A Collection of Problems in Illustration of the Principles of Theoretical MechanicsDeighton, Bell, 1876 - 667 páginas |
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Página 64
... solution is rather more brief than the pre- ceding one . The component of W in the plane zAx at right angles to AB is equal to Wsin B , and the component of Wsin B at right angles to the door is Wsin ẞ sin a : hence the moment of W ...
... solution is rather more brief than the pre- ceding one . The component of W in the plane zAx at right angles to AB is equal to Wsin B , and the component of Wsin B at right angles to the door is Wsin ẞ sin a : hence the moment of W ...
Página 67
William Walton. review of Stone's work by John Bernoulli ' , the solution given by Stone was declared to be erroneous ... solutions given by Bernoulli and by Couplet , and even down to very late years numerous memoirs have appeared on the ...
William Walton. review of Stone's work by John Bernoulli ' , the solution given by Stone was declared to be erroneous ... solutions given by Bernoulli and by Couplet , and even down to very late years numerous memoirs have appeared on the ...
Página 106
... solution of the same problem : - Let the forces P , Q , R , ...... be represented in magnitude by the lines 2AB , 2BC , 2CD , ...... , to which they are proportional , Instead of the force 2AB acting at the middle point of the side AB ...
... solution of the same problem : - Let the forces P , Q , R , ...... be represented in magnitude by the lines 2AB , 2BC , 2CD , ...... , to which they are proportional , Instead of the force 2AB acting at the middle point of the side AB ...
Página 107
... , as in the former solution , that the sides of the polygon must be so arranged that its angular points may all lie in the circumference of a single circle . ( 3 ) A quadrilateral ABCD , ( fig . EQUILIBRIUM OF SEVERAL BODIES . 107.
... , as in the former solution , that the sides of the polygon must be so arranged that its angular points may all lie in the circumference of a single circle . ( 3 ) A quadrilateral ABCD , ( fig . EQUILIBRIUM OF SEVERAL BODIES . 107.
Página 109
... solution of the same problem . For the equilibrium of the rod AB there is , taking moments about B , N. BD . sin BDA = P. BO . sin △ BOC ; L L and for the equilibrium of the rod CD , taking moments about C , N. CA. sin CAD = Q . CO ...
... solution of the same problem . For the equilibrium of the rod AB there is , taking moments about B , N. BD . sin BDA = P. BO . sin △ BOC ; L L and for the equilibrium of the rod CD , taking moments about C , N. CA. sin CAD = Q . CO ...
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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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