A Collection of Problems in Illustration of the Principles of Theoretical MechanicsDeighton, Bell, 1876 - 667 páginas |
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Página 134
... dt ) sin Okp — t - = ( t + dt ) cos y — t , - or , retaining infinitesimals of the first order only , Fmds cos + = dt ; and therefore , ds cos & being equal to dr , Fmdr = dt ... From the equation ( a ) , since . ( b ) . dr P and sin o ...
... dt ) sin Okp — t - = ( t + dt ) cos y — t , - or , retaining infinitesimals of the first order only , Fmds cos + = dt ; and therefore , ds cos & being equal to dr , Fmdr = dt ... From the equation ( a ) , since . ( b ) . dr P and sin o ...
Página 139
... dt = the tension at p . = = Then for the equilibrium of the element Pp of the string we have , resolving forces at right angles to the tangent at P , ( t + dt ) sin d0 = pds = ppd0 , and therefore , retaining infinitesimals of the first ...
... dt = the tension at p . = = Then for the equilibrium of the element Pp of the string we have , resolving forces at right angles to the tangent at P , ( t + dt ) sin d0 = pds = ppd0 , and therefore , retaining infinitesimals of the first ...
Página 140
... dt ) sin do = pds = ppd0 ; and therefore in the limit td0 = ppd0 , t = pp ...... again , resolving forces parallel to the tangent at P , ( t + dt ) cos de- t + μpds = 0 , and therefore in the limit and consequently by ( 1 ) dt + μpds ...
... dt ) sin do = pds = ppd0 ; and therefore in the limit td0 = ppd0 , t = pp ...... again , resolving forces parallel to the tangent at P , ( t + dt ) cos de- t + μpds = 0 , and therefore in the limit and consequently by ( 1 ) dt + μpds ...
Página 223
... dt ' dt > are both equal to zero initially , μ ' dx dx ' + μ dt dt - 0 ; integrating again , μα + μα = μα + μα .... a , a ' , being the initial values of x , x ' . Again , subtracting ( 1 ) from ( 2 ) , da dta ( x − x ) + ( μ + μ ) ( x ...
... dt ' dt > are both equal to zero initially , μ ' dx dx ' + μ dt dt - 0 ; integrating again , μα + μα = μα + μα .... a , a ' , being the initial values of x , x ' . Again , subtracting ( 1 ) from ( 2 ) , da dta ( x − x ) + ( μ + μ ) ( x ...
Página 236
... dt dt d dx2 dt dt2 = - == = μ μ 3 + k dx2 x3 x3 dť2 ' 2μ dx + adt d 1 - k dt x - 2k dx3 x3 dt dx d 1 dt2 dt x2 dx2 1 Assume = w and = z ; then 236 RECTILINEAR MOTION OF A PARTICLE .
... dt dt d dx2 dt dt2 = - == = μ μ 3 + k dx2 x3 x3 dť2 ' 2μ dx + adt d 1 - k dt x - 2k dx3 x3 dt dx d 1 dt2 dt x2 dx2 1 Assume = w and = z ; then 236 RECTILINEAR MOTION OF A PARTICLE .
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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
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