Key And Supplement to Elementary Mechanics
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In our eReader you can find the full English version of the book. Read Key And Supplement to Elementary Mechanics Online - link to read the book on full screen. Our eReader also allows you to upload and read Pdf, Txt, ePub and fb2 books. In the Mini eReder on the page below you can quickly view all pages of the book - Read Book Key And Supplement to Elementary Mechanics
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99 To find AE '; in the right-angled triangle DAB we have A D tang B = = i = 0-666 + ; . -. B = 33 41'. Then AE = AB sin B = 3-327 + inches.
Substituting above gives 24 x 20 2. Taking the origin of moments at D we have P ^ ~ 20 x 24 10A . , : - = 120 Ibs.
3. Taking the origin of moments at B we have W. BC^F. AB, or PAGE 116.
4. If t = TF, we have 100 KEY AND SUPPLEMENT or = AC.
From the right-angled triangle AEB we have sin 45 C substituting, AC 5. Taking the origin of moments at A we have t x ...AC= BE x W. But from the example DB=2AB, 6 = 45, TF= Ibs. From the figure BE- AB sin 45 = $V2 AB = 0-7071AB.
BE AB sin 45 . -. 9> = 20 42' 17". To find A C we have from the figure AC=AB&\nABC = J# sin (0 - cp) TO ELEMENTARY MECHANICS. 101 = AB sin 24 17' 43" = 04115 AB.
Substituting in the first equation above gives Q. 7071 x 50 04115 = 85-97 + Ibs.
To get the compression on the bar, take the origin of moments at D, in which case the moment of the tension will be zero. The perpendicular from D upon AB prolonged will be AD sin #, and from D perpendicular upon the vertical through B will equal BE = AB sin 6 ; hence we have calling c the compression c.
What to read after Key And Supplement to Elementary Mechanics?
You can find similar books in the "Read Also" column, or choose other free books by De Volson Wood to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
99 To find AE '; in the right-angled triangle DAB we have A D tang B = = i = 0-666 + ; . -. B = 33 41'. Then AE = AB sin B = 3-327 + inches.
Substituting above gives 24 x 20 2. Taking the origin of moments at D we have P ^ ~ 20 x 24 10A . , : - = 120 Ibs.
3. Taking the origin of moments at B we have W. BC^F. AB, or PAGE 116.
4. If t = TF, we have 100 KEY AND SUPPLEMENT or = AC.
From the right-angled triangle AEB we have sin 45 C substituting, AC 5. Taking the origin of moments at A we have t x ...AC= BE x W. But from the example DB=2AB, 6 = 45, TF= Ibs. From the figure BE- AB sin 45 = $V2 AB = 0-7071AB.
BE AB sin 45 . -. 9> = 20 42' 17". To find A C we have from the figure AC=AB&\nABC = J# sin (0 - cp) TO ELEMENTARY MECHANICS. 101 = AB sin 24 17' 43" = 04115 AB.
Substituting in the first equation above gives Q. 7071 x 50 04115 = 85-97 + Ibs.
To get the compression on the bar, take the origin of moments at D, in which case the moment of the tension will be zero. The perpendicular from D upon AB prolonged will be AD sin #, and from D perpendicular upon the vertical through B will equal BE = AB sin 6 ; hence we have calling c the compression c.
What to read after Key And Supplement to Elementary Mechanics?
You can find similar books in the "Read Also" column, or choose other free books by De Volson Wood to read onlineMoreLess
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