CHAPTER 5 - PROBLEM 5.1 Locate the centroid of the plane...

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PROBLEM 5.1 Locate the centroid of the plane area shown. SOLUTION 2 , in A , in. x , in. y 3 , in xA 3 , in yA 1 86 4 8 ×= 4 9 192 432 2 16 12 192 8 6 1536 1152 Σ 240 1344 1584 Then 3 2 1344 in 240 in xA X A Σ == Σ or 5.60 in. X = W and 3 2 1584 in 240 in yA Y A Σ Σ or 6.60 in. Y = W
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PROBLEM 5.2 Locate the centroid of the plane area shown. SOLUTION 2 , mm A , mm x , mm y 3 , mm xA 3 , mm yA 1 1 60 75 2250 2 ××= 40 25 90 000 56 250 2 105 75 7875 ×= 112.5 37.5 885 900 295 300 Σ 10 125 975 900 351 600 Then 3 2 975 900 mm 10 125 mm xA X A Σ == Σ or 96.4 mm X = W and 3 2 351 600 mm 10 125 mm yA Y A Σ Σ or 34.7 mm Y = W
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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 () 2 24 in. 3 Y = or 16.00 in. Y = W 2 , in A , in. x 3 , in xA 1 1 24 10 120 2 ××= 2 10 6.667 3 = 800 2 1 24 16 192 2 1 10 16 15.333 3 += 2944 Σ 312 3744 Then 3 2 3744 in 312 in xA X A Σ == Σ or 12.00 in. X = W
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PROBLEM 5.4 Locate the centroid of the plane area shown. SOLUTION 2 , mm A , mm x , mm y 3 , mm xA 3 , mm yA 1 21 22 462 ×= 1.5 11 693 5082 2 () 1 69 2 7 2 −= 6 2 162 54 3 ( ) 1 61 2 3 6 2 8 2 288 72 Σ 399 567 4956 Then 3 2 567 mm 399 mm xA X A Σ == Σ or 1.421 mm X = W and 3 2 4956 mm 399 mm yA Y A Σ Σ or 12.42 mm Y = W
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PROBLEM 5.5 Locate the centroid of the plane area shown. SOLUTION 2 , mm A , mm x , mm y 3 , mm xA 3 , mm yA 1 120 200 24 000 ×= 60 120 1 440 000 2 880 000 2 () 2 60 5654.9 2 π −= 94.5 120 534 600 678 600 Σ 18 345 905 400 2 201 400 Then 3 2 905 400 mm 18 345 mm xA X A Σ == Σ or 49.4 mm X = W and 3 2 2 201 400 mm 18 345 mm yA Y A Σ Σ or 93.8 mm Y = W
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PROBLEM 5.6 Locate the centroid of the plane area shown. SOLUTION 2 , in A , in. x , in. y 3 , in xA 3 , in yA 1 () 2 9 63.617 4 π = ( ) 49 3.8917 3 =− 3.8917 243 243 2 ( ) 1 15 9 67.5 2 = 5 3 337.5 202.5 Σ 131.1 94.5 445.5 Then 3 2 94.5 in 131.1 in xA X A Σ == Σ or 0.721 in. X = W and 3 2 445.5 in 131.1 in yA Y A Σ Σ or 3.40 in. Y = W
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PROBLEM 5.7 Locate the centroid of the plane area shown. SOLUTION First note that symmetry implies XY = 2 , mm A , mm x 3 , mm xA 1 40 40 1600 × = 20 32 000 2 2 (40) 1257 4 π −= 16.98 21 330 Σ 343 10 667 Then 3 2 10 667 mm 343 mm xA X A Σ == Σ or 31.1 mm X = W and 31 .1 mm YX W
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PROBLEM 5.8 Locate the centroid of the plane area shown. SOLUTION First note that symmetry implies 0 X = W 2 , in A , in. y 3 , in yA 1 () 2 4 25.13 2 π −= 1.6977 42.67 2 2 6 56.55 2 = 2.546 144 Σ 31.42 101.33 Then 3 2 101.33 in 31.42 in yA Y A Σ == Σ or 3.23 in. Y = W
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PROBLEM 5.9 For the area of Problem 5.8, determine the ratio 21 / rr so that 1 3/ 4 . yr = SOLUTION A y yA 1 2 1 2 r π 1 4 3 r 3 1 2 3 r 2 2 2 2 r 2 4 3 r 3 2 2 3 r Σ ( ) 22 2 ( ) 33 2 3 Then YA y A Σ= Σ or ( ) ( ) 12 1 2 1 32 42 3 r r r ×− =− 23 11 9 16       Let 2 1 r p r = [] 2 9 ( 1)( 1) ( 16 pp p + −=− + + or 2 16 (16 9 ) 9 ) 0 ππ + −+ =
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PROBLEM 5.9 CONTINUED Then 2 (16 9 ) 4(16)(16 2(16) p π ππ −− ± = or 0.5726 1.3397 pp =− = Taking the positive root 2 1 1.340 r r = W
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