ) - (19 m )( 529.7 ) = 0 or X = 10.48 m X = 10.48 m to the right of A (c) Consider water section BCD as the free body. Have F = 0 Then -R = 1675 kN 18.43 18.43 or R = 1675 kN Alternative solution to part (c) Consider the face BC of the dam. Have BC = 62 + 182 = 18.9737 m tan = and 6 18 = 18.43 p = ( g ) h = 1000 kg/m3 9.81 m/s 2 (18 m ) = 176.6 kN/m 2 ( )( ) Then R= 1 1 Ap = (18.97 m )(1 m ) 176.6 kN/m 2 2 2 = 1675 kN ( ) R = 1675 kN 18.43 PROBLEM 5.76 The cross section of a concrete dam is as shown. For a dam section of unit width, determine (a) the reaction forces exerted by the ground on the base AB of the dam, (b) the point of application of the resultant of the reaction forces of part a, (c) the resultant of the pressure forces exerted by the water on the face BC of the dam. SOLUTION The free body shown consists of a 1-ft thick section of the dam and the parabolic section of water above (and behind) the dam. Note x1 = 5 (16 ft ) = 10 ft 8 1 x2 = 16 + ( 6 ) ft = 19 ft 2 1 x3 = 22 + (12 ) ft = 25 ft 4 5 x4 = 22 + (12 ) ft = 29.5 ft 8 PROBLEM 5.76 CONTINUED Now W = V 2 W1 = 150 lb/ft 3 (16 ft )( 24 ft ) (1 ft ) = 38, 400 lb 3 ( ) W2 = 150 lb/ft 3 ( 6 ft )( 24 ft ) (1 ft ) = 21, 600 lb 1 W3 = 150 lb/ft 3 (12 ft )(18 ft ) (1 ft ) = 10,800 lb 3 2 W4 = 62.4 lb/ft 3 (12 ft )(18 ft ) (1 ft ) = 8985.6 lb 3 Also (a) Then P= 1 1 Ap = (18 1) ft 2 62.4 lb/ft 3 18 ft = 10,108.8 lb 2 2 ( ) ( ) ( ) ( ) Fx = 0: H - 10,108.8 lb = 0 or H = 10.11 kips Fy = 0: V - 38, 400 lb - 21, 600 lb - 10,800 lb - 8995.6 lb = 0 or (b) V = 79,785.6 V = 79.8 kips M A = 0: X ( 79, 785.6 lb ) - ( 6 ft )( 38, 400 lb ) - (19 ft )( 21,600 lb ) - ( 25 ft )(10,800 lb ) - ( 29.5 ft )( 8985.6 lb ) + ( 6 ft )(10,108.8 lb ) = 0 or X = 15.90 ft The point of application of the resultant is 15.90 ft to the right of A (c) Consider the water section BCD as the free body. Have F = 0 R = 13.53 kips = 41.6 On the face BD of the dam R = 13.53 kips 41.6 PROBLEM 5.77 The 9 12-ft side AB of a tank is hinged at its bottom A and is held in place by a thin rod BC. The maximum tensile force the rod can withstand without breaking is 40 kips, and the design specifications require the force in the rod not exceed 20 percent of this value. If the tank is slowly filled with water, determine the maximum allowable depth of water d in the tank. SOLUTION Consider the free-body diagram of the side. Have Now Then, for d max: P= 1 1 Ap = A ( d ) 2 2 M A = 0: ( 9 ft ) T - d P=0 3 ( 9 ft ) ( 0.2 ) ( 40 103 lb ) - or or d max 1 3 (12 ft ) ( d max ) 62.4 lb/ft d max = 0 3 2 ( ) 3 216 103 ft 3 = 374.4 d max 3 d max = 576.92 ft 3 d max = 8.32 ft PROBLEM 5.78 The 9 12-ft side of an open tank is hinged at its bottom A and is held in place by a thin rod. The tank is filled with glycerine, whose specific weight is 80 lb/ft 3. Determine the force T in the rod and the reactions at the hinge after the tank is filled to a depth of 8 ft. SOLUTION Consider the free-body diagram of the side. Have P= 1 1 Ap = A ( d ) 2 2 1 ( 8 ft )(12 ft ) 80 lb/ft 3 ( 8 ft ) = 30, 720 lb 2 Fy = 0: Ay = 0 = ( ) Then M A = 0: ( 9 ft ) T 8 - ft ( 30, 720 lb ) = 0 3 or T = 9102.22 lb T = 9.10 kips Fx = 0: Ax + 30,720 lb - 9102.22 lb = 0 or A = -21, 618 lb A = 21.6 kips PROBLEM 5.79 The friction force between a 2 2-m square sluice gate AB and its guides is equal to 10 percent of the resultant of the pressure forces exerted by the water on the face of the gate. Determine the initial force needed to lift the gate that its mass is 500 kg. SOLUTION Consider the free-body diagram of the gate. Now PI = 1 1 ApI = ( 2 2 ) m 2 103 kg/m3 9.81 m/s 2 ( 3 m ) 2 2 ( )( ) = 58.86 kN PII = 1 1 ApII = ( 2 2 ) m 2 103 kg/m3 9.81 m/s 2 ( 5 m ) 2 2 F = 0.1P = 0.1( PI + PII ) ( )( ) = 98.10 kN Then = 0.1( 58.86 + 98.10 ) kN = 15.696 kN Finally Fy = 0: T - 15.696 kN - ( 500 kg ) 9.81 m/s 2 = 0 ( ) or T = 20.6 kN PROBLEM 5.80 The dam for a lake is designed to withstand the additional force caused by silt which has settled on the lake bottom. Assuming that silt is equivalent to a liquid of density s = 1.76 103 kg/m3 and considering a 1-m-wide section of dam, determine the percentage increase in the force acting on the dam face for a silt accumulation of depth 1.5 m. SOLUTION First, determine the force on the dam face without the silt. Have Pw = 1 1 Apw = A ( gh ) 2 2 1 ( 6 m )(1 ) m 103 kg/m3 9.81 m/s 2 ( 6 m ) 2 = ( )( ) = 176.58 kN Next, determine the force on the dam face with silt. Have Pw = 1 ( 4.5 m )(1m ) 103 kg/m3 9.81 m/s 2 ( 4.5 m ) 2 ( )( ) = 99.326 kN ( Ps )I = (1.5 m )(1 m ) 103 kg/m3 9.81 m/s 2 ( 4.5 m ) = 66.218 kN ( )( ) ( Ps )II = 1 (1.5 m )(1 m ) 1.76 103 kg/m3 9.81 m/s 2 (1.5 m ) 2 P = Pw + ( Ps )I + ( Ps )II = 184.97 kN ( )( ) = 19.424 kN Then The percentage increase, % inc., is then given by % inc. = (184.97 - 176.58) 100% = 4.7503% P - Pw 100% = 176.58 Pw % inc. = 4.75% PROBLEM 5.81 The base of a dam for a lake is designed to resist up to 150 percent of the horizontal force of the water. After construction, it is found that silt (which is equivalent to a liquid of density s = 1.76 103 kg/m3 ) is settling on the lake bottom at a rate of 20 mm/y. Considering a 1-m-wide section of dam, determine the number of years until the dam becomes unsafe. SOLUTION From Problem 5.80, the force on the dam face before the silt is deposited, is Pw = 176.58 kN. The maximum allowable force Pallow on the dam is then: Pallow = 1.5Pw = (1.5 )(176.58 kN ) = 264.87 kN Next determine the force P on the dam face after a depth d of silt has settled. Have Pw = 1 ( 6 - d ) m (1 m ) 103 kg/m3 9.81 m/s 2 ( 6 - d ) m 2 2 ( )( ) = 4.905 ( 6 - d ) kN ( Ps )I = d (1 m ) 103 kg/m3 9.81 m/s 2 ( 6 - d ) m = 9.81 6d - d 2 kN ( )( ) ( ) ( Ps )II = 1 d (1 m ) 1.76 103 kg/m3 9.81 m/s 2 ( d ) m 2 ( )( ) ) = 8.6328d 2 kN P = Pw + ( Ps )I + ( Ps )II = 4.905 36 - 12d + d 2 + 9.81 6d - d 2 + 8.6328d 2 kN = 3.7278d 2 + 176.58 kN ( ( ) PROBLEM 5.81 CONTINUED Now required that P = Pallow to determine the maximum value of d. or Finally ( 3.7278d 2 + 176.58 kN = 264.87 kN d = 4.8667 m ) 4.8667 m = 20 10-3 m N year or N = 243 years PROBLEM 5.82 The square gate AB is held in the position shown by hinges along its top edge A and by a shear pin at B. For a depth of water d = 3.5 m, determine the force exerted on the gate by the shear pin. SOLUTION First consider the force of the water on the gate. Have Then PI = P= 1 1 Ap = A ( gh ) 2 2 1 (18 m )2 103 kg/m3 9.81 m/s2 (1.7 m ) 2 ( )( ) = 26.99 kN PII = 1 (18 m )2 103 kg/m3 9.81 m/s2 (1.7 1.8cos 30 ) m 2 ( )( ) = 51.74 kN Now or or M A = 0: 1 2 ( LAB ) PI + ( LAB ) PII - LAB FB = 0 3 3 1 2 ( 26.99 kN ) + ( 51.74 kN ) - FB = 0 3 3 FB = 43.49 kN FB = 4.35 kN 30.0 PROBLEMS 5.83 AND 5.84 Problem 5.83: A temporary dam is constructed in a 5-ft-wide fresh water channel by nailing two boards to pilings located at the sides of the channel and propping a third board AB against the pilings and the floor of the channel. Neglecting friction, determine the reactions at A and B when rope BC is slack. Problem 5.84: A temporary dam is constructed in a 5-ft-wide fresh water channel by nailing two boards to pilings located at the sides of the channel and propping a third board AB against the pilings and the floor of the channel. Neglecting friction, determine the magnitude and direction of the minimum tension required in rope BC to move board AB. SOLUTION First, consider the force of the water on the gate. Have So that P= PI = 1 1 Ap = A ( h ) 2 2 1 (1.5 ft )( 5 ft ) 62.4 lb/ft 3 (1.8 ft ) 2 ( ) = 421.2 lb PII = 1 (1.5 ft )( 5 ft ) 62.4 lb/ft 3 ( 3 ft ) 2 ( ) = 702 lb 5.83 Find the reactions at A and B when rope is slack. M A = 0: - ( 0.9 ft ) B + ( 0.5 ft )( 421.2 lb ) + (1.0 ft )( 702 lb ) = 0 or B = 1014 lb B = 1014 lb Fx = 0: 2 Ax + or 4 4 ( 421.2 lb ) + ( 702 lb ) = 0 5 5 Ax = -449.28 lb Note that the factor 2 (2 Ax ) is included since Ax is the horizontal force exerted by the board on each piling. Fy = 0: 1014 lb - or 3 3 ( 421.2 lb ) - ( 702 lb ) + Ay = 0 5 5 Ay = -340.08 lb A = 563 lb 37.1 PROBLEMS 5.83 AND 5.84 CONTINUED 5.84 Note that there are two ways to move the board: 1. Pull upward on the rope fastened at B so that the board rotates about A. For this case B 0 and TBC is perpendicular to AB for minimum tension. 2. Pull horizontally at B so that the edge B of the board moves to the left. For this case Ay 0 and the board remains against the pilings because of the force of the water. M A = 0: -1.5TBC + ( 0.5 ft )( 421.2 lb ) + (1.0 ft )( 702 lb ) = 0 or TBC = 608.4 lb Case (1) Case (2) M B = 0: - (1.2 ft ) ( 2 Ax ) - ( 0.5 ft )( 702 lb ) - (1.0 ft )( 421.2 lb ) = 0 or 2 Ax = - 643.5 lb Fx = 0: - TBC - 643.5 lb + + 4 ( 421.2 lb ) 5 4 ( 702 lb ) = 0 5 or TBC = 255.06 lb ( TBC )min = 255 lb PROBLEMS 5.85 AND 5.86 Problem 5.85: A 2 3-m gate is hinged at A and is held in position by rod CD. End D rests against a spring whose constant is 12 kN/m. The spring is undeformed when the gate is vertical. Assuming that the force exerted by rod CD on the gate remains horizontal, determine the minimum depth of water d for which the bottom B of the gate will move to the end of the cylindrical portion of the floor. Problem 5.86: Solve Problem 5.85 if the mass of the gate is 500 kg. SOLUTION First, determine the forces exerted on the gate by the spring and the water when B is at the end of the cylindrical portion of the floor. Have Then sin = 1 2 = 30 xsp = (1.5 m ) tan 30 Fsp = kxsp = (12 kN/m )(1.5 m ) tan 30 = 10.39 kN and Assume Have Then PI = P= d 2m 1 1 Ap = A ( g ) h 2 2 1 ( 2 m )( 3 m ) 103 kg/m3 9.81 m/s 2 ( d - 2 ) m 2 ( )( ) = 29.43 ( d - 2 ) kN PII = 1 ( 2 m )( 3 m ) 103 kg/m3 9.81 m/s 2 ( d - 2 + 2cos 30 ) m 2 ( )( ) = 29.43 ( d - 0.2679 ) kN PROBLEMS 5.85 AND 5.86 CONTINUED 5.85 Find d min so that gate opens, W = 0. 2 M A = 0: m 29.43 ( d - 2 ) kN 3 4 + m 29.43 ( d - 0.2679 ) kN 3 - (1.5 m )(10.39 kN ) = 0 Using the above free-body diagrams of the gate, we have or 19.62 ( d - 2 ) + 39.24 ( d - 0.2679 ) = 15.585 58.86d = 65.3374 or d = 1.1105 m d = 1.110 m 5.86 Find d min so that the gate opens. W = 9.81 m/s 2 ( 500 kg ) = 4.905 kN ( ) Using the above free-body diagrams of the gate, we have 2 M A = 0: m 29.43 ( d - 2 ) kN 3 4 + m 29.43 ( d - 0.2679 ) kN 3 - (1.5 m )(10.39 kN ) + - ( 0.5 m )( 4.905 kN ) = 0 or or 19.62 ( d - 2 ) + 39.24 ( d - 0.2679 ) = 18.0375 d = 1.15171 m d = 1.152 m PROBLEMS 5.87 AND 5.88 Problem 5.87: The gate at the end of a 3-ft-wide fresh water channel is fabricated from three 240-lb, rectangular steel plates. The gate is hinged at A and rests against a frictionless support at D. Knowing that d = 2.5 ft, determine the reactions at A and D. Problem 5.88: The gate at the end of a 3-ft-wide fresh water channel is fabricated from three 240-lb, rectangular steel plates. The gate is hinged at A and rests against a frictionless support at D. Determine the depth of water d for which the gate will open. SOLUTION 5.87 Thus, at Note that in addition to the weights of the gate segments, the water exerts pressure on all submerged surfaces ( p = h ) . h = 0.5 ft, p0.5 = 62.4 lb/ft 3 ( 0.5 ) ft = 31.2 lb/ft 2 h = 2.5 ft, p2.5 = 6.24 lb/ft 3 ( 2.5 ) ft = 156.0 lb/ft 2 ( ) ( ) Then P = 1 1 ( 0.5 ft )( 3 ft ) 31.2 lb/ft 2 = 23.4 lb 2 ( ) P2 = ( 2 ft )( 3 ft ) 31.2 lb/ft 2 = 187.2 lb P3 = P4 = and 1 ( 2 ft )( 3 ft ) 31.2 lb/ft 2 = 93.6 lb 2 1 ( 2 ft )( 3 ft ) 156 lb/ft 2 = 468 lb 2 ( ) ( ) ( ) 1 M A = 0: - 4 D + ( 2 ft )( 240 lb ) + (1 ft )( 240 lb ) - 2 + 0.5 ft ( 23.4 lb ) - (1 ft )(187.2 lb ) 3 2 1 - ( 2 ft )( 93.6 lb ) - ( 2 ft )( 468 lb ) = 0 3 3 D = 11.325 lb D = 11.33 lb or PROBLEMS 5.87 AND 5.88 CONTINUED Fx = 0: Ax + 11.32 + 23.4 + 93.6 + 468 = 0 or Ax = -596.32 lb Fy = 0: Ay - 240 - 240 - 240 + 187.2 = 0 or Ay = 532.8 lb At h = ( d - 2 ) ft, pd - 2 = ( d - 2 ) lb/ft 2 h = d ft, pd = ( d ) lb/ft 2 Then P = 1 1 1 3 2 A1 pd - 2 = ( d - 2 ) ft ( 3 ft ) lb/ft 3 ( d - 2 ) ft = ( d - 2 ) lb 2 2 2 where A = 800 lb 41.8 5.88 = 62.4 lb/ft 3 (Note: For simplicity, the numerical value of the density will be substituted into the equilibrium equations below, rather than at this level of the calculations.) P2 = A2 pd - 2 = ( 2 ft )( 3 ft ) ( d - 2 ) ft = 6 ( d - 2 ) lb P3 = P4 = 1 1 A3 pd - 2 = ( 2 ft )( 3 ft ) ( d - 2 ) ft = 3 ( d - 2 ) lb 2 2 1 1 A4 pd = ( 2 ft )( 3 ft ) ( d ft ) = ( 3 d ) lb = 3 ( d - 2 ) + 6 lb 2 2 As the gate begins to open, D 0 M A = 0: ( 2 ft )( 240 lb ) + (1 ft )( 240 lb ) - 2 ft + ( d - 2 ) ft 1 3 3 2 ( d - 2 ) lb + 2 2 - (1 ft ) 6 ( d - 2 ) lb - ( 2 ft ) 3 ( d - 2 ) lb 3 1 - ( 2 ft ) 3 ( d - 2 ) lb + 6 lb = 0 3 or 1 720 -4 ( d - 2 )3 + 3 ( d - 2 )2 + 12 ( d - 2 ) = 2 = 720 -4 62.4 d = 2.55 ft = 7.53846 Solving numerically yields PROBLEM 5.89 A rain gutter is supported from the roof of a house by hangers that are spaced 0.6 m apart. After leaves clog the gutter's drain, the gutter slowly fills with rainwater. When the gutter is completely filled with water, determine (a) the resultant of the pressure force exerted by the water on the 0.6-m section of the curved surface of the gutter, (b) the force-couple system exerted on a hanger where it is attached to the gutter. SOLUTION (a) Consider a 0.6 m long parabolic section of water. Then P= 1 1 Ap = A ( gh ) 2 2 1 ( 0.08 m )( 0.6 m ) 103 kg/m3 9.81 m/s2 ( 0.08 m ) 2 = ( )( ) = 18.84 N Ww = gV 2 = 103 kg/m3 9.81 m/s 2 ( 0.12 m )( 0.08 m )( 0.6 m ) 3 = 37.67 N Now So that R = ( )( ) F = 0: P 2 + Ww2 , ( -R ) + P + Ww tan = Ww P =0 = 42.12 N, = 63.4 R = 42.1 N 63.4 (b) Consider the free-body diagram of a 0.6 m long section of water and gutter. Then Fx = 0: Bx = 0 Fy = 0: By - 37.67 N = 0 or By = 37.67 N M B = 0: M B + ( 0.06 - 0.048 ) m ( 37.67 N ) = 0 or M B = -0.4520 N m 37.7 N , 0.452 N m The force-couple system exerted on the hanger is then PROBLEM 5.90 The composite body shown is formed by removing a hemisphere of radius r from a cylinder of radius R and height 2R. Determine (a) the y coordinate of the centroid when r = 3R/4, (b) the ratio r/R for which y = -1.2 R. SOLUTION Note, for the axes shown V y 3 yV 1 2 ( R ) ( 2R ) = 2 R 2 -R -2 R 4 1 4 r 4 2 - r3 3 3 - r 8 r3 2 R3 - 3 r4 -2 R 4 - 8 Then 1 R4 - r 4 yV 8 Y = =- 1 V 3 R - r3 3 1- = 1 r 8 R 4 1 r 1- 3 R 1- 3 (a ) r = 3 R: y = - 4 1 3 3 4 4 1 3 1- 3 4 3 R or y = -1.118R 1- (b) y = -1.2 R : - 1.2 R = - 1 r 8 R 4 1 r 1- 3 R 3 3 R or r r - 3.2 + 1.6 = 0 R R r = 0.884 R 4 Solving numerically PROBLEM 5.91 Determine the y coordinate of the centroid of the body shown. SOLUTION First note that the values of Y will be the same for the given body and the body shown below. Then V Cone Cylinder y yV 1 - a 2h 2 12 1 2 2 a b 8 - 1 2 a h 3 1 a - b = - a 2b 2 4 2 1 - h 4 1 - b 2 12 a 2 ( 4h - 3b ) 24 a 2 2h 2 - 3b 2 ( ) Have Y V = yV Then Y a 2 ( 4h - 3b ) = - a 2 2h 2 - 3b 2 24 12 ( ) or Y = - 2h 2 - 3b 2 2 ( 4h - 3b ) PROBLEM 5.92 Determine the z coordinate of the centroid of the body shown. (Hint: Use the result of Sample Problem 5.13.) SOLUTION First note that the body can be formed by removing a "half-cylinder" from a "half-cone," as shown. Half-Cone Half-Cylinder V 1 2 a h 6 - - z a * zV 1 3 - ah 6 1 3 ab 12 - 1 3 a ( 2h - b ) 12 a 24 2 2 b=- ab 22 8 - 4 a 2a =- 3 2 3 a 2 ( 4h - 3b ) From Sample Problem 5.13 Have Then Z V = zV 1 Z a 2 ( 4h - 3b ) = - a3 ( 2h - b ) 24 12 or Z = - 2a 2h - b 4h - 3b PROBLEM 5.93 Consider the composite body shown. Determine (a) the value of x when h = L/2 , (b) the ratio h/L for which x = L . SOLUTION V Rectangular prism Pyramid Then Now Lab 1 b a h 3 2 x 1 L 2 1 L+ h 4 xV = xV 1 2 L ab 2 1 1 abh L + h 6 4 1 V = ab L + h 6 1 2 1 ab 3L + h L + h 6 4 so that X V = xV 1 1 1 X ab L + h = ab 3L2 + hL + h 2 6 6 4 or (a) X = ? when h = h 1 h2 1 h 1 X 1 + = L3 + + L 4 L2 6 L 6 (1) 1 L 2 h 1 Substituting = into Eq. (1) L 2 2 1 1 1 1 11 X 1 + = L 3 + + 6 2 6 2 4 2 or (b) h = ? when X = L L Substituting into Eq. (1) X = 57 L 104 X = 0.548L h 1 h2 1 h 1 = L3 + + L 1 + L 4 L2 6 L 6 or or 1+ 1h 1 1h 1 h2 = + + 6L 2 6 L 24 L2 h2 = 12 L2 h =2 3 L PROBLEMS 5.94 AND 5.95 Problem 5.94: For the machine element shown, determine the x coordinate of the center of gravity. Problem 5.95: For the machine element shown, determine the y coordinate of the center of gravity. SOLUTIONS First, assume that the machine element is homogeneous so that its center of gravity coincides with the centroid of the corresponding volume. I II III IV V , in 3 (4)(3.6)(0.75) = 10.8 (2.4)(2.0)(0.6) = 2.88 x , in. 2.0 3.7 4.2 1.2 y , in. 0.375 1.95 2.15 0.375 xV , in 4 21.6 10.656 1.0688 yV , in 4 4.05 5.616 0.54711 (0.45)2 (0.4) = 0.2545 -(0.5) 2 (0.75) = - 0.5890 13.3454 -0.7068 32.618 -0.22089 9.9922 5.94 Have X V = x V X 13.3454 in 3 = 32.618 in 4 ( ) or X = 2.44 in. 5.95 Have Y V = yV Y 13.3454 in 3 = 9.9922 in 4 ( ) or Y = 0.749 in. PROBLEMS 5.96 AND 5.97 Problem 5.96: For the machine element shown, locate the x coordinate of the center of gravity. Problem 5.97: For the machine element shown, locate the y coordinate of the center of gravity. SOLUTIONS First, assume that the machine element is homogeneous so that its center of gravity coincides with the centroid of the corresponding volume. V , mm3 1 2 3 4 x , mm y , mm xV , mm 4 12 441 600 5 443 200 3 534 114 -2 316 233 19 102 681 yV , mm 4 1 399 680 4 354 560 185 508 -130 288 5 809 460 (160 )( 54 )(18) = 155 520 1 (120 )( 42 )( 54 ) = 136 080 2 80 40 160 + 160 36 9 32 2 (27) 2 (18) = 6561 9 9 -(16)2 (18) = -4608 297 736 5.96 Have X V = x V X 297 736 mm3 = 19 102 681 mm 4 ( ) or X = 64.2 mm 5.97 Have Y V = yV Y 297 736 mm3 = 5 809 460 mm 4 ( ) or Y = 19.51 mm PROBLEMS 5.98 AND 5.99 Problem 5.98: For the stop bracket shown, locate the x coordinate of the center of gravity. Problem 5.99: For the stop bracket shown, locate the z coordinate of the center of gravity. SOLUTIONS First, assume that the bracket is homogeneous so that its center of gravity coincides with the centroid of the corresponding volume. Have.. Z II = 24 mm + Z III = 24 mm + X III = 68 mm + 1 (90 + 86) mm = 112 mm 2 1 (102) mm = 58 mm 3 1 ( 20) mm = 78 mm 2 2 ( 90) mm = 170 mm 3 2 (132) mm = 156 mm 3 Z IV = 110 mm + X IV = 60 mm + V , mm3 I II III IV x , mm z , mm xV , mm 4 84 480 000 84 480 000 9 865 440 -22 239 360 156 586 080 zV , mm 4 1 013 760 94 617 600 733 840 -24 235 200 8 785 584 ( 200 )(176 )( 24 ) = 844 800 ( 200 )( 24 )(176 ) = 844 800 1 ( 20 )(124 )(102 ) = 126 480 2 1 - ( 90 )(132 )( 24 ) = -142 560 2 1 673 520 100 100 78 156 12 112 58 170 5.98 Have X V = xV X 1 673 520 mm3 = 156 586 080 mm 4 ( ) or X = 93.6 mm 5.99 Have Z V = zV Z 1 673 520 mm3 = 8 785 584 mm 4 ( ) or Z = 52.5 mm PROBLEM 5.100 Locate the center of gravity of the sheet-metal form shown. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity coincides with the centroid of the corresponding area. A, mm 2 1 1 ( 90 )( 60 ) 2 = 2700 ( 90 )( 200 ) = 18 000 - ( 45 )(100 ) = -4500 x , mm y , mm z , mm xA, mm3 81 000 yA, mm3 378 000 zA, mm3 0 30 120 + 20 = 140 0 2 3 45 22.5 60 30 80 120 160 + 810 000 -101 250 1 080 000 -135 000 1 440 000 -540 000 4 Have 2 = 1012.5 19 380.9 ( 45)2 ( 4 )( 45) 3 = 179.1 45 0 143 139 932 889 0 1 323 000 569 688 1 469 688 X A = xA: X 19 380.9 mm 2 = 932 889 mm3 ( ) or = 48.1 mm X = 48.1 mm Y A = yA Y 19 380.9 mm 2 = 1 323 000 mm3 or Y = 68.3 mm Z A = zA Z 19 380.9 mm 2 = 1 469 688 mm3 or Z = 75.8 mm Z = 75.8 mm Y = 68.3 mm ( ) ( ) PROBLEM 5.101 A mounting bracket for electronic components is formed from sheet metal of uniform thickness. Locate the center of gravity of the bracket. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the bracket coincides with the centroid of the corresponding area. Then (see diagram) zV = 22.5 - 4 ( 6.25 ) 3 = 19.85 mm AV = - 2 ( 6.25)2 = -61.36 mm 2 A, mm 2 I II III IV V Have x , mm y , mm z , mm xA, mm3 18 750 18 750 12 937.5 -3750 -613.6 46 074 yA, mm3 0 -4687.5 -5625 0 0 -10 313 zA, mm3 45 000 22 500 13 500 -14 062.5 -1218.0 65 720 ( 25)( 60) = 1500 (12.5)( 60 ) = 750 ( 7.5)( 60 ) = 450 - (12.5 )( 30 ) = -375 -61.36 2263.64 12.5 25 28.75 10 10 0 -6.25 -12.5 0 0 30 30 30 37.5 19.85 X A = xA X 2263.64 mm 2 = 46 074 mm3 Y A = yA Y 2263.64 mm 2 = -10 313 mm3 Z A = zA ( ) or X = 20.4 mm ( ) or Y = -4.55 mm Z 2263.64 mm 2 = 65 720 mm3 ( ) or Z = 29.0 mm PROBLEM 5.102 Locate the center of gravity of the sheet-metal form shown. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the form coincides with the centroid of the corresponding area. yI = 6 + zI = 4 = 7.333 in. 3 1 ( 6) = 2 in. 3 1 xII = yII = xIV = 11 - ( 2)( 6) = 3.8197 in. 1 ( 4)(1.5) = 10.363 in. 3 I II III IV A, in 2 12 56.55 30 -3.534 95.01 x , in. 0 3.8197 8.5 10.363 y , in. 7.333 3.8197 0 0 z , in. 2 3 3 3 x A, in 3 0 216 255 -36.62 434.4 y A, in 3 88 216 0 0 304 z A, in 3 24 169.65 90 -10.603 273.0 Have X A = x A X 95.01 in 2 = 434 in 3 Y A = y A ( ) or X = 4.57 in. Y 95.01 in 2 = 304.0 in 3 Z A = z A Z 95.01 in 2 = 273.0 in 3 ( ) or Y = 3.20 in. ( ) or Z = 2.87 in. PROBLEM 5.103 An enclosure for an electronic device is formed from sheet metal of uniform thickness. Locate the center of gravity of the enclosure. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the form coincides with the centroid of the corresponding area. Consider the division of the back, sides, and top into eight segments according to the sketch. Note that symmetry implies and A8 = A2 A7 = A 3 A6 = A5 Z = 6.00 in. Thus A, in 2 1 2 3 x , in. 0 5.6 y , in. 4.5 4.5 xA, in 3 0 569.5 yA, in 3 486 457.6 (12 )( 9 ) = 108 (11.3)( 9 ) = 101.7 1 (11.3)( 2.4 ) = 13.56 2 (12 )(11.454 ) = 137.45 1 (1.5)(11.454 ) = 8.591 2 8.591 13.56 101.7 493.2 7.467 5.6 7.467 7.467 7.467 5.6 9.8 10.2 10.6 10.6 9.8 4.5 101.25 769.72 64.15 64.15 101.25 569.5 2239.5 132.89 1402.0 91.06 91.06 132.89 457.6 3251.1 or X = 4.54 in. or Y = 6.59 in. 4 5 6 7 8 Have X A = xA: X 493.2 in 2 = 2239.5 in 3 2 3 ( ) Y A = yA: Y ( 493.2 in ) = 3251.1 in PROBLEM 5.104 A 200-mm-diameter cylindrical duct and a 100 200-mm rectangular duct are to be joined as indicated. Knowing that the ducts are fabricated from the same sheet metal, which is of uniform thickness, locate the center of gravity of the assembly. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the centroid of the corresponding area. Also note that symmetry implies Z =0 A, m 2 x, m y, m xA, m3 yA, m3 1 2 ( 0.2 )( 0.3) = 0.1885 - 0 2 ( 0.1) 0.15 0.25 0 -0.02000 0.028274 -0.007854 2 ( 0.2 )( 0.1) = -0.0314 ( 0.1) = 0.01571 2 = 0.06366 3 4 5 6 7 8 2 -4 ( 0.1) = -0.04244 3 0.15 0.15 4 ( 0.1) 3 0.15 0.15 0.30 0.30 0.20 0.20 0.25 0.25 -0.000667 0.00900 0.00900 -0.000667 0.004500 0.004500 0.023667 0.004712 0.001800 0.001200 -0.003142 0.007500 0.007500 0.066991 ( 0.3)( 0.2 ) = 0.060 ( 0.3)( 0.2 ) = 0.060 - 2 ( 0.1)2 = -0.1571 ( 0.3)( 0.1) = 0.030 ( 0.3)( 0.1) = 0.030 0.337080 Have or X A = xA: X (0.337080 mm 2 ) = 0.023667 mm3 X = 0.0702 m X = 70.2 mm Y A = yA: Y 0.337080 mm 2 = 0.066991 mm3 ( ) or Y = 0.19874 m Y = 198.7 mm PROBLEM 5.105 An elbow for the duct of a ventilating system is made of sheet metal of uniform thickness. Locate the center of gravity of the elbow. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the duct coincides with the centroid of the corresponding area. Also, note that the shape of the duct implies Y = 1.5 in. Note that xI = zI = 16 in. - xII = 16 in. - zII = 12 in. - 2 2 (16 in.) = 5.81408 in. 2 (8 in.) = 10.9070 in. (8 in.) = 6.9070 in. 4 (16 in.) = 9.2094 in. 3 xIV = zIV = 16 in. - xV = 16 in. - 4 (8 in.) = 12.6047 in. 3 4 zV = 12 in. - (8 in.) = 8.6047 in. 3 Also note that the corresponding top and bottom areas will contribute equally when determining x and z . Thus I II III IV V VI A, in 2 x , in. 5.81408 10.9070 8 9.2094 12.6047 12 z , in. 5.81408 6.9070 14 9.2094 8.6047 14 xA, in 3 438.37 411.18 96.0 3703.32 -1267.16 -768.0 2613.71 zA, in 3 438.37 260.39 168.0 3703.32 -865.04 -896.0 2809.04 2 (16 )( 3) = 75.3982 (8)( 3) = 37.6991 4 ( 3) = 12 2 2 2 (16) = 402.1239 4 2 -2 ( 8 ) = -100.5309 4 -2 ( 4 )( 8 ) = -64 362.69 Have X A = xA: X 362.69 in 2 = 2613.71 in 3 or X = 7.21 in. ( ) Z A = zA: Z 362.69 in 2 = 2809.04 in 3 or Z = 7.74 in. ( ) PROBLEM 5.106 A window awning is fabricated from sheet metal of uniform thickness. Locate the center of gravity of the awning. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the awning coincides with the centroid of the corresponding area. yII = yVI = 80 + zII = zVI = yIV = 80 + zIV = ( 4 )( 500 ) 3 = 292.2 mm ( 4 )( 500 ) 3 ( 2 )( 500 ) = 212.2 mm = 398.3 mm ( 2 )( 500 ) 4 = 318.3 mm = 196 350 mm 2 AII = AVI = AIV = ( 500 )2 2 ( 500 )( 680 ) = 534 071 mm2 yA, mm3 zA, mm3 I A, mm 2 (80)(500) = 40 000 y , mm z , mm 40 250 1.6 106 57.4 106 0.2176 106 212.7 106 1.6 106 57.4 106 332.9 106 10 106 41.67 106 27.2 106 170 106 10 106 41.67 106 300.5 106 X = 340 mm II III IV V VI 196 350 (80)(680) = 54 400 534 071 (80)(500) = 40 000 196 350 1.061 106 292.2 40 398.3 40 292.2 212.2 500 318.3 250 212.2 Now, symmetry implies and Y A = yA: Y 1.061 10 mm ( 6 2 ) = 332.9 10 ) 6 mm 3 or Y = 314 mm Z A = zA: Z 1.061 106 mm 2 = 300.5 106 mm3 or Z = 283 mm ( PROBLEM 5.107 The thin, plastic front cover of a wall clock is of uniform thickness. Locate the center of gravity of the cover. SOLUTION First, assume that the plastic is homogeneous so that the center of gravity of the cover coincides with the centroid of the corresponding area. Next, note that symmetry implies X = 150.0 mm A, mm 2 y , mm z , mm y A, mm3 0 350 000 375 000 350 000 0 z , mm3 11 760 000 1 960 000 0 1 960 000 1 2 3 4 5 ( 300 )( 280 ) = 84 000 0 25 25 25 0 140 140 0 140 130 ( 280 )( 50 ) = 14 000 ( 300 )( 50 ) = 15 000 ( 280 )( 50 ) = 14 000 - (100 ) 2 = -31 416 -4 084 070 -174 783 6 7 - ( 30 )2 4 = -706.86 - ( 30 )2 4 = -706.86 94 170 0 260 - ( 4 )( 30 ) 3 = 247.29 0 0 260 - ( 4 )( 30 ) 3 = 247.29 0 1 075 000 -174 783 11 246 363 Have Y A = yA: Y 94 170 mm 2 = 1 075 000 mm3 or Y = 11.42 mm ( ) Z A = zA: Z 94 170 mm 2 = 11 246 363 mm3 or Z = 119.4 mm ( ) PROBLEM 5.108 A thin steel wire of uniform cross section is bent into the shape shown, where arc BC is a quarter circle of radius R. Locate its center of gravity. SOLUTION First, assume that the wire is homogeneous so that its center of gravity coincides with the centroid of the corresponding line. L, in 2 x , in. y , in. z ,in. x L, in 2 y L, in 2 z L, in 2 1 2 3 15 14 13 0 7 5 9 12 = 11.5 3 2 15 5 = 5.73 7.5 0 0 30 24 0 0 6 0 98 149.5 112.5 0 0 0 0 78 4 (15) 2 = 23.56 65.56 = 9.549 = 7.639 135.0 382.5 225.0 337.5 180.0 258.0 or X = 5.83 in. or Y = 5.15 in. or Z = 3.94 in. Have X L = x L: X ( 65.56 in.) = 382.5 in 2 Y L = y L: Y ( 65.56 in.) = 337.5 in 2 Z L = z L: Z ( 65.56 in.) = 258.0 in 2 PROBLEM 5.109 A thin steel wire of uniform cross section is bent into the shape shown, where arc BC is a quarter circle of radius R. Locate its center of gravity. SOLUTION First, assume that the wire is homogeneous so that its center of gravity coincides with the centroid of the corresponding line Have x2 = z2 = L1 = ( 2 )(8) = 16 ft 82 + 32 = 8.5440 ft 8 L 2 = 2 = 4 ft 4 L, ft 1 2 3 4 8.5440 4 8 3 32.110 x , ft 4 16 0 0 y , ft 1.5 0 0 1.5 z , ft 0 16 x L, ft 2 34.176 64.0 0 0 98.176 y L, ft 2 12.816 0 0 4.5 17.316 z L, ft 2 0 64.0 32 0 96.0 4 0 Have X L = x L: X ( 32.110 ft ) = 98.176 ft 2 Y L = y L: Y ( 32.110 ft ) = 17.316 ft 2 Z L = z L: Z ( 32.110 ft ) = 96.0 ft 2 or X = 3.06 ft or Y = 0.539 ft or Z = 2.99 ft PROBLEM 5.110 The frame of a greenhouse is constructed from uniform aluminum channels. Locate the center of gravity of the portion of the frame shown. SOLUTION First, assume that the channels are homogeneous so that the center of gravity of the frame coincides with the centroid of the corresponding line. Note x8 = x9 = ( 2 )( 0.9 ) = 0.57296 m y8 = y9 = 1.5 + L 7 = L8 = ( 2 )( 0.9 ) = 2.073 m 2 ( 0.9 ) = 1.4137 m L, m x, m y, m z, m 1 2 3 4 5 6 7 8 9 10 Have 0.6 0.9 1.5 1.5 2.4 0.6 0.9 1.4137 1.4137 0.6 11.827 0.9 0.45 0.9 0.9 0 0.9 0.45 0.573 0.573 0 0 0 0.75 0.75 1.2 1.5 1.5 2.073 2.073 2.4 0.3 0.6 0 0.6 0.6 0.3 0.6 0 0.6 0.3 xL, m 2 0.540 0.4050 1.350 1.350 0 0.540 0.4050 0.8100 0.8100 0 6.210 yL, m 2 0 0 1.125 1.125 2.880 0.9 1.350 2.9306 2.9306 1.440 14.681 zL, m 2 0.18 0.54 0 0.9 1.44 0.18 0.54 0 0.8482 0.18 4.8082 or X = 0.525 m or Y = 1.241 m or Z = 0.406 m X L = xL: X (11.827 m ) = 6.210 m 2 Y L = yL: Y (11.827 m ) = 14.681 m 2 Z L = zL: Z (11.827 m ) = 4.8082 m 2 * PROBLEM 5.111 The decorative metalwork at the entrance of a store is fabricated from uniform steel structural tubing. Knowing that R = 1.2 m, locate the center of gravity of the metalwork. SOLUTION First, assume that the tubes are homogeneous so that the center of gravity of the metalwork coincides with the centroid of the corresponding line. Note that symmetry implies Z =0 L, m 1 2 3 4 3 3 x, m y, m xL, m 2 2.5456 2.5456 0 yL, m 2 4.5 4.5 14.1897 (1.2) cos 45 = 0.8485 (1.2) cos 45 = 0.8485 ( 2 )(1.2 ) ( 2 )(1.2 ) 0 = 0.7639 = 0.7639 1.5 1.5 3.7639 1.2 1.2 0.6 15.425 3 3.7639 2.88 1.44 9.4112 11.3097 7.0949 41.594 or X = 0.610 m or Y = 2.70 m 5 Have X L = xL: X (15.425 m ) = 9.4112 m 2 Y L = yL: Y (15.425 m ) = 41.594 m 2 PROBLEM 5.112 A scratch awl has a plastic handle and a steel blade and shank. Knowing that the specific weight of plastic is 0.0374 lb/in 3 and of steel is 0.284 lb/in 3, locate the center of gravity of the awl. SOLUTION First, note that symmetry implies Y =Z =0 xI = 5 2 ( 0.5 in.) = 0.3125 in., WI = 0.0374 lb/in 3 ( 0.5 in.)3 = 0.009791 lb 8 3 ( ) xII = 1.6 in. + 0.5 in. = 2.1 in. WII = 0.0374 lb/in 3 ( )( 0.5 in.) ( 3.2 in.) = 0.093996 lb 2 2 xIII = 3.7 in. - 1 in. = 2.7 in., WIII = - 0.0374 lb/in 3 ( 0.12 in.) ( 2 in.) = -0.000846 lb 4 ( ) ( ) 2 2 xIV = 7.3 in. - 2.8 in. = 4.5 in., WIV = 0.284 lb/in 3 ( 0.12 in.) ( 5.6 in.) = 0.017987 lb 4 ( ) xV = 7.3 in. + 1 ( 0.4 in.) = 7.4 in., WV = 0.284 lb/in 3 ( 0.06 in.)2 ( 0.4 in.) = 0.000428 lb 4 3 ( ) I II III IV V W , lb 0.009791 0.093996 -0.000846 0.017987 0.000428 0.12136 x , in. 0.3125 2.1 2.7 4.5 7.4 xW , in lb 0.003060 0.197393 -0.002284 0.080942 0.003169 0.28228 Have X W = xW : X ( 0.12136 lb ) = 0.28228 in. lb or X = 2.33 in. PROBLEM 5.113 A bronze bushing is mounted inside a steel sleeve. Knowing that the density of bronze is 8800 kg/m3 and of steel is 7860 kg/m3 , determine the center of gravity of the assembly. SOLUTION First, note that symmetry implies X =Z =0 Now W = ( g )V yI = 4 mm , WI = 7860 kg/m3 9.81 m/s2 0.0362 - 0.0152 m 2 ( 0.008 m ) 4 ( )( ) ( ) = 0.51887 N yII = 18 mm, WII = 7860 kg/m3 9.81 m/s 2 0.02252 - 0.052 m 2 ( 0.02 m ) 4 ( )( ) ( ) = 0.34065 N yIII = 14 mm, WIII = 8800 kg/m3 9.81 m/s2 0.152 - 0.102 m 2 ( 0.028 m ) 4 ( )( ) ( ) = 0.23731 N Have Y = Y W = yW ( 4 mm)( 0.5189 N ) + (18 mm)( 0.3406 N ) + (14 mm )( 0.2373 N ) 0.5189 N + 0.3406 N + 0.2373 N or Y = 10.51 mm ( above base ) PROBLEM 5.114 A marker for a garden path consists of a truncated regular pyramid carved from stone of specific weight 160 lb/ft3. The pyramid is mounted on a steel base of thickness h. Knowing that the specific weight of steel is 490 lb/ft3 and that steel plate is available in 1 in. increments, specify the 4 minimum thickness h for which the center of gravity of the marker is approximately 12 in. above the top of the base. SOLUTION First, locate the center of gravity of the stone. Assume that the stone is homogeneous so that the center of gravity coincides with the centroid of the corresponding volume. Have y1 = 3 ( 56 in.) = 42 in., 4 3 ( 28 in.) = 21 in., 4 V1 = 1 (12 in.)(12 in.)( 56 in.) 3 1 ( 6 in.)( 6 in.)( 28 in.) 3 = 2688 in 3 y2 = V2 = - = -366 in 3 Then Vstone = 2688 in 3 - 366 in 3 = 2352 in 3 and Y = = ( 42 in.) ( 2688 in 3 ) + ( 21 in.) ( -366 in 3 ) 2352 in 3 yV V = 45 in. Therefore, the center of gravity of the stone is ( 45 - 28 ) in. = 17 in. above the base. Now Wstone = stoneVstone = 160 lb/ft = 217.78 lb Wsteel = steelVsteel 1 ft = 490 lb/ft (12 in.)(12 in.) h 12 in. 3 ( 3 )( 1 ft 2352 in 12 in. 3 ) 3 ( 3 ) = ( 40.833h ) 1b PROBLEM 5.114 CONTINUED Then Ymarker = yW = 12 in. W = or (17 in.)( 217.78 lb ) + - h in. ( 40.833 h ) lb 2 ( 217.78 + 40.833h ) lb h 2 + 24h - 53.334 = 0 With positive solution h = 2.0476 in. specify h = 2 in. PROBLEM 5.115 The ends of the park bench shown are made of concrete, while the seat and back are wooden boards. Each piece of wood is 36 120 1180 mm. Knowing that the density of concrete is 2320 kg/m3 and of wood is 470 kg/m3 , determine the x and y coordinates of the center of gravity of the bench. SOLUTION First, note that we will account for the two concrete ends by counting twice the weights of components 1, 2, and 3. W1 = ( c g )V1 = 2320 kg/m3 9.81 m/s 2 ( 0.480 m )( 0.408 m )( 0.072 m ) = 320.9 N W2 = - ( c g )V2 = - 2320 kg/m3 9.81 m/s 2 ( 0.096 m )( 0.048 m )( 0.072 m ) = -7.551 N W3 = ( c g )V3 = 2320 kg/m3 9.81 m/s 2 ( 0.096 m )( 0.384 m )( 0.072 m ) = 60.41 N W4 = W5 = W6 = W7 = wVboard = 470 kg/m3 9.81 m/s 2 ( 0.120 m )( 0.036 m )(1.180 m ) = 23.504 N ( )( ) ( )( ) ( )( ) ( )( ) PROBLEM 5.115 CONTINUED W, N 1 2 3 4 5 6 7 8 Have 2 ( 320.4 ) = 641.83 2 ( -7.551) = -15.10 2 ( 60.41) = 120.82 23.504 23.504 23.504 23.504 23.504 865.06 x , mm y , mm x W, mm N 200 251.4 -4711.8 10 148.5 5358.8 8461.3 10 388.5 2930.9 3762.9 236 590 y W, mm N -130 933.6 5799.1 23 196.5 423.1 423.1 423.1 7716.2 3281.1 -89 671 312 312 84 228 360 442 124.7 160.1 -204 -384 192 18 18 18 328.3 139.6 X W = xW : X ( 865.06 N ) = 236 590 mm N or X = 274 mm Y W = yW : Y ( 865.06 N ) = -89 671 mm N or Y = -103.6 mm PROBLEM 5.116 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Fig. 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A hemisphere. SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then dV = r 2dx, xEL = x The equation of the generating curve is x 2 + y 2 = a 2 so that r 2 = a 2 - x 2 and then dV = a 2 - x 2 dx Component 1 V1 = = a/2 0 a/2 ( ) ( x3 a - x dx = a 2 x - 3 0 2 2 ) 11 3 a 24 and a/2 2 2 1 xEL dV = 0 x ( a - x ) dx x2 x4 = a 2 - 4 0 2 = Now 7 a4 64 a/2 7 11 x1V1 = 1 xEL dV : x1 a3 = a4 24 64 or x1 = Component 2 V2 = a a /2 a 21 a 88 ( x3 a - x dx = a 2 x - 3 a/2 2 2 ) a3 a = a 2 ( a ) - - a2 - 3 2 5 = a 3 24 (a) 2 3 3 PROBLEM 5.116 CONTINUED and 2 x2 x4 a 2 2 2 xELdV = a/2 x a - x dx = a 2 - 4 a/2 ( ) a 2 2 ( a )4 - a 2 a - 2 (a) 2 a = - 2 4 2 9 = a4 64 ( ) (a) 2 4 4 Now 9 5 x2V2 = 2 xELdV : x2 a3 = a4 24 64 or x2 = 27 a 40 PROBLEM 5.117 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Fig. 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A semiellipsoid of revolution. SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then dV = r 2dx, xEL = x The equation of the generating curve is r2 = a2 2 h - x 2 and then h2 x2 y2 + 2 = 1 so that h2 a ( ) dV = Component 1 V1 = = h/2 0 a2 2 h - x 2 dx h2 ( ) a2 2 a2 x3 h - x 2 dx = 2 h 2 x - 3 0 h2 h ( ) h/2 11 2 a h 24 and h/2 2 2 1 xEL dV = 0 x h 2 ( h - x ) dx a2 a2 = 2 h = 2 x2 x4 - h 4 0 2 h/2 Now 7 a 2h 2 64 7 11 x1V1 = 1 xEL dV : x1 a 2h = a 2h 2 24 64 or x1 = 21 h 88 PROBLEM 5.117 CONTINUED Component 2 V2 = h/2 h a2 2 a2 x3 h - x 2 dx = 2 h 2 x - 2 3 h/2 h h ( ) h 3 a 2 2 ( h )3 - h 2 h - h 2 = 2 h ( h ) - 3 2 3 h 5 = a 2 h 24 () and h 2 2 2 xELdV = h/2 x h2 ( h - x ) dx a2 = a2 h2 2 x2 x4 - h 4 h/2 2 h 2 2 ( h )4 - h 2 h - a2 2 ( h ) 2 = 2 h - 2 4 2 h 9 = a 2h 2 64 ( ) (h) 2 4 4 Now 9 5 x2V2 = 2 xEL dV : x2 a 2h = a 2h 2 24 64 or x2 = 27 h 40 PROBLEM 5.118 Determine by direct integration the values of x for the two volumes obtained by passing a vertical cutting plane through the given shape of Fig. 5.21. The cutting plane is parallel to the base of the given shape and divides the shape into two volumes of equal height. A paraboloid of revolution. SOLUTION Choose as the element of volume a disk of radius r and thickness dx. Then dV = r 2dx, xEL = x The equation of the generating curve is x = h - r2 = a2 ( h - x ) and then h dV = Component 1 V1 = 0 = = and h/2 h 2 y so that a2 a2 ( h - x ) dx h a2 ( h - x ) dx h x2 hx - 2 0 h/2 a2 h 3 2 a h 8 2 h/2 a xEL dV = 0 x ( h - x ) dx 1 h a 2 x 2 x3 = - h 3 0 h 2 = 1 a 2h 2 12 h/2 Now 1 3 x1V1 = 1 xEL dV : x1 a 2h = a 2h 2 8 12 or x1 = 2 h 9 PROBLEM 5.118 CONTINUED Component 2 V2 = h/2 a2 = h = h a2 a2 x2 ( h - x ) dx = hx - h h 2 h/2 ( h )2 - h h - h ( h ) - 2 2 h (h) 2 2 2 1 2 a h 8 h and a2 a 2 x 2 x3 h 2 xELdV = h/2 x h ( h - x ) dx = h h 2 - 3 h/2 2 2 ( h )3 - h h - a2 ( h ) h 2 = - 3 2 h 2 1 = a 2h 2 12 ( ) (h) 2 3 3 Now 1 1 x2V2 = 2 xEL dV : x2 a 2h = a 2h2 8 12 or x2 = 2 h 3 PROBLEM 5.119 Locate the centroid of the volume obtained by rotating the shaded area about the x axis. SOLUTION First note that symmetry implies y =0 z =0 Choose as the element of volume a disk of radius r and thickness dx. Then dV = r 2dx, xEL = x x2 Now r = b1 - 2 so that a x2 dV = b 1 - 2 dx a 2 2 Then x2 2x2 x4 a a V = 0 b 2 1 - 2 dx = 0 b 2 1 - 2 + 4 dx a a a 2 x5 2 x3 = b x - 2 + 4 3a 5a 0 2 1 = ab 2 1 - + 3 5 8 = ab 2 15 and 2 a 2x2 x4 a xEL dV = 0 b 2 x 1 - 2 + 4 dx a a x2 2x4 x6 = b - 2 + 4 2 4a 6a 2 a 0 1 1 1 = a 2b 2 - + 2 2 6 = 1 2 2 a b 6 PROBLEM 5.119 CONTINUED Then 1 8 xV = xEL dV : x ab 2 = a 2b 2 15 16 or x = 15 a 6 PROBLEM 5.120 Locate the centroid of the volume obtained by rotating the shaded area about the x axis. SOLUTION First, note that symmetry implies y =0 z =0 Choose as the element of volume a disk of radius r and thickness dx. Then dV = r 2dx, xEL = x Now r = 1 - 1 so that x 1 dV = 1 - dx x 2 1 = 1 - + 2 dx x x Then 2 1 1 3 V = 1 1 - + 2 dx = x - 2 ln x - x x x 1 1 1 = 3 - 2 ln3 - - 1 - 2 ln 1 - 3 1 = ( 0.46944 ) m3 3 2 and x ELdV = 3 x 1 x2 2 1 1 - + 2 dx = - 2 x + ln x x x 2 1 3 32 13 = - 2 ( 3) + ln 3 - - 2 (1) + ln1 2 2 = (1.09861 ) m Now xV = x EL dV : X 0.46944 m3 = 1.09861 m 4 or x = 2.34 m ( ) PROBLEM 5.121 Locate the centroid of the volume obtained by rotating the shaded area about the line x = h. SOLUTION First, note that symmetry implies x =h z =0 Choose as the element of volume a disk of radius r and thickness dx. Then dV = r 2dy, yEL = y Now x 2 = Then and Let Then h2 2 h 2 a - y2 a - y 2 so that r = h - 2 a a 2 h2 dV = 2 a - a 2 - y 2 dy a ( ) ( ) V = 0 a h2 a - a2 - y2 2 a ( ) dy 2 y = a sin dy = a cos d V = = h 2 /2 2 2 2 a - a - a sin a2 0 ( ) a cos d 2 h 2 /2 2 2 2 2 a - 2a ( a cos ) + a - a sin a cos d a2 0 ( ) = ah 2 0 /2 ( 2 cos - 2 cos - sin cos ) d 2 2 sin 2 1 3 = ah 2 2sin - 2 + - sin 4 3 2 0 = ah 2 2 - 1 2 2 - 2 3 /2 = 0.095870 ah 2 PROBLEM 5.121 CONTINUED and a 2 2 y ELdV = 0 y a 2 a - a - y dy h2 ( ) 2 = = = = Now h2 a 2a 2 y - 2ay a 2 - y 2 - y 3 dy 2 0 a h2 a2 2 2 2 2 2 a y + 3 a a - y ( ) ( ) 3/2 - 1 4 y 4 0 a 1 2 h2 2 2 a ( a ) - a4 - a a2 2 4 3 a ( ) 3/2 1 a 2h 2 12 yV = y EL dV : y 0.095870 ah 2 = ( ) 1 a 2h 2 12 or y = 0.869a PROBLEM 5.122 Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the x axis. SOLUTION First, note that symmetry implies y =0 z =0 Choose as the element of volume a disk of radius r and thickness dx. Then dV = r 2dx, xEL = x Now so that Then r = b sin x 2a dV = b 2 sin 2 2a x 2a dx V = a b 2 sin 2 2 x 2a dx 2a x x sin a 2 = b - 2 2 a a = b 2 22a - 1 = ab 2 2 ( ) ( a ) 2 x and 2a 2 2 x ELdV = a x b sin 2a dx Use integration by parts with u = x du = dx dV = sin 2 V = x - 2 2 a x 2a a sin x PROBLEM 5.122 CONTINUED Then x sin x x EL dV = b x - 2 a 2 a 2 sin x 2a x - a - 2 a 2 a a 2a dx 2a x a2 2a a 1 = b 2 2a - a - x 2 + cos 2 2 2 4 a a 2 1 a2 a2 2 2 3 1 = b 2 a 2 - ( 2a ) + - (a) + 2 2 4 2 2 2 4 1 3 = a 2b 2 - 2 4 Now = 0.64868 a 2b 2 1 xV = x EL dV : x ab 2 = 0.64868 a 2b 2 2 or x = 1.297a PROBLEM 5.123 Locate the centroid of the volume generated by revolving the portion of the sine curve shown about the y axis. (Hint: Use a thin cylindrical shell of radius r and thickness dr as the element of volume.) SOLUTION First note that symmetry implies Choose as the element of volume a cylindrical shell of radius r and thickness dr. 1 dV = ( 2r )( y )( dr ) , yEL = y Then 2 Now so that Then y = b sin x =0 z =0 r 2a dV = 2br sin V = a 2br sin 2a r 2a dr r 2a dr Use integration by parts with u =r du = dr dv = sin v=- 2a r 2a dr cos r 2a Then 2a 2a r r 2a 2a V = 2b ( r ) - cos - a cos dr 2a a 2a 2a 2a 4a 2 r ( 2a ) ( -1) + 2 sin = 2b - 2a a 2 2 4a 4a - 2 V = 2b 1 = 8a 2b 1 - = 5.4535 a 2b Also 2a y ELdV = a 1 b sin 2a 2br sin 2a dr 2 r r = b 2 a r sin 2 2a r dr 2a PROBLEM 5.123 CONTINUED Use integration by parts with u =r du = dr dv = sin 2 v= r - 2 dr 2a sin ar 2 a r Then r sin ar y EL dV = b ( r ) 2 - 2 a 2 sin r 2a r - a - 2 a 2 a a 2a dr 2a a2 r 2a a r 2 cos = b 2 ( 2a ) - ( a ) - + 2 2 4 a a 2 2 3 ( 2a ) 2 ( a )2 + a 2 a2 = b2 a 2 - + - 2 4 2 2 2 2 4 1 3 = a 2b 2 - 2 4 = 2.0379a 2b 2 Now yV = y EL dV : y 5.4535a 2b = 2.0379a 2b 2 or y = 0.374b ( ) PROBLEM 5.124 Show that for a regular pyramid of height h and n sides ( n = 3, 4, ... ) the centroid of the volume of the pyramid is located at a distance h / 4 above the base. SOLUTION Choose as the element of a horizontal slice of thickness dy. For any number N of sides, the area of the base of the pyramid is given by Abase = kb 2 where k = k ( N ) ; see note below. Using similar triangles, have s h-y = b h b s = (h - y) h b2 2 dV = Aslicedy = ks 2dy = k 2 ( h - y ) dy h V = 0 k = h or Then and b2 b2 ( h - y )2 dy = k 2 h2 h 3 1 - 3 ( h - y ) 0 h 1 2 kb h 3 Also yEL = y 2 b2 h h b 2 so then y EL dV = 0 y k 2 ( h - y ) dy = k 2 0 h 2 y - 2hy 2 + y 3 dy h h ( ) =k 1 2 2 b2 1 2 2 2 3 1 4 h y - hy + y = kb h 2 3 4 0 12 h 2 h Now 1 2 2 1 yV = y EL dV : y kb 2h = kb h 3 12 or y = 1 h Q.E.D. 4 Note 1 Abase = N b 2 = N b2 4 tan N b 2 tan N = k ( N ) b2 PROBLEM 5.125 Determine by direct integration the location of the centroid of one-half of a thin, uniform hemispherical shell of radius R. SOLUTION First note that symmetry implies x =0 The element of area dA of the shell shown is obtained by cutting the shell with two planes parallel to the xy plane. Now dA = ( r )( Rd ) yEL = - 2r where so that r = R sin dA = R 2 sin d yEL = - 2R sin Then 2 A = 02 R 2 sin d = R 2 [ - cos ]0 = R2 and 2 y ELdA = 02 - sin ( R sin d ) 2R sin 2 2 = -2 R - 4 0 2 =- 3 2 R3 Now yA = y EL dA: y R 2 = - ( ) 2 R3 1 or y = - R 2 Symmetry implies z = y z =- 1 R 2 PROBLEM 5.126 The sides and the base of a punch bowl are of uniform thickness t. If t << R and R = 350 mm, determine the location of the center of gravity of (a) the bowl, (b) the punch. SOLUTION (a) Bowl First note that symmetry implies x =0 z =0 for the coordinate axes shown below. Now assume that the bowl may be treated as a shell; the center of gravity of the bowl will coincide with the centroid of the shell. For the walls of the bowl, an element of area is obtained by rotating the arc ds about the y axis. Then dAwall = ( 2 R sin )( Rd ) and ( yEL ) wall = - R cos Then Awall = /6 2 R 2 sin d = 2 R 2 [ - cos ] /6 = 3R 2 ywall Awall = ( yEL )wall dA /2 /2 /2 and = /6 ( - R cos ) 2 R 2 sin d = R3 cos 2 3 = - R3 4 By observation Now or /2 /6 ( ) Abase = 4 R2, ybase = - 3 R 2 y A = yA 3 3 y 3R 2 + R 2 = - R3 + R 2 - R 4 4 4 2 y = -0.48763R R = 350 mm y = -170.7 mm or (b) Punch First note that symmetry implies x =0 z =0 and that because the punch is homogeneous, its center of gravity will coincide with the centroid of the corresponding volume. Choose as the element of volume a disk of radius x and thickness dy. Then dV = x 2dy, yEL = y PROBLEM 5.126 CONTINUED Now Then x2 + y 2 = R2 V = so that 0 - 3/2 R dV = R 2 - y 2 dy ( 1 R 2 - y 2 dy = R 2 y - y 3 3 - 3 ) ( ) 0 3/2 R 3 1 3 3 = - R 2 - = 3R3 R - - R 2 3 2 8 and yEL dV = - 0 1 1 y R 2 - y 2 dy = R 2 y 2 - y 4 3/2 R ( ) 2 4 - ( ) 0 3/2 R 2 4 1 R 2 - 3 R - 1 - 3 R = - 15 R 4 = - 2 2 4 2 64 Now or 15 3 yV = yEL dV : y 3 R3 = - R 4 8 64 y =- 5 8 3 R R = 350 mm y = -126.3 mm PROBLEM 5.127 After grading a lot, a builder places four stakes to designate the corners of the slab for a house. To provide a firm, level base for the slab, the builder places a minimum of 60 mm of gravel beneath the slab. Determine the volume of gravel needed and the x coordinate of the centroid of the volume of the gravel. (Hint: The bottom of the gravel is an oblique plane, which can be represented by the equation y = a + bx + cz.) SOLUTION First, determine the constants a, b, and c. At x = 0, z = 0 : y = -60 mm -60 mm = a; a = -60 mm x = 7200 mm, z = 0 : y = -100 mm At - 100 mm = -60 + b ( 7200) 1 180 x = 0 , z = 12 000 mm: y = -120 mm b=- - 120 mm = -60 mm + c (12000 ) c=- 1 200 At It follows that y = -60 - 1 1 x- z where all dimensions 180 120 are in mm. Choose as the element of volume a filament of base dx dz and height | y |. Then dV = y dxdz, xEL = x or dV = -60 - 1 1 x- z dxdz 180 200 Then V = 0 12000 7200 60 0 + 1 1 x+ z dxdz 180 200 7200 = 0 = 12000 1 2 1 60 x + 360 x + 200 xz 0 dz 12000 0 ( 60 )( 7200 ) ( 7200 )2 + 360 12000 + ( 7200 ) z dz 200 36 2 z = 576000 z + 2 0 = 9.504 109 mm3 = 9.50 m3 V = 9.50 m3 PROBLEM 5.127 CONTINUED and 12000 7200 x ELdV = 0 0 x 60 + 180 x + 200 z dxdz 1 1 = 0 = 0 12000 60 12000 ( 2246.4 10 2 x2 + 1 3 1 2 x + x z 540 400 0 6 7200 dz + 129 600 z dz 12000 ) 129 600 2 z = 2246.4 106 z - 2 0 = 2.695 1013 + 0.933 1013 = 3.63 1013 mm 4 = 36.3 m 4 Now xV = x EL dV : x 9.50 m3 = 36.3 m 4 ( ) or x = 3.82 m PROBLEM 5.128 Determine by direct integration the location of the centroid of the volume between the xz plane and the portion shown of the surface 16h ax - x 2 bz - z 2 y = . a 2b 2 ( )( ) SOLUTION First note that symmetry implies x = a 2 b z = 2 Choose as the element of volume a filament of base dx dz and height y. Then dV = ydxdz, yEL = or Then V = = and 1 y 2 dV = 16h ax - x 2 bz - z 2 dxdz a 2b 2 b a ( )( ) V = 0 0 16h ax - x 2 bz - z 2 dxdz 2 2 ab ( )( ) 16h b 1 a bz - z 2 x 2 - x3 dz 2 2 0 3 0 ab z 16h a 2 1 3 b 1 8ah b 2 1 3 4 a - ( a ) z 2 - z 3 = 2 ( b ) - ( b ) = abh 2 2 3 3 0 3b 2 3 a b 2 2 9 1 16h 2 ( ) a ( ) b b a 2 2 2 2 yELdV = 0 0 2 a 2b2 ( ax - x )( bz - z ) a 2b2 ( ax - x )( bz - z ) dxdz 16h = = 128h b a 2 2 a x - 2ax3 + x 4 b 2 z 2 - 2bz 3 + z 4 dxdz 4 4 0 0 ab 2 128h 2 b 2 2 a 4 1 5 3 4 a 3 0 b z - 2bz + z 3 x - 2 x + 5 x dz a 2b 4 0 ( )( ) ( ) a = = 128h 2 a 4b 4 a2 a 1 5 b2 3 b 4 1 5 3 4 (a) - (a) + (a) Z - Z + Z 5 0 Z 2 5 3 3 b 64ah 2 b3 1 32 b abh 2 ( b )3 - ( b )4 + ( b )5 = 4 2 5 225 15b 3 PROBLEM 5.128 CONTINUED Now 32 4 yV = y EL dV : y abh = abh 2 9 225 or y = 8 h 25 PROBLEM 5.129 Locate the centroid of the section shown, which was cut from a circular cylinder by an inclined plane. SOLUTION First note that symmetry implies x =0 Choose as the element of volume a vertical slice of width 2x, thickness dz, and height y. Then 1 dV = 2 xy dz, yEL = y, z EL = z 2 h h h z Now y = - z = 1 - x = a2 - z 2 and a 2 2a 2 So z dV = h a 2 - z 2 1 - dz a V = = = a h 0 Then z 1 z 1 2 a - z 1 - dz = h z a 2 - z 2 + a 2 sin -1 + a - z2 a 3a a 2 2 2 ( ) 3/2 a -a 1 2 -1 a h sin (1) - sin -1 ( -1) 2 2 a 2h PROBLEM 5.129 CONTINUED Then h z a 1 2 2 yEL dV = -a 2 2 1 - a h a - z 1 - = h2 a z z2 2 2 -a a - z 1 - 2 a + a 2 dz 4 2 2 h2 1 -1 z 2 2 2 a - z2 z a - z + a sin + 4 2 a 3a z dz a = ( ) 3 2 a 1 + 2 a z 2 2 - a - z 4 ( ) 3 2 a2z 2 a4 z + a - z2 + sin -1 8 8 a -a = 5h 2a 2 -1 sin (1) - sin -1 ( -1) 32 Then a 2 5h 2a 2 yV = yEL dV : y 2 h = 32 ( ) or y = z a 2 2 zELdV = -a z h a - z 1 - a dz 1 = h - a 2 - z 2 3 5 h 16 and ( ) 3 2 1 z - - a 2 - z 2 a 4 ( ) 3 2 a2 z 2 a4 z + a - z2 + sin -1 8 8 a a -a =- a3h -1 -1 sin (1) - sin ( -1) 8 a 2h a3h zV = z EL dV : z =- 2 8 or z = - a 4 PROBLEM 5.130 Locate the centroid of the plane area shown. SOLUTION A, in 2 x , in. y , in. xA, in 3 yA, in 3 1 2 14 20 = 280 - ( 4 ) = -16 2 7 10 1960 -301.59 2800 -603.19 6 12 229.73 X A = xA X 229.73 in 2 = 1658.41 in 3 1658.41 2196.8 Then ( ) or X = 7.22 in. and Y A = yA Y 229.73 in 2 = 2196.8 in 3 or Y = 9.56 in. ( ) PROBLEM 5.131 For the area shown, determine the ratio a/b for which x = y. SOLUTION A 1 2 x 3 a 8 1 a 3 y xA ab 4 a 2b - 6 a 2b 12 2 yA 2ab 2 5 ab 2 - 3 ab 2 15 2 ab 3 1 - ab 2 1 ab 6 3 b 5 2 b 3 Then X A = xA 1 a 2b X ab = 6 12 or 1 a 2 Y A = yA X = 1 ab 2 Y ab = 6 15 2 Y = b 5 1 2 X =Y a= b 2 5 or Now or 4 a = b 5 PROBLEM 5.132 Locate the centroid of the plane area shown. SOLUTION Dimensions in mm A, mm 2 1 2 x , mm y , mm xA, mm3 yA, mm3 3 126 54 = 6804 1 126 30 = 1890 2 1 72 = 1728 2 10 422 9 30 48 27 64 -16 61 236 56 700 82 944 200 880 183 708 120 960 -27 648 277 020 Then X A = xA X 10 422 m 2 = 200 880 mm 2 or X = 19.27 mm ( ) and Y A = yA Y 10 422 m 2 = 270 020 mm3 or Y = 26.6 mm ( ) PROBLEM 5.133 Determine by direct integration the centroid of the area shown. Express your answer in terms of a and h. SOLUTION By observation y1 = - h x+h a x = h 1 - a or For y2: At x = 0, y = h: h = k (1 - 0) k =h 1 a2 At Then x = a, y = 0: 0 = h 1 - ca 2 ( ) or C = x2 y2 = h 1 - 2 a Now x2 x dA = ( y2 - y1 ) dx = h 1 - 2 - 1 - dx a a x x2 = h - 2 dx a a 1 h x x2 = ( y1 - y2 ) = 1 - + 1 - 2 2 2 a a = A = dA = = 1 ah 6 a x h 0 a h 0 xEL = x yEL h x x2 2 - - 2 a a 2 a Then x2 x x2 x3 - 2 dx = h - 2 a a 2a 3a 0 and x ELdA = = x 3 x x2 x4 - 2 dx = h - 2 a a 3a 4a 0 a 1 2 a h 12 PROBLEM 5.133 CONTINUED y EL dA = 0 = a h x x2 x x2 2 - - 2 h - 2 dx a a a a 2 h2 a x x2 x4 0 2 a - 3 2 + 4 dx 2 a a a h2 x2 x3 x5 1 2 = - 2 + 4 = ah 2 a a 5a 0 10 1 2 1 xA = x EL dA: x ah = a h 6 12 1 2 1 yA = y EL dA: y ah = a h 6 10 x = y = 1 a 2 3 h 5 PROBLEM 5.134 Member ABCDE is a component of a mobile and is formed from a single piece of aluminum tubing. Knowing that the member is supported at C and that l = 2 m, determine the distance d so that portion BCD of the member is horizontal. SOLUTION First note that for equilibrium, the center of gravity of the component must lie on a vertical line through C. Further, because the tubing is uniform, the center of gravity of the component will coincide with the centroid of the corresponding line. Thus, X = 0 So that Then xL = 0 0.75 cos 55 m ( 0.75 m ) - d - 2 + ( 0.75 - d ) m (1.5 m ) 1 + (1.5 - d ) m - 2 m cos 55 ( 2 m ) = 0 2 or ( 0.75 + 1.5 + 2 ) d 2 1 = ( 0.75 ) - 2 cos 55 + ( 0.75 )(1.5 ) + 3 2 or d = 0.739 m PROBLEM 5.135 A cylindrical hole is drilled through the center of a steel ball bearing shown here in cross section. Denoting the length of the hole by L, show that the volume of the steel remaining is equal to the volume of a sphere of diameter L. SOLUTION Calculate volumes by rotating cross sections about a line and using Theorem II of Pappus-Guldinus For the sector: y AA = h= A= = 2R sin 3 2 A = R2 y AA = 2 1 h= 4R 2 - L2 , 3 3 For the triangle: 1 L R2 - = 4R 2 - L2 2 2 1 ( L )( h ) 2 1 L 4R 2 - L2 4 Using Theorem II of Pappus-Guldinus Vball = 2 ( y AA )1 A1 - 2 ( y AA )2 A2 2 R sin 1 1 R 2 - 4R 2 - L2 L 4R 2 - L2 = 2 3 3 4 ( ) L 2 4 R 2 - L2 = 2 R3 sin - 12 3 ( ) Now Then R sin = L 2 2 L 1 1 3 V = 2 R 2 - LR 2 + L 3 12 3 2 = 6 L3 Note Vsphere = L If r = , then 2 4 r where r is the radius 3 Vsphere 4 L = = L3 3 2 6 Vball = Vsphere = 3 Therefore, 6 L3 Q.E.D. PROBLEM 5.136 For the beam and loading shown, determine (a) the magnitude and location of the resultant of the distributed load, (b) the reactions at the beam supports. SOLUTION (a) Have RI = ( 4 m )( 200 N/m ) = 800 N RII = 2 ( 4 m )( 600 N/m ) = 1600 N 3 Fy : -R = - RI - RII Then or and or (b) Reactions R = 800 + 1600 = 2400 N M A: - X ( 2400 ) = -2 ( 800 ) - 2.5 (1600 ) X = 7 m 3 R = 2400 N , X = 2.33 m Fx = 0: Ax = 0 7 M A = 0: ( 4 m ) By - m ( 2400 N ) = 0 3 or By = 1400 N Fy = 0: Ay + 1400 N - 2400 N = 0 or Ay = 1000 N A = 1000 N , B = 1400 N PROBLEM 5.137 Determine the reactions at the beam supports for the given loading. SOLUTION Have RI = RII = 1 ( 3 ft )( 480 lb/ft ) = 720 lb 2 1 ( 6 ft )( 600 lb/ft ) = 1800 lb 2 RIII = ( 2 ft )( 600 lb/ft ) = 1200 lb Then M B = 0: Fx = 0: Bx = 0 ( 2 ft )( 720 lb ) - ( 4 ft )(1800 lb ) + ( 6 ft ) Cy - ( 7 ft )(1200 lb ) = 0 C = 2360 lb or C y = 2360 1b Fy = 0: - 720 lb + By - 1800 lb + 2360 lb - 1200 lb = 0 or By = 1360 lb B = 1360 lb PROBLEM 5.138 Locate the center of gravity of the sheet-metal form shown. SOLUTION First, assume that the sheet metal is homogeneous so that the center of gravity of the form will coincide with the centroid of the corresponding area. yI = - zI = xIII 1 (1.2) = -0.4 m 3 1 (3.6) = 1.2 m 3 4 (1.8) 2.4 m =- =- 3 A, m 2 I II III x, m y, m -0.4 z, m xA, m3 3.24 4.59 -3.888 yA, m3 -0.864 zA, m3 2.592 11.016 9.1609 22.769 1 ( 3.6) (1.2) = 2.16 2 (3.6)(1.7) = 6.12 1.5 0.75 - 2.4 1.2 1.8 1.8 0.4 0.8 2.448 4.0715 5.6555 2 (1.8) 2 = 5.0894 13.3694 3.942 Have X V = xV : X 13.3694 m 2 = 3.942 m3 ( ) Y V = yV : Y 13.3694 m 2 = 5.6555 m3 or Y = 0.423 m ( ) or X = 0.295 m Z V = zV : Z 13.3694 m 2 = 22.769 m3 or Z = 1.703 m ( ) PROBLEM 5.139 The composite body shown is formed by removing a semiellipsoid of revolution of semimajor axis h and semiminor axis a from a hemisphere of 2 radius a. Determine (a) the y coordinate of the centroid when h = a/2, (b) the ratio h/a for which y = -0.4a. SOLUTION V y yV 1 - a 4 4 + 1 a 2 h 2 16 Hemisphere Semiellipsoid 2 3 a 3 2 a 1 - h = - a 2 h 3 2 6 2 3 - a 8 3 - h 8 Then Now V = 6 a 2 ( 4a - h ) Y V = yV yV = - 16 a 2 4a 2 - h 2 ( ) so that or (a) Y = ? when h = Substituting a 2 Y a 2 ( 4a - h ) = - a 2 4a 2 - h 2 16 6 ( ) (1) 2 h 3 h Y 4 - = - a 4 - a a 8 h 1 = into Eq. (1) a 2 2 1 3 1 Y 4 - = - a 4 - 2 8 2 or Y =- 45 a 112 Y = -0.402a PROBLEM 5.139 CONTINUED (b) h = ? when Y = -0.4a a Substituting into Eq. (1) ( -0.4a ) 4 - 2 2 h 3 h = - a 4 - a 8 a or h h 3 - 3.2 + 0.8 = 0 a a 2 h 3.2 ( -3.2 ) - 4 ( 3)( 0.8 ) = a 2 ( 3) Then = 3.2 0.8 6 or h 2 h 2 = and = a 5 a 3 PROBLEM 5.140 A thin steel wire of uniform cross section is bent into the shape shown. Locate its center of gravity. SOLUTION First assume that the wire is homogeneous so that its center of gravity will coincide with the centroid of the corresponding line. x1 = 0.3sin 60 = 0.15 3 m z1 = 0.3cos 60 = 0.15 m 0.6sin 30 x2 = sin 30 6 0.9 = m 0.6sin 30 z2 = cos 30 6 0.9 = 3m L2 = ( 0.6 ) = ( 0.2 ) m 3 L, m 1 2 3 4 1.0 0.2 x, m 0.15 3 y, m z, m xL, m 2 0.25981 0.18 0 0 0.43981 yL, m 2 0.4 0 0.32 0.48 1.20 zL, m 2 0.15 0.31177 0.48 0.18 1.12177 0.4 0 0.4 0.8 0.15 0.9 3 0.9 0 0 0.6 0.3 0.8 0.6 3.0283 Have X L = x L: X ( 3.0283 m ) = 0.43981 m 2 Y L = yL: Y ( 3.0283 m ) = 1.20 m 2 Z L = z L: Z ( 3.0283 m ) = 1.12177 m 2 or X = 0.1452 m or Y = 0.396 m or Z = 0.370 m PROBLEM 5.141 Locate the centroid of the volume obtained by rotating the shaded area about the x axis. SOLUTION First note that symmetry implies y = k ( X - h) x = 0, y = a: a = k ( -h ) 2 y =0 and z = 0 Have At 2 or k = a h2 Choose as the element of volume a disk of radius r and thickness dx. Then dV = r 2dx, X EL = x a 2 r = 2 ( x - h) h dV = V = 0 = h Now so that Then a2 ( x - h)4 dx h4 h a2 a2 ( x - h )4 dx = 4 ( x - h )5 4 0 5h h 1 2 a h 5 and 4 h x EL dV = 0 x h 4 ( x - h ) dx a2 = = = a2 h 5 x - 4hx 4 + 6h 2 x3 - 4h3 x 2 + h 4 x dx 4 0 h ( ) a2 1 6 4 5 3 2 4 4 3 3 1 4 2 x - hx + h x - h x + h x 5 2 3 2 h4 6 0 h 1 a 2h 2 30 Now 2 2 xV = x EL dV : x a 2h = a h 5 30 or x = 1 h 6

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