Mechanics of engineeringComprising statics and dynamics of solids: and the mechanics of the materials of constructions, or strength and elasticity of beams, columns, arches, shafts, etc . — Z- Pdv — -x—n , = chord X radius -f- length of wire. For a semicircular wire, this reduces to x = 2r -f- n. Problem 2. Centre of gravity of trapezoidal (and trian-gular) thin plates, homogeneous, etc.—Prolong the non-parallelsides of the trapezoid to intersect at 0, which take as an origin,,making the axis X perpendicular to the bases b and lx. ~Wemay here use equations (4), § 23, and may take a vertica
Mechanics of engineeringComprising statics and dynamics of solids: and the mechanics of the materials of constructions, or strength and elasticity of beams, columns, arches, shafts, etc . — Z- Pdv — -x—n , = chord X radius -f- length of wire. For a semicircular wire, this reduces to x = 2r -f- n. Problem 2. Centre of gravity of trapezoidal (and trian-gular) thin plates, homogeneous, etc.—Prolong the non-parallelsides of the trapezoid to intersect at 0, which take as an origin,,making the axis X perpendicular to the bases b and lx. ~Wemay here use equations (4), § 23, and may take a vertical stripfor our element of area, dF, in determining x; for each pointof such a strip has the same x. Now dF = (y -f- y)dx, and PARALLEL FORCES AND THE CENTRE OP GRAVITY. 21 from similar triangles y -f- y = j x. Hence F = <r (bh — bji^) can be written — 7 (A2 2Ab Ph A/), and x = - ^ becomes x*dx -f lb 2 A (A2 - h?) = 3 A2 - A, for the trapezoid. For a triangle Ax = 0, and we Lave x— A; that is, the o centre of gravity of a triangle is one third the altitude from the base. The centre of gravity is finally determined by knowing •—* •x y b ( ^^-----^- -X-i y 1 i. Fig. 17. Fig. 18. that a line joining the middles of b and bl is a line of gravity;or joining O and the middle of b in the case of a triangle. Problem 3. Sector of a circle. Thin plate, etc.—Let thenotation, axes, etc., be as in Fig. 18. Angle of sector = 2a;x = ? Using polar co-ordinates, the element of area dF (asmall rectangle) = pdcp. dp, and its x = p cos cp; hence thetotal area = , F = r2a. From equations (4), § 23, we havex = yfxdF — yJ J cos <PP*dpdp — j?J_a [_COS(pJ0 Pdpj dcp. 22 MECHANICS OF ENGINEERING. (Note on double integration,—The quantitycos cp J p* dp \d<p, is that portion of the summation / / cos cppdpdcp which belongs to a single elementary sector (triangle), since all its-elements (rectangles), from centre to circumference, have thesame cp and dcp.)That is, 1 r
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Keywords: ., bookcentury1800, bookdecade1880, booksubjectenginee, bookyear1888
