A Collection of Problems in Illustration of the Principles of Theoretical MechanicsDeighton, Bell, 1876 - 667 páginas |
Dentro del libro
Resultados 1-5 de 80
Página 4
... hence F = Sy dx fa but 25 ( a - x ) and therefore ;。( a - x )。( ax ) da ( 1 ) ; dx $ dx = − 2.x3 ( a− x ) 1 +5 √x3 ( a — x ) 1 dx , ra 0 ( a — x ) 1 - ra · - dx = 5 | ̃x3 ( a − x ) 3 dx 0 - = 5a " a 。( α — x ) 0 dx - 5 hence from ( ...
... hence F = Sy dx fa but 25 ( a - x ) and therefore ;。( a - x )。( ax ) da ( 1 ) ; dx $ dx = − 2.x3 ( a− x ) 1 +5 √x3 ( a — x ) 1 dx , ra 0 ( a — x ) 1 - ra · - dx = 5 | ̃x3 ( a − x ) 3 dx 0 - = 5a " a 。( α — x ) 0 dx - 5 hence from ( ...
Página 5
... hence from ( 2 ) we have Again , and 0 a Σxy dx = fa3 sin a .. Σ Jxy da a cos a x tan a dx = a2 sin a cosa , [ a 4 ( a2 — x2 ) 3 dx = § ( a2a — a2 sin a cos a ) ; a cos a hence from ( 3 ) we have Σfy dx = 1a'a . From the relations ( 1 ) ...
... hence from ( 2 ) we have Again , and 0 a Σxy dx = fa3 sin a .. Σ Jxy da a cos a x tan a dx = a2 sin a cosa , [ a 4 ( a2 — x2 ) 3 dx = § ( a2a — a2 sin a cos a ) ; a cos a hence from ( 3 ) we have Σfy dx = 1a'a . From the relations ( 1 ) ...
Página 7
... Hence , from ( 1 ) , ( 2 ) , ( 3 ) , we get ( 3 ) . sins a x = fa a- sin a cos a This result may be obtained as easily by rectangular co- ordinates ; thus , putting a cos a = a ' , but hence a a [ exy de a dx y dx y2 = a2 — x2 ; a ...
... Hence , from ( 1 ) , ( 2 ) , ( 3 ) , we get ( 3 ) . sins a x = fa a- sin a cos a This result may be obtained as easily by rectangular co- ordinates ; thus , putting a cos a = a ' , but hence a a [ exy de a dx y dx y2 = a2 — x2 ; a ...
Página 12
... hence Some r2 cos ( π — 0 ) de dr - { m3 · = - and therefore but = cos 10 ̄ ̄ 1 + tan2 10 cos 0 10 hence 0 cos 10 do , COS r2 cos ( π − 0 ) de dr = code ; - COS 0 1 - tan210 1 1 - tan210 = - 3 Ja = = ( 1 - tan310 ) sec * 10 ; cos cos ...
... hence Some r2 cos ( π — 0 ) de dr - { m3 · = - and therefore but = cos 10 ̄ ̄ 1 + tan2 10 cos 0 10 hence 0 cos 10 do , COS r2 cos ( π − 0 ) de dr = code ; - COS 0 1 - tan210 1 1 - tan210 = - 3 Ja = = ( 1 - tan310 ) sec * 10 ; cos cos ...
Página 22
... hence [ * [ * dx dy dz = [ * da dy { ( a2 — xa — and therefore r + a rx ' 0 2a - 1 y1 ) 1 — — 4 ( a2 — x2 — y ' ) } 2a = dæ { } π ( aa — x3 ) — 1 } / 1⁄2 ( a3 — 2 " ) ® } , dx r + a За 1 - [ * [ * [ " dx — dx dy dz = [ ** dx { \ π ( aa— ...
... hence [ * [ * dx dy dz = [ * da dy { ( a2 — xa — and therefore r + a rx ' 0 2a - 1 y1 ) 1 — — 4 ( a2 — x2 — y ' ) } 2a = dæ { } π ( aa — x3 ) — 1 } / 1⁄2 ( a3 — 2 " ) ® } , dx r + a За 1 - [ * [ * [ " dx — dx dy dz = [ ** dx { \ π ( aa— ...
Contenido
1 | |
9 | |
15 | |
25 | |
26 | |
42 | |
48 | |
56 | |
89 | |
103 | |
153 | |
169 | |
179 | |
189 | |
206 | |
219 | |
234 | |
248 | |
263 | |
279 | |
295 | |
298 | |
473 | |
491 | |
507 | |
523 | |
535 | |
542 | |
560 | |
577 | |
588 | |
602 | |
630 | |
643 | |
664 | |
Otras ediciones - Ver todas
Términos y frases comunes
absolute force acted angular velocity attraction axes axis beam body catenary centre of force centre of gravity circle co-ordinates coefficient of friction cos² curve cycloid cylinder denote density descends determine distance dt dt dt² dx dy dx² elastic ellipse equal equation Euler find the centre fixed point hence horizontal plane inclined plane inertia integrating James Bernoulli John Bernoulli lamina mass middle point moment of inertia motion natural length oscillation parabola perpendicular position of equilibrium pressure radius of gyration radius vector resolving forces rest right angles SECT sin² smooth sphere square straight line string supposing surface taking moments tangent tension triangle tube uniform rod varying inversely vertex vertical plane Vis Viva weight
Pasajes populares
Página 203 - A centre of force attracting inversely as the square of the distance is at the centre of a spherical cavity within an infinite mass of liquid, the pressure on which at an infinite distance is...