PROBLEM 5.15 PROBLEM mm SOLUTION SOLUTION. X ( mm 2 ) = mm 3 or X = 149 mm t. YS A = S ya

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1 PROBLEM. Locte the centroid of the plne re shown. PROBLEM. Locte the centroid of the plne re shown. mm mm 7 mm 7 mm A, mm, mm y, mm A, mm ya, mm mm Ê ˆ Á 7 mm Ë Ê 4 7 ˆ Á + Ë mm p nd A 97 9 mm X A mm ya 6 mm Y A mm or X 96.4 mm or Y 4.7 mm A, mm, mm y, mm A, mm 4 ya, mm 4 () (7) = p 4 (7) = Â XA A X ( mm ) = 9. 4 mm or X = 49 mm t nd YS A = S ya Y ( mm ) = mm or Y = 6 mm t

2 PROBLEM.7 The horizontl is is drwn through the centroid C of the re shown nd divides the re into two component res A nd A. Determine the first moment of ech component re with respect to the is, nd eplin the results obtined. PROBLEM. A thin, homogeneous wire is bent to form the perimeter of the figure indicted. Locte the center of grvity of the wire figure thus formed. Note tht Q m 6 m Q ya or. mm nd Q Q. m 9.m. m 6.m Now Q Q Q This result is epected since is centroidl is thus y nd Q ya YA y Q or. mm Q First note tht becuse the wire is homogeneous, its center of grvity will coincide with the centroid of the corresponding line. L, mm, mm y, mm L, mm yl, mm XL L nd YL yl X (77. mm). mm or X.44 mm Y (77. mm) 98.7 mm or Y.7 mm

PROBLEM. PROBLEM.44 Determine by direct integrtion the centroid of the re shown. A homogeneous wire is bent into the shpe shown. Determine by direct integrtion the coordinte of its centroid.

3 PROBLEM. PROBLEM.44 Determine by direct integrtion the centroid of the re shown. A homogeneous wire is bent into the shpe shown. Determine by direct integrtion the coordinte of its centroid. First note tht symmetry implies For the element (EL) shown y Rcos, Rsin d R cos d First note tht becuse the wire is homogeneous, its center of grvity coincides with the centroid of the corresponding line Now Where cos nd EL dl d dy cos : d cos sind da yd R cos d y sin : dy sin cosd Hence sin A da R cos d R R sin 4 R yelda cos R cos d R cos sin sin R cos sin sin But ya yelda so or Alterntively, y R cos y R sin sin cos sin sin R sin sin y Rsin sin / / d dl cos sin d sin cos d cos sin cos sin cossind / / L dl cossind sin / / cos nd ELdL cos cossind Hence L ELdL:

4 PROBLEM.44 CONTINUED Alterntive solution cos cos y y sin sin / / y / / / or y dy / / / / d Now EL / dy / / / / nd dl d d d / / PROBLEM. Determine the volume of the solid generted by rotting the semiellipticl re shown bout () the is AA, (b) the is BB, (c) the y is. nd Hence / / / / L dl d / / / / ELdL d L ELdL: Applying the second theorem of Pppus-Guldinus, we hve () Rottion bout is AA: (b) Rottion bout is BB: b Volume ya b b Volume ya b V b V b (c) Rottion bout y-is: 4 b Volume ya b V b

PROBLEM.6 The reflector of smll flshlight hs the prbolic shpe shown. Determine the surfce re of the inside of the reflector. z 9 mm 4 mm 8 mm mm y r =. mm r =. mm mm 6 mm mm mm mm mm mm mm PROBLEMS.

5 PROBLEM.6 The reflector of smll flshlight hs the prbolic shpe shown. Determine the surfce re of the inside of the reflector. z 9 mm 4 mm 8 mm mm y r =. mm r =. mm mm 6 mm mm mm mm mm mm mm PROBLEMS.94 AND.9 Problem.94: For the mchine element shown, determine the coordinte of the center of grvity. Problem.9: For the mchine element shown, determine the y coordinte of the center of grvity. First note tht the required surfce re A cn be generted by rotting the prbolic cross section through rdins bout the is. Applying the first theorem of Pppus-Guldinus, we hve A yl Now, since ky, t : k 7. or 6. k () At mm: k. or 6.k () Eq. () 6.k : or 8.47 mm Eq. () 6.k Eq. () k. mm d. y nd.y dy d Now dl dy.9y dy dy So A yl nd yl ydl. 7. A y.9y dy.9y.8 mm. / 7. or A mm S First, ssume tht the mchine element is homogeneous so tht its center of grvity coincides with the centroid of the corresponding volume..94 Hve XV V.9 Hve V, mm 4, mm y, mm V, mm 4 4 yv, mm 4 4 I () (9) (9) = II (6) () () = III p (.) () = IV p (.) (9) = Â X (.8 4 mm ) = mm 4 or X = 6 mm t YS V = S yv Y (.8 4 mm ) = 4.96 mm 4 or Y = 9 mm t

8 mm r = mm mm z mm mm y mm mm mm mm mm PROBLEM.7 The thin, plstic front cover of wll clock is of uniform thickness. Locte the center of grvity of the cover. mm y PROBLEM.

6 8 mm r = mm mm z mm mm y mm mm mm mm mm PROBLEM.7 The thin, plstic front cover of wll clock is of uniform thickness. Locte the center of grvity of the cover. mm y PROBLEM.8 A thin steel wire of uniform cross section is bent into the shpe shown, where rc BC is qurter circle of rdius R. Locte its center of grvity. r = mm First, ssume tht the plstic is homogeneous so tht the center of grvity of the cover coincides with the centroid of the corresponding re. Net, note tht symmetry implies X. mm z mm mm First, ssume tht the wire is homogeneous so tht its center of grvity coincides with the centroid of the corresponding line A, mm y,mm z,mm ya, mm z,mm Hve Y A ya Y : 94 7 mm 7 mm ZA za: Z 94 7 mm 46 6 mm or or Y.4 mm Z 9.4 mm L,mm, mm y, mm z, mm L, mm yl, mm zl, mm F H G I K J = 87. F HG I K J b g F HG p 7 4 (7) p p p = = 4.9 = 8.7 = Â Hve XS L = S L: X (69.49 mm) = 9.6 mm or X = 4.9 mm t I K J YS L = S yl: Y (69.49 mm) =.98 mm or Y = 8.7 mm t ZS L= SzL: Z (69.49 mm) = 6. mm or Z = 98.4 mm t

7 PROBLEM.9 Locte the centroid of the volume obtined by rotting the shded re bout the is. PROBLEM.9 CONTINUED 8 V ELdV: b b 6 or 6 First note tht symmetry implies y z Choose s the element of volume disk of rdius r nd thickness d. Now r b so tht dv r d, EL dv b d 4 V b d b d 4 b 4 b 8 b nd 4 ELdV b d b b 6 6 b

STATICS. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Centroids and Centers of Gravity.

STATICS. Vector Mechanics for Engineers: Statics VECTOR MECHANICS FOR ENGINEERS: Centroids and Centers of Gravity. 5 Distributed CHAPTER VECTOR MECHANICS FOR ENGINEERS: STATICS Ferdinnd P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Wlt Oler Texs Tech Universit Forces: Centroids nd Centers of Grvit Contents Introduction